MATERI ALJABAR LINEAR INGGRIS

Linear Algebra
Jim Hefferon
¡2
1 ¢
¡1
3 ¢
¯¯¯¯
1 2
3 1¯¯¯¯
¡2
1 ¢
x1 ¢ ¡1
3 ¢
¯¯¯¯
x1 ¢ 1 2
x1 ¢ 3 1¯¯¯¯
¡2
1 ¢
¡6
8¢
¯¯¯¯
6 2
8 1¯¯¯¯
Notation
R real numbers
N natural numbers: f0; 1; 2; : : :g C complex numbers
f: : :¯¯ : : :g set of . . . such that . . .
h: : :i sequence; like a set but order matters
V;W;U vector spaces
~v; ~w vectors
~0, ~0V zero vector, zero vector of V
B;D bases
En = h~e1; : : : ; ~eni standard basis for Rn
~¯;~± basis vectors
RepB(~v) matrix representing the vector
Pn set of n-th degree polynomials
Mn£m set of n£m matrices
[S] span of the set S
M © N direct sum of subspaces
V »= W isomorphic spaces
h; g homomorphisms, linear maps
H;G matrices
t; s transformations; maps from a space to itself
T; S square matrices
RepB;D(h) matrix representing the map h
hi;j matrix entry from row i, column j
jTj determinant of the matrix T
R(h);N (h) rangespace and nullspace of the map h
R1(h);N1(h) generalized rangespace and nullspace
Lower case Greek alphabet
name character name character name character
alpha ® iota ¶ rho ½
beta ¯ kappa · sigma ¾
gamma ° lambda ¸ tau ¿
delta ± mu ¹ upsilon À
epsilon ² nu º phi Á
zeta ³ xi » chi Â
eta ´ omicron o psi Ã
theta µ pi ¼ omega !
Cover. This is Cramer’s Rule for the system x1 + 2x2 = 6, 3x1 + x2 = 8. The size of
the first box is the determinant shown (the absolute value of the size is the area). The
size of the second box is x1 times that, and equals the size of the final box. Hence, x1
is the final determinant divided by the first determinant.
Contents
Chapter One: Linear Systems 1
I Solving Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . 1
1 Gauss’ Method . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2 Describing the Solution Set . . . . . . . . . . . . . . . . . . . . 11
3 General = Particular + Homogeneous . . . . . . . . . . . . . . 20
II Linear Geometry of n-Space . . . . . . . . . . . . . . . . . . . . . 32
1 Vectors in Space . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2 Length and Angle Measures¤ . . . . . . . . . . . . . . . . . . . 38
III Reduced Echelon Form . . . . . . . . . . . . . . . . . . . . . . . . 46
1 Gauss-Jordan Reduction . . . . . . . . . . . . . . . . . . . . . . 46
2 Row Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . 52
Topic: Computer Algebra Systems . . . . . . . . . . . . . . . . . . . 62
Topic: Input-Output Analysis . . . . . . . . . . . . . . . . . . . . . . 64
Topic: Accuracy of Computations . . . . . . . . . . . . . . . . . . . . 68
Topic: Analyzing Networks . . . . . . . . . . . . . . . . . . . . . . . . 72
Chapter Two: Vector Spaces 79
I Definition of Vector Space . . . . . . . . . . . . . . . . . . . . . . 80
1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . 80
2 Subspaces and Spanning Sets . . . . . . . . . . . . . . . . . . . 91
II Linear Independence . . . . . . . . . . . . . . . . . . . . . . . . . 102
1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . 102
III Basis and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . 113
1 Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
2 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
3 Vector Spaces and Linear Systems . . . . . . . . . . . . . . . . 124
4 Combining Subspaces¤ . . . . . . . . . . . . . . . . . . . . . . . 131
Topic: Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
Topic: Crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
Topic: Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . 147
vii
Chapter Three: Maps Between Spaces 155
I Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . 155
2 Dimension Characterizes Isomorphism . . . . . . . . . . . . . . 164
II Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
2 Rangespace and Nullspace . . . . . . . . . . . . . . . . . . . . . 179
III Computing Linear Maps . . . . . . . . . . . . . . . . . . . . . . . 191
1 Representing Linear Maps with Matrices . . . . . . . . . . . . . 191
2 Any Matrix Represents a Linear Map¤ . . . . . . . . . . . . . . 201
IV Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . 208
1 Sums and Scalar Products . . . . . . . . . . . . . . . . . . . . . 208
2 Matrix Multiplication . . . . . . . . . . . . . . . . . . . . . . . 210
3 Mechanics of Matrix Multiplication . . . . . . . . . . . . . . . . 218
4 Inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227
V Change of Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
1 Changing Representations of Vectors . . . . . . . . . . . . . . . 234
2 Changing Map Representations . . . . . . . . . . . . . . . . . . 238
VI Projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246
1 Orthogonal Projection Into a Line¤ . . . . . . . . . . . . . . . . 246
2 Gram-Schmidt Orthogonalization¤ . . . . . . . . . . . . . . . . 250
3 Projection Into a Subspace¤ . . . . . . . . . . . . . . . . . . . . 256
Topic: Line of Best Fit . . . . . . . . . . . . . . . . . . . . . . . . . . 265
Topic: Geometry of Linear Maps . . . . . . . . . . . . . . . . . . . . 270
Topic: Markov Chains . . . . . . . . . . . . . . . . . . . . . . . . . . 277
Topic: Orthonormal Matrices . . . . . . . . . . . . . . . . . . . . . . 283
Chapter Four: Determinants 289
I Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290
1 Exploration¤ . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290
2 Properties of Determinants . . . . . . . . . . . . . . . . . . . . 295
3 The Permutation Expansion . . . . . . . . . . . . . . . . . . . . 299
4 Determinants Exist¤ . . . . . . . . . . . . . . . . . . . . . . . . 308
II Geometry of Determinants . . . . . . . . . . . . . . . . . . . . . . 315
1 Determinants as Size Functions . . . . . . . . . . . . . . . . . . 315
III Other Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322
1 Laplace’s Expansion¤ . . . . . . . . . . . . . . . . . . . . . . . . 322
Topic: Cramer’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 327
Topic: Speed of Calculating Determinants . . . . . . . . . . . . . . . 330
Topic: Projective Geometry . . . . . . . . . . . . . . . . . . . . . . . 333
Chapter Five: Similarity 345
I Complex Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . 345
1 Factoring and Complex Numbers; A Review¤ . . . . . . . . . . 346
2 Complex Representations . . . . . . . . . . . . . . . . . . . . . 347
II Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349
viii
1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . 349
2 Diagonalizability . . . . . . . . . . . . . . . . . . . . . . . . . . 351
3 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . 355
III Nilpotence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
1 Self-Composition¤ . . . . . . . . . . . . . . . . . . . . . . . . . 363
2 Strings¤ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366
IV Jordan Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377
1 Polynomials of Maps and Matrices¤ . . . . . . . . . . . . . . . . 377
2 Jordan Canonical Form¤ . . . . . . . . . . . . . . . . . . . . . . 384
Topic: Method of Powers . . . . . . . . . . . . . . . . . . . . . . . . . 397
Topic: Stable Populations . . . . . . . . . . . . . . . . . . . . . . . . 401
Topic: Linear Recurrences . . . . . . . . . . . . . . . . . . . . . . . . 403
Appendix A-1
Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A-1
Quantifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A-3
Techniques of Proof . . . . . . . . . . . . . . . . . . . . . . . . . . A-5
Sets, Functions, and Relations . . . . . . . . . . . . . . . . . . . . . A-7
¤Note: starred subsections are optional.
ix

Chapter Two
Vector Spaces
The first chapter began by introducing Gauss’ method and finished with a fair
understanding, keyed on the Linear Combination Lemma, of how it finds the
solution set of a linear system. Gauss’ method systematically takes linear combinations
of the rows. With that insight, we now move to a general study of
linear combinations.
We need a setting for this study. At times in the first chapter, we’ve combined
vectors from R2, at other times vectors from R3, and at other times vectors
from even higher-dimensional spaces. Thus, our first impulse might be to work
in Rn, leaving n unspecified. This would have the advantage that any of the
results would hold for R2 and for R3 and for many other spaces, simultaneously.
But, if having the results apply to many spaces at once is advantageous then
sticking only to Rn’s is overly restrictive. We’d like the results to also apply to
combinations of row vectors, as in the final section of the first chapter. We’ve
even seen some spaces that are not just a collection of all of the same-sized
column vectors or row vectors. For instance, we’ve seen a solution set of a
homogeneous system that is a plane, inside of R3. This solution set is a closed
system in the sense that a linear combination of these solutions is also a solution.
But it is not just a collection of all of the three-tall column vectors; only some
of them are in this solution set.
We want the results about linear combinations to apply anywhere that linear
combinations are sensible. We shall call any such set a vector space. Our results,
instead of being phrased as “Whenever we have a collection in which we can
sensibly take linear combinations . . . ”, will be stated as “In any vector space
. . . ”.
Such a statement describes at once what happens in many spaces. The step
up in abstraction from studying a single space at a time to studying a class
of spaces can be hard to make. To understand its advantages, consider this
analogy. Imagine that the government made laws one person at a time: “Leslie
Jones can’t jay walk.” That would be a bad idea; statements have the virtue of
economy when they apply to many cases at once. Or, suppose that they ruled,
“Kim Ke must stop when passing the scene of an accident.” Contrast that with,
“Any doctor must stop when passing the scene of an accident.” More general
statements, in some ways, are clearer.
79
80 Chapter Two. Vector Spaces
I Definition of Vector Space
We shall study structures with two operations, an addition and a scalar multiplication,
that are subject to some simple conditions. We will reflect more on
the conditions later, but on first reading notice how reasonable they are. For
instance, surely any operation that can be called an addition (e.g., column vector
addition, row vector addition, or real number addition) will satisfy all the
conditions in (1) below.
I.1 Definition and Examples
1.1 Definition A vector space (over R) consists of a set V along with two
operations ‘+’ and ‘¢’ such that
(1) if ~v; ~w 2 V then their vector sum ~v + ~w is in V and
² ~v + ~w = ~w +~v
² (~v + ~w) + ~u = ~v + ( ~w + ~u) (where ~u 2 V )
² there is a zero vector ~0 2 V such that ~v +~0 = ~v for all ~v 2 V
² each ~v 2 V has an additive inverse ~w 2 V such that ~w +~v = ~0
(2) if r; s are scalars (members of R) and ~v; ~w 2 V then each scalar multiple
r ¢ ~v is in V and
² (r + s) ¢ ~v = r ¢ ~v + s ¢ ~v
² r ¢ (~v + ~w) = r ¢ ~v + r ¢ ~w
² (rs) ¢ ~v = r ¢ (s ¢ ~v)
² 1 ¢ ~v = ~v.
1.2 Remark Because it involves two kinds of addition and two kinds of multiplication,
that definition may seem confused. For instance, in ‘(r + s) ¢ ~v =
r ¢ ~v + s ¢ ~v ’, the first ‘+’ is the real number addition operator while the ‘+’ to
the right of the equals sign represents vector addition in the structure V . These
expressions aren’t ambiguous because, e.g., r and s are real numbers so ‘r + s’
can only mean real number addition.
The best way to go through the examples below is to check all of the conditions
in the definition. That check is written out in the first example. Use
it as a model for the others. Especially important are the two: ‘~v + ~w is in V ’
and ‘r ¢ ~v is in V ’. These are the closure conditions. They specify that the
addition and scalar multiplication operations are always sensible—they must
be defined for every pair of vectors, and every scalar and vector, and the result
of the operation must be a member of the set (see Example 1.4).
Section I. Definition of Vector Space 81
1.3 Example The set R2 is a vector space if the operations ‘+’ and ‘¢’ have
their usual meaning.
µx1
x2¶+ µy1
y2¶= µx1 + y1
x2 + y2¶ r ¢ µx1
x2¶= µrx1
rx2¶
We shall check all of the conditions in the definition.
There are five conditions in item (1). First, for closure of addition, note that
for any v1; v2;w1;w2 2 R the result of the sum
µv1
v2¶+ µw1
w2¶= µv1 + w1
v2 + w2¶
is a column array with two real entries, and so is in R2. Second, to show that
addition of vectors commutes, take all entries to be real numbers and compute
µv1
v2¶+ µw1
w2¶= µv1 + w1
v2 + w2¶= µw1 + v1
w2 + v2¶= µw1
w2¶+ µv1
v2¶
(the second equality follows from the fact that the components of the vectors
are real numbers, and the addition of real numbers is commutative). The third
condition, associativity of vector addition, is similar.
(µv1
v2¶+ µw1
w2¶) + µu1
u2¶= µ(v1 + w1) + u1
(v2 + w2) + u2¶
= µv1 + (w1 + u1)
v2 + (w2 + u2)¶
= µv1
v2¶+ (µw1
w2¶+ µu1
u2¶)
For the fourth we must produce a zero element—the vector of zeroes is it.
µv1
v2¶+ µ00
¶= µv1
v2¶
Fifth, to produce an additive inverse, note that for any v1; v2 2 R we have
µ¡v1
¡v2¶+ µv1
v2¶= µ0
0¶
so the first vector is the desired additive inverse of the second.
The checks for the five conditions in item (2) are just as routine. First, for
closure under scalar multiplication, where r; v1; v2 2 R,
r ¢ µv1
v2¶= µrv1
rv2¶
is a column array with two real entries, and so is in R2. This checks the second
condition.
(r + s) ¢ µv1
v2¶= µ(r + s)v1
(r + s)v2¶= µrv1 + sv1
rv2 + sv2¶= r ¢ µv1
v2¶+ s ¢ µv1
v2¶
82 Chapter Two. Vector Spaces
For the third condition, that scalar multiplication distributes from the left over
vector addition, the check is also straightforward.
r ¢ (µv1
v2¶+ µw1
w2¶) = µr(v1 + w1)
r(v2 + w2)¶= µrv1 + rw1
rv2 + rw2¶= r ¢ µv1
v2¶+ r ¢ µw1
w2¶
The fourth
(rs) ¢ µv1
v2¶= µ(rs)v1
(rs)v2¶= µr(sv1)
r(sv2)¶= r ¢ (s ¢ µv1
v2¶)
and fifth conditions are also easy.
1 ¢ µv1
v2¶= µ1v1
1v2¶= µv1
v2¶
In a similar way, each Rn is a vector space with the usual operations of
vector addition and scalar multiplication. (In R1, we usually do not write the
members as column vectors, i.e., we usually do not write ‘(¼)’. Instead we just
write ‘¼’.)
1.4 Example This subset of R3 that is a plane through the origin
P = f0@
x
y
z1A
¯¯
x + y + z = 0g
is a vector space if ‘+’ and ‘¢’ are interpreted in this way.
0@
x1
y1
z1
1A
+0@
x2
y2
z2
1A
=0@
x1 + x2
y1 + y2
z1 + z2
1A
r ¢0@
x
y
z1A =0@
rx
ry
rz1A
The addition and scalar multiplication operations here are just the ones of R3,
reused on its subset P. We say that P inherits these operations from R3. This
example of an addition in P
0@
11
¡21A +0@
¡1
01
1A
=0@
01
¡11A
illustrates that P is closed under addition. We’ve added two vectors from P —
that is, with the property that the sum of their three entries is zero—and the
result is a vector also in P. Of course, this example of closure is not a proof of
closure. To prove that P is closed under addition, take two elements of P
0@
x1
y1
z1
1A
0@
x2
y2
z2
1A
Section I. Definition of Vector Space 83
(membership in P means that x1 + y1 + z1 = 0 and x2 + y2 + z2 = 0), and
observe that their sum 0@
x1 + x2
y1 + y2
z1 + z2
1A
is also in P since its entries add (x1 + x2) + (y1 + y2) + (z1 + z2) = (x1 + y1 +
z1) + (x2 + y2 + z2) to 0. To show that P is closed under scalar multiplication,
start with a vector from P 0@
x
y
z1A (so that x + y + z = 0) and then for r 2 R observe that the scalar multiple
r ¢0@
x
y
z1A =0@
rx
ry
rz1A
satisfies that rx + ry + rz = r(x + y + z) = 0. Thus the two closure conditions
are satisfied. The checks for the other conditions in the definition of a vector
space are just as straightforward.
1.5 Example Example 1.3 shows that the set of all two-tall vectors with real
entries is a vector space. Example 1.4 gives a subset of an Rn that is also a
vector space. In contrast with those two, consider the set of two-tall columns
with entries that are integers (under the obvious operations). This is a subset
of a vector space, but it is not itself a vector space. The reason is that this set is
not closed under scalar multiplication, that is, it does not satisfy requirement (2)
in the definition. Here is a column with integer entries, and a scalar, such that
the outcome of the operation
0:5 ¢ µ4
3¶= µ 2
1:5¶
is not a member of the set, since its entries are not all integers.
1.6 Example The singleton set
f0BB@
0000
1CCA
g
is a vector space under the operations
0BB@
0000
1CCA
+0BB@
0000
1CCA
=0BB@
0000
1CCA
r ¢0BB@
0000
1CCA
=0BB@
0000
1CCA
that it inherits from R4.
84 Chapter Two. Vector Spaces
A vector space must have at least one element, its zero vector. Thus a
one-element vector space is the smallest one possible.
1.7 Definition A one-element vector space is a trivial space.
Warning! The examples so far involve sets of column vectors with the usual
operations. But vector spaces need not be collections of column vectors, or even
of row vectors. Below are some other types of vector spaces. The term ‘vector
space’ does not mean ‘collection of columns of reals’. It means something more
like ‘collection in which any linear combination is sensible’.
1.8 Example Consider P3 = fa0 + a1x + a2x2 + a3x3¯¯ a0; : : : ; a3 2 Rg, the
set of polynomials of degree three or less (in this book, we’ll take constant
polynomials, including the zero polynomial, to be of degree zero). It is a vector
space under the operations
(a0 + a1x + a2x2 + a3x3) + (b0 + b1x + b2x2 + b3x3)
= (a0 + b0) + (a1 + b1)x + (a2 + b2)x2 + (a3 + b3)x3
and
r ¢ (a0 + a1x + a2x2 + a3x3) = (ra0) + (ra1)x + (ra2)x2 + (ra3)x3
(the verification is easy). This vector space is worthy of attention because these
are the polynomial operations familiar from high school algebra. For instance,
3 ¢ (1 ¡ 2x + 3x2 ¡ 4x3) ¡ 2 ¢ (2 ¡ 3x + x2 ¡ (1=2)x3) = ¡1 + 7x2 ¡ 11x3.
Although this space is not a subset of any Rn, there is a sense in which we
can think of P3 as “the same” as R4. If we identify these two spaces’s elements
in this way
a0 + a1x + a2x2 + a3x3 corresponds to 0BB@
a0
a1
a2
a3
1CCA
then the operations also correspond. Here is an example of corresponding additions.
1 ¡ 2x + 0x2 + 1x3
+ 2 + 3x + 7x2 ¡ 4x3
3 + 1x + 7x2 ¡ 3x3
corresponds to 0BB@
1
¡2
01
1CCA
+0BB@
237
¡4
1CCA
=0BB@
317
¡3
1CCA
Things we are thinking of as “the same” add to “the same” sum. Chapter Three
makes precise this idea of vector space correspondence. For now we shall just
leave it as an intuition.
Section I. Definition of Vector Space 85
1.9 Example The set ff¯¯ f : N ! Rg of all real-valued functions of one natural
number variable is a vector space under the operations
(f1 + f2) (n) = f1(n) + f2(n) (r ¢ f) (n) = r f(n)
so that if, for example, f1(n) = n2 + 2 sin(n) and f2(n) = ¡sin(n) + 0:5 then
(f1 + 2f2) (n) = n2 + 1.
We can view this space as a generalization of Example 1.3 by thinking of
these functions as “the same” as infinitely-tall vectors:
n f(n) = n2 + 1
0 1
1 2
2 5
3 10
...
...
corresponds to
0BBBBB@
125
10
...
1CCCCCA
with addition and scalar multiplication are component-wise, as before. (The
“infinitely-tall” vector can be formalized as an infinite sequence, or just as a
function from N to R, in which case the above correspondence is an equality.)
1.10 Example The set of polynomials with real coefficients
fa0 + a1x + ¢ ¢ ¢ + anxn¯¯ n 2 N and a0; : : : ; an 2 Rg
makes a vector space when given the natural ‘+’
(a0 + a1x + ¢ ¢ ¢ + anxn) + (b0 + b1x + ¢ ¢ ¢ + bnxn)
= (a0 + b0) + (a1 + b1)x + ¢ ¢ ¢ + (an + bn)xn
and ‘¢’.
r ¢ (a0 + a1x + : : : anxn) = (ra0) + (ra1)x + : : : (ran)xn
This space differs from the space P3 of Example 1.8. This space contains not just
degree three polynomials, but degree thirty polynomials and degree three hundred
polynomials, too. Each individual polynomial of course is of a finite degree,
but the set has no single bound on the degree of all of its members.
This example, like the prior one, can be thought of in terms of infinite-tuples.
For instance, we can think of 1 + 3x +5x2 as corresponding to (1; 3; 5; 0; 0; : : :).
However, don’t confuse this space with the one from Example 1.9. Each member
of this set has a bounded degree, so under our correspondence there are no
elements from this space matching (1; 2; 5; 10; : : : ). The vectors in this space
correspond to infinite-tuples that end in zeroes.
1.11 Example The set ff¯¯ f : R ! Rg of all real-valued functions of one real
variable is a vector space under these.
(f1 + f2) (x) = f1(x) + f2(x) (r ¢ f) (x) = r f(x)
The difference between this and Example 1.9 is the domain of the functions.
86 Chapter Two. Vector Spaces
1.12 Example The set F = fa cos µ+b sin µ¯¯ a; b 2 Rg of real-valued functions
of the real variable µ is a vector space under the operations
(a1 cos µ + b1 sin µ) + (a2 cos µ + b2 sin µ) = (a1 + a2) cos µ + (b1 + b2) sin µ
and
r ¢ (a cos µ + b sin µ) = (ra) cos µ + (rb) sin µ
inherited from the space in the prior example. (We can think of F as “the same”
as R2 in that a cos µ + b sin µ corresponds to the vector with components a and
b.)
1.13 Example The set
ff : R ! R¯¯ d2f
dx2 + f = 0g
is a vector space under the, by now natural, interpretation.
(f + g) (x) = f(x) + g(x) (r ¢ f) (x) = r f(x)
In particular, notice that closure is a consequence:
d2(f + g)
dx2 + (f + g) = (d2f
dx2 + f) + ( d2g
dx2 + g)
and
d2(rf)
dx2 + (rf) = r(d2f
dx2 + f)
of basic Calculus. This turns out to equal the space from the prior example—
functions satisfying this differential equation have the form a cos µ + b sin µ—
but this description suggests an extension to solutions sets of other differential
equations.
1.14 Example The set of solutions of a homogeneous linear system in n variables
is a vector space under the operations inherited from Rn. For closure
under addition, if
~v =0B@
v1
...
vn
1CA
~w =0B@
w1
...
wn
1CA
both satisfy the condition that their entries add to zero then ~v+ ~w also satisfies
that condition: c1(v1 +w1)+¢ ¢ ¢+cn(vn +wn) = (c1v1 +¢ ¢ ¢+cnvn)+(c1w1 +
¢ ¢ ¢ + cnwn) = 0. The checks of the other conditions are just as routine.
As we’ve done in those equations, we often omit the multiplication symbol ‘¢’.
We can distinguish the multiplication in ‘c1v1’ from that in ‘r~v ’ since if both
multiplicands are real numbers then real-real multiplication must be meant,
while if one is a vector then scalar-vector multiplication must be meant.
The prior example has brought us full circle since it is one of our motivating
examples.
Section I. Definition of Vector Space 87
1.15 Remark Now, with some feel for the kinds of structures that satisfy the
definition of a vector space, we can reflect on that definition. For example, why
specify in the definition the condition that 1 ¢ ~v = ~v but not a condition that
0 ¢ ~v = ~0?
One answer is that this is just a definition—it gives the rules of the game
from here on, and if you don’t like it, put the book down and walk away.
Another answer is perhaps more satisfying. People in this area have worked
hard to develop the right balance of power and generality. This definition has
been shaped so that it contains the conditions needed to prove all of the interesting
and important properties of spaces of linear combinations, and so that it
does not contain extra conditions that only bar as examples spaces where those
properties occur. As we proceed, we shall derive all of the properties natural to
collections of linear combinations from the conditions given in the definition.
The next result is an example. We do not need to include these properties
in the definition of vector space because they follow from the properties already
listed there.
1.16 Lemma In any vector space V ,
(1) 0 ¢ ~v = ~0
(2) (¡1 ¢ ~v) +~v = ~0
(3) r ¢ ~0 = ~0
for any ~v 2 V and r 2 R.
Proof. For the first item, note that ~v = (1 + 0) ¢ ~v = ~v + (0 ¢ ~v). Add to both
sides the additive inverse of ~v, the vector ~w such that ~w +~v = ~0.
~w +~v = ~w +~v + 0 ¢ ~v
~0 = ~0 + 0 ¢ ~v
~0 = 0 ¢ ~v
The second item is easy: (¡1 ¢ ~v) + ~v = (¡1 + 1) ¢ ~v = 0 ¢ ~v = ~0 shows that
we can write ‘¡~v ’ for the additive inverse of ~v without worrying about possible
confusion with (¡1) ¢ ~v.
For the third one, this r ¢ ~0 = r ¢ (0 ¢ ~0) = (r ¢ 0) ¢ ~0 = ~0 will do. QED
We finish this subsection with a recap, and a comment.
Chapter One studied Gaussian reduction. That led us to study collections
of linear combinations. We have named any such structure a ‘vector space’. In
a phrase, the point of this material is that vector spaces are the right context
in which to study linearity.
Finally, a comment. From the fact that it forms a whole chapter, and especially
because that chapter is the first one, a reader could come to think that
the study of linear systems is our purpose. The truth is, we will not so much
use vector spaces in the study of linear systems as we will instead have linear
88 Chapter Two. Vector Spaces
systems start us on the study of vector spaces. The wide variety of examples
from this subsection shows that the study of vector spaces is interesting and important
in its own right, aside from how it helps us understand linear systems.
Linear systems won’t go away. But from now on our primary objects of study
will be vector spaces.
Exercises
1.17 Give the zero vector from each of these vector spaces.
(a) The space of degree three polynomials under the natural operations
(b) The space of 2£4 matrices
(c) The space ff : [0::1] ! R¯¯ f is continuousg
(d) The space of real-valued functions of one natural number variable
X 1.18 Find the additive inverse, in the vector space, of the vector.
(a) In P3, the vector ¡3 ¡ 2x + x2
(b) In the space of 2£2 matrices with real number entries under the usual matrix
addition and scalar multiplication,µ1 ¡1
0 3 ¶
(c) In faex + be¡x¯¯ a; b 2 Rg, a space of functions of the real variable x under
the natural operations, the vector 3ex ¡ 2e¡x.
X 1.19 Show that each of these is a vector space.
(a) The set of linear polynomials P1 = fa0 + a1x¯¯ a0; a1 2 Rg under the usual
polynomial addition and scalar multiplication operations.
(b) The set of 2£2 matrices with real entries under the usual matrix operations.
(c) The set of three-component row vectors with their usual operations.
(d) The set
L = f0B@
x
y
z
w
1CA
2 R4¯¯ x + y ¡ z + w = 0g
under the operations inherited from R4.
X 1.20 Show that each of these is not a vector space. (Hint. Start by listing two
members of each set.)
(a) Under the operations inherited from R3, this set
fÃx
y
z!2 R3¯¯ x + y + z = 1g
(b) Under the operations inherited from R3, this set
fÃx
y
z!2 R3¯¯ x2 + y2 + z2 = 1g
(c) Under the usual matrix operations,
fµa 1
b c¶¯¯ a; b; c 2 Rg
(d) Under the usual polynomial operations,
fa0 + a1x + a2x2¯¯ a0; a1; a2 2 R+g
where R+ is the set of reals greater than zero
Section I. Definition of Vector Space 89
(e) Under the inherited operations,
fµx
y¶2 R2¯¯ x + 3y = 4 and 2x ¡ y = 3 and 6x + 4y = 10g
1.21 Define addition and scalar multiplication operations to make the complex
numbers a vector space over R.
X 1.22 Is the set of rational numbers a vector space over R under the usual addition
and scalar multiplication operations?
1.23 Show that the set of linear combinations of the variables x; y; z is a vector
space under the natural addition and scalar multiplication operations.
1.24 Prove that this is not a vector space: the set of two-tall column vectors with
real entries subject to these operations.
µx1
y1¶+ µx2
y2¶= µx1 ¡ x2
y1 ¡ y2¶ r ¢ µx
y¶= µrx
ry¶
1.25 Prove or disprove that R3 is a vector space under these operations.
(a) Ãx1
y1
z1!+ Ãx2
y2
z2!= Ã0
0
0! and rÃx
y
z!= Ãrx
ry
rz!
(b) Ãx1
y1
z1!+ Ãx2
y2
z2!= Ã0
0
0! and rÃx
y
z!= Ã0
0
0!
X 1.26 For each, decide if it is a vector space; the intended operations are the natural
ones.
(a) The diagonal 2£2 matrices
fµa 0
0 b¶¯¯ a; b 2 Rg
(b) This set of 2£2 matrices
fµ x x + y
x + y y ¶¯¯ x; y 2 Rg
(c) This set
f0B@
x
y
z
w
1CA
2 R4¯¯ x + y + w = 1g
(d) The set of functions ff : R ! R¯¯ df=dx + 2f = 0g
(e) The set of functions ff : R ! R¯¯ df=dx + 2f = 1g
X 1.27 Prove or disprove that this is a vector space: the real-valued functions f of
one real variable such that f(7) = 0.
X 1.28 Show that the set R+ of positive reals is a vector space when ‘x + y’ is interpreted
to mean the product of x and y (so that 2+3 is 6), and ‘r ¢ x’ is interpreted
as the r-th power of x.
1.29 Is f(x; y)¯¯ x; y 2 Rg a vector space under these operations?
(a) (x1; y1) + (x2; y2) = (x1 + x2; y1 + y2) and r(x; y) = (rx; y)
(b) (x1; y1) + (x2; y2) = (x1 + x2; y1 + y2) and r ¢ (x; y) = (rx; 0)
1.30 Prove or disprove that this is a vector space: the set of polynomials of degree
greater than or equal to two, along with the zero polynomial.
90 Chapter Two. Vector Spaces
1.31 At this point “the same” is only an intuition, but nonetheless for each vector
space identify the k for which the space is “the same” as Rk.
(a) The 2£3 matrices under the usual operations
(b) The n£m matrices (under their usual operations)
(c) This set of 2£2 matrices
fµa 0
b c¶¯¯ a; b; c 2 Rg
(d) This set of 2£2 matrices
fµa 0
b c¶¯¯ a + b + c = 0g
X 1.32 Using ~+ to represent vector addition and ~¢ for scalar multiplication, restate
the definition of vector space.
X 1.33 Prove these.
(a) Any vector is the additive inverse of the additive inverse of itself.
(b) Vector addition left-cancels: if ~v; ~s;~t 2 V then ~v + ~s = ~v +~t implies that
~s =~t.
1.34 The definition of vector spaces does not explicitly say that ~0+~v = ~v (it instead
says that ~v +~0 = ~v). Show that it must nonetheless hold in any vector space.
X 1.35 Prove or disprove that this is a vector space: the set of all matrices, under
the usual operations.
1.36 In a vector space every element has an additive inverse. Can some elements
have two or more?
1.37 (a) Prove that every point, line, or plane thru the origin in R3 is a vector
space under the inherited operations.
(b) What if it doesn’t contain the origin?
X 1.38 Using the idea of a vector space we can easily reprove that the solution set of
a homogeneous linear system has either one element or infinitely many elements.
Assume that ~v 2 V is not ~0.
(a) Prove that r ¢ ~v = ~0 if and only if r = 0.
(b) Prove that r1 ¢ ~v = r2 ¢ ~v if and only if r1 = r2.
(c) Prove that any nontrivial vector space is infinite.
(d) Use the fact that a nonempty solution set of a homogeneous linear system is
a vector space to draw the conclusion.
1.39 Is this a vector space under the natural operations: the real-valued functions
of one real variable that are differentiable?
1.40 A vector space over the complex numbers C has the same definition as a vector
space over the reals except that scalars are drawn from C instead of from R. Show
that each of these is a vector space over the complex numbers. (Recall how complex
numbers add and multiply: (a0 + a1i) + (b0 + b1i) = (a0 + b0) + (a1 + b1)i and
(a0 + a1i)(b0 + b1i) = (a0b0 ¡ a1b1) + (a0b1 + a1b0)i.)
(a) The set of degree two polynomials with complex coefficients
(b) This set
fµ0 a
b 0¶¯¯ a; b 2 C and a + b = 0 + 0ig
1.41 Find a property shared by all of the Rn’s not listed as a requirement for a
vector space.
Section I. Definition of Vector Space 91
X 1.42 (a) Prove that a sum of four vectors ~v1; : : : ; ~v4 2 V can be associated in any
way without changing the result.
((~v1 +~v2) +~v3) +~v4 = (~v1 + (~v2 +~v3)) +~v4
= (~v1 +~v2) + (~v3 +~v4)
= ~v1 + ((~v2 +~v3) +~v4)
= ~v1 + (~v2 + (~v3 +~v4))
This allows us to simply write ‘~v1 +~v2 +~v3 +~v4’ without ambiguity.
(b) Prove that any two ways of associating a sum of any number of vectors give
the same sum. (Hint. Use induction on the number of vectors.)
1.43 For any vector space, a subset that is itself a vector space under the inherited
operations (e.g., a plane through the origin inside of R3) is a subspace.
(a) Show that fa0 + a1x + a2x2¯¯ a0 + a1 + a2 = 0g is a subspace of the vector
space of degree two polynomials.
(b) Show that this is a subspace of the 2£2 matrices.
fµa b
c 0¶¯¯ a + b = 0g
(c) Show that a nonempty subset S of a real vector space is a subspace if and only
if it is closed under linear combinations of pairs of vectors: whenever c1; c2 2 R
and ~s1; ~s2 2 S then the combination c1~v1 + c2~v2 is in S.
I.2 Subspaces and Spanning Sets
One of the examples that led us to introduce the idea of a vector space was the
solution set of a homogeneous system. For instance, we’ve seen in Example 1.4
such a space that is a planar subset of R3. There, the vector space R3 contains
inside it another vector space, the plane.
2.1 Definition For any vector space, a subspace is a subset that is itself a
vector space, under the inherited operations.
2.2 Example The plane from the prior subsection,
P = f0@
x
y
z1A
¯¯
x + y + z = 0g
is a subspace of R3. As specified in the definition, the operations are the ones
that are inherited from the larger space, that is, vectors add in P3 as they add
in R3 0@
x1
y1
z1
1A
+0@
x2
y2
z2
1A
=0@
x1 + x2
y1 + y2
z1 + z2
1A
and scalar multiplication is also the same as it is in R3. To show that P is a
subspace, we need only note that it is a subset and then verify that it is a space.
92 Chapter Two. Vector Spaces
Checking that P satisfies the conditions in the definition of a vector space is
routine. For instance, for closure under addition, just note that if the summands
satisfy that x1 + y1 + z1 = 0 and x2 + y2 + z2 = 0 then the sum satisfies that
(x1 + x2) + (y1 + y2) + (z1 + z2) = (x1 + y1 + z1) + (x2 + y2 + z2) = 0.
2.3 Example The x-axis in R2 is a subspace where the addition and scalar
multiplication operations are the inherited ones.
µx1
0 ¶+ µx2
0 ¶= µx1 + x2
0 ¶ r ¢ µx
0 ¶= µrx
0 ¶
As above, to verify that this is a subspace, we simply note that it is a subset
and then check that it satisfies the conditions in definition of a vector space.
For instance, the two closure conditions are satisfied: (1) adding two vectors
with a second component of zero results in a vector with a second component
of zero, and (2) multiplying a scalar times a vector with a second component of
zero results in a vector with a second component of zero.
2.4 Example Another subspace of R2 is
fµ00
¶g
its trivial subspace.
Any vector space has a trivial subspace f~0 g. At the opposite extreme, any
vector space has itself for a subspace. These two are the improper subspaces.
Other subspaces are proper.
2.5 Example The condition in the definition requiring that the addition and
scalar multiplication operations must be the ones inherited from the larger space
is important. Consider the subset f1g of the vector space R1. Under the operations
1+1 = 1 and r ¢1 = 1 that set is a vector space, specifically, a trivial space.
But it is not a subspace of R1 because those aren’t the inherited operations, since
of course R1 has 1 + 1 = 2.
2.6 Example All kinds of vector spaces, not just Rn’s, have subspaces. The
vector space of cubic polynomials fa + bx + cx2 + dx3¯¯ a; b; c; d 2 Rg has a subspace
comprised of all linear polynomials fm + nx¯¯ m; n 2 Rg.
2.7 Example Another example of a subspace not taken from an Rn is one
from the examples following the definition of a vector space. The space of all
real-valued functions of one real variable f : R ! R has a subspace of functions
satisfying the restriction (d2 f=dx2) + f = 0.
2.8 Example Being vector spaces themselves, subspaces must satisfy the closure
conditions. The set R+ is not a subspace of the vector space R1 because
with the inherited operations it is not closed under scalar multiplication: if
~v = 1 then ¡1 ¢ ~v 62 R+.
Section I. Definition of Vector Space 93
The next result says that Example 2.8 is prototypical. The only way that a
subset can fail to be a subspace (if it is nonempty and the inherited operations
are used) is if it isn’t closed.
2.9 Lemma For a nonempty subset S of a vector space, under the inherited
operations, the following are equivalent statements.¤
(1) S is a subspace of that vector space
(2) S is closed under linear combinations of pairs of vectors: for any vectors
~s1; ~s2 2 S and scalars r1; r2 the vector r1~s1 + r2~s2 is in S
(3) S is closed under linear combinations of any number of vectors: for any
vectors ~s1; : : : ; ~sn 2 S and scalars r1; : : : ; rn the vector r1~s1 + ¢ ¢ ¢ + rn~sn is
in S.
Briefly, the way that a subset gets to be a subspace is by being closed under
linear combinations.
Proof. ‘The following are equivalent’ means that each pair of statements are
equivalent.
(1) () (2) (2) () (3) (3) () (1)
We will show this equivalence by establishing that (1) =) (3) =) (2) =) (1). This strategy is suggested by noticing that (1) =) (3) and (3) =) (2)
are easy and so we need only argue the single implication (2) =) (1).
For that argument, assume that S is a nonempty subset of a vector space V
and that S is closed under combinations of pairs of vectors. We will show that
S is a vector space by checking the conditions.
The first item in the vector space definition has five conditions. First, for
closure under addition, if ~s1; ~s2 2 S then ~s1 +~s2 2 S, as ~s1 +~s2 = 1 ¢ ~s1 +1 ¢~s2.
Second, for any ~s1; ~s2 2 S, because addition is inherited from V , the sum ~s1+~s2
in S equals the sum ~s1+~s2 in V , and that equals the sum ~s2+~s1 in V (because
V is a vector space, its addition is commutative), and that in turn equals the
sum ~s2+~s1 in S. The argument for the third condition is similar to that for the
second. For the fourth, consider the zero vector of V and note that closure of S
under linear combinations of pairs of vectors gives that (where ~s is any member
of the nonempty set S) 0 ¢ ~s + 0 ¢ ~s = ~0 is in S; showing that ~0 acts under the
inherited operations as the additive identity of S is easy. The fifth condition is
satisfied because for any ~s 2 S, closure under linear combinations shows that
the vector 0 ¢ ~0 + (¡1) ¢ ~s is in S; showing that it is the additive inverse of ~s
under the inherited operations is routine.
The checks for item (2) are similar and are saved for Exercise 32. QED
We usually show that a subset is a subspace with (2) =) (1).
2.10 Remark At the start of this chapter we introduced vector spaces as
collections in which linear combinations are “sensible”. The above result speaks
to this.
¤ More information on equivalence of statements is in the appendix.
94 Chapter Two. Vector Spaces
The vector space definition has ten conditions but eight of them, the ones
stated there with the ‘²’ bullets, simply ensure that referring to the operations
as an ‘addition’ and a ‘scalar multiplication’ is sensible. The proof above checks
that if the nonempty set S satisfies statement (2) then inheritance of the operations
from the surrounding vector space brings with it the inheritance of these
eight properties also (i.e., commutativity of addition in S follows right from
commutativity of addition in V ). So, in this context, this meaning of “sensible”
is automatically satisfied.
In assuring us that this first meaning of the word is met, the result draws
our attention to the second meaning. It has to do with the two remaining
conditions, the closure conditions. Above, the two separate closure conditions
inherent in statement (1) are combined in statement (2) into the single condition
of closure under all linear combinations of two vectors, which is then extended
in statement (3) to closure under combinations of any number of vectors. The
latter two statements say that we can always make sense of an expression like
r1~s1 + r2~s2, without restrictions on the r’s—such expressions are “sensible” in
that the vector described is defined and is in the set S.
This second meaning suggests that a good way to think of a vector space
is as a collection of unrestricted linear combinations. The next two examples
take some spaces and describe them in this way. That is, in these examples we
paramatrize, just as we did in Chapter One to describe the solution set of a
homogeneous linear system.
2.11 Example This subset of R3
S = f0@
x
y
z1A
¯¯
x ¡ 2y + z = 0g
is a subspace under the usual addition and scalar multiplication operations of
column vectors (the check that it is nonempty and closed under linear combinations
of two vectors is just like the one in Example 2.2). To paramatrize, we
can take x ¡ 2y + z = 0 to be a one-equation linear system and expressing the
leading variable in terms of the free variables x = 2y ¡ z.
S = f0@
2y ¡ z
y
z 1A
¯¯
y; z 2 Rg = fy0@
210
1A
+ z0@
¡1
01
1A
¯¯
y; z 2 Rg
Now the subspace is described as the collection of unrestricted linear combinations
of those two vectors. Of course, in either description, this is a plane
through the origin.
2.12 Example This is a subspace of the 2£2 matrices
L = fµa 0
b c¶¯¯ a + b + c = 0g
Section I. Definition of Vector Space 95
(checking that it is nonempty and closed under linear combinations is easy). To
paramatrize, express the condition as a = ¡b ¡ c.
L = fµ¡b ¡ c 0
b c¶¯¯ b; c 2 Rg = fbµ¡1 0
1 0¶+ cµ¡1 0
0 1¶¯¯ b; c 2 Rg
As above, we’ve described the subspace as a collection of unrestricted linear
combinations (by coincidence, also of two elements).
Paramatrization is an easy technique, but it is important. We shall use it
often.
2.13 Definition The span (or linear closure) of a nonempty subset S of a
vector space is the set of all linear combinations of vectors from S.
[S] = fc1~s1 + ¢ ¢ ¢ + cn~sn¯¯ c1; : : : ; cn 2 R and ~s1; : : : ; ~sn 2 Sg
The span of the empty subset of a vector space is the trivial subspace.
No notation for the span is completely standard. The square brackets used here
are common, but so are ‘span(S)’ and ‘sp(S)’.
2.14 Remark In Chapter One, after we showed that the solution set of a
homogeneous linear system can written as fc1~¯1 + ¢ ¢ ¢ + ck ~¯k¯¯ c1; : : : ; ck 2 Rg,
we described that as the set ‘generated’ by the ~¯’s. We now have the technical
term; we call that the ‘span’ of the set f~¯1; : : : ;~¯kg.
Recall also the discussion of the “tricky point” in that proof. The span of
the empty set is defined to be the set f~0g because we follow the convention that
a linear combination of no vectors sums to ~0. Besides, defining the empty set’s
span to be the trivial subspace is a convienence in that it keeps results like the
next one from having annoying exceptional cases.
2.15 Lemma In a vector space, the span of any subset is a subspace.
Proof. Call the subset S. If S is empty then by definition its span is the trivial
subspace. If S is not empty then by Lemma 2.9 we need only check that the
span [S] is closed under linear combinations. For a pair of vectors from that
span, ~v = c1~s1+¢ ¢ ¢+cn~sn and ~w = cn+1~sn+1+¢ ¢ ¢+cm~sm, a linear combination
p ¢ (c1~s1 + ¢ ¢ ¢ + cn~sn) + r ¢ (cn+1~sn+1 + ¢ ¢ ¢ + cm~sm)
= pc1~s1 + ¢ ¢ ¢ + pcn~sn + rcn+1~sn+1 + ¢ ¢ ¢ + rcm~sm
(p, r scalars) is a linear combination of elements of S and so is in [S] (possibly
some of the ~si’s forming ~v equal some of the ~sj ’s from ~w, but it does not
matter). QED
The converse of the lemma holds: any subspace is the span of some set,
because a subspace is obviously the span of the set of its members. Thus a
subset of a vector space is a subspace if and only if it is a span. This fits the
96 Chapter Two. Vector Spaces
intuition that a good way to think of a vector space is as a collection in which
linear combinations are sensible.
Taken together, Lemma 2.9 and Lemma 2.15 show that the span of a subset
S of a vector space is the smallest subspace containing all the members of S.
2.16 Example In any vector space V , for any vector ~v, the set fr ¢ ~v¯¯ r 2 Rg is a subspace of V . For instance, for any vector ~v 2 R3, the line through the
origin containing that vector, fk~v¯¯ k 2 Rg is a subspace of R3. This is true even
when ~v is the zero vector, in which case the subspace is the degenerate line, the
trivial subspace.
2.17 Example The span of this set is all of R2.
fµ11
¶;µ 1
¡1¶g
Tocheck this we must show that any member of R2 is a linear combination of
these two vectors. So we ask: for which vectors (with real components x and y)
are there scalars c1 and c2 such that this holds?
c1 µ1
1¶+ c2 µ 1
¡1¶= µx
y¶
Gauss’ method
c1 + c2 = x
c1 ¡ c2 = y
¡½1+½2 ¡!
c1 + c2 = x
¡2c2 = ¡x + y
with back substitution gives c2 = (x ¡ y)=2 and c1 = (x + y)=2. These two
equations show that for any x and y that we start with, there are appropriate
coefficients c1 and c2 making the above vector equation true. For instance, for
x = 1 and y = 2 the coefficients c2 = ¡1=2 and c1 = 3=2 will do. That is, any
vector in R2 can be written as a linear combination of the two given vectors.
Since spans are subspaces, and we know that a good way to understand a
subspace is to paramatrize its description, we can try to understand a set’s span
in that way.
2.18 Example Consider, in P2, the span of the set f3x ¡ x2; 2xg. By the
definition of span, it is the subspace of unrestricted linear combinations of the
two fc1(3x ¡ x2) + c2(2x)¯¯ c1; c2 2 Rg. Clearly polynomials in this span must
have a constant term of zero. Is that necessary condition also sufficient?
We are asking: for which members a2x2 +a1x+a0 of P2 are there c1 and c2
such that a2x2 + a1x + a0 = c1(3x ¡ x2) + c2(2x)? Since polynomials are equal
if and only if their coefficients are equal, we are looking for conditions on a2,
a1, and a0 satisfying these.
¡c1 = a2
3c1 + 2c2 = a1
0 = a0
Section I. Definition of Vector Space 97
Gauss’ method gives that c1 = ¡a2, c2 = (3=2)a2 + (1=2)a1, and 0 = a0. Thus
the only condition on polynomials in the span is the condition that we knew
of—as long as a0 = 0, we can give appropriate coefficients c1 and c2 to describe
the polynomial a0 +a1x+a2x2 as in the span. For instance, for the polynomial
0 ¡ 4x + 3x2, the coefficients c1 = ¡3 and c2 = 5=2 will do. So the span of the
given set is fa1x + a2x2¯¯ a1; a2 2 Rg.
This shows, incidentally, that the set fx; x2g also spans this subspace. A
space can have more than one spanning set. Two other sets spanning this subspace
are fx; x2;¡x + 2x2g and fx; x + x2; x + 2x2; : : : g. (Naturally, we usually
prefer to work with spanning sets that have only a few members.)
2.19 Example These are the subspaces of R3 that we now know of, the trivial
subspace, the lines through the origin, the planes through the origin, and the
whole space (of course, the picture shows only a few of the infinitely many
subspaces). In the next section we will prove that R3 has no other type of
subspaces, so in fact this picture shows them all.
fx³1
0
0´+ y³0
1
0 ´+ z³0
0
1´g
»»»»»»»»»»
fx³1
0
0 ´+ y³0
1
0 ´g
³³³³³³
fx³1
0
0 ´+ z³0
0
1´g
¡
¡
fx³1
1
0 ´+ z³0
0
1 ´g . . .
¤
¤
³³³³³
fx³1
0
0 ´g
A
A
fy³0
1
0 ´g
HHHH
fy³2
1
0 ´g
¡ ¡
fy³1
1
1 ´g . . .
XXXXXXXXXXXX
PPPPPPPP
HHHHH
@
@
f³0
0
0 ´g
The subsets are described as spans of sets, using a minimal number of members,
and are shown connected to their supersets. Note that these subspaces fall
naturally into levels—planes on one level, lines on another, etc.—according to
how many vectors are in a minimal-sized spanning set.
So far in this chapter we have seen that to study the properties of linear
combinations, the right setting is a collection that is closed under these combinations.
In the first subsection we introduced such collections, vector spaces,
and we saw a great variety of examples. In this subsection we saw still more
spaces, ones that happen to be subspaces of others. In all of the variety we’ve
seen a commonality. Example 2.19 above brings it out: vector spaces and subspaces
are best understood as a span, and especially as a span of a small number
of vectors. The next section studies spanning sets that are minimal.
Exercises
X 2.20 Which of these subsets of the vector space of 2£2 matrices are subspaces
under the inherited operations? For each one that is a subspace, paramatrize its
98 Chapter Two. Vector Spaces
description. For each that is not, give a condition that fails.
(a) fµa 0
0 b¶¯¯ a; b 2 Rg
(b) fµa 0
0 b¶¯¯ a + b = 0g
(c) fµa 0
0 b¶¯¯ a + b = 5g
(d) fµa c
0 b¶¯¯ a + b = 0; c 2 Rg
X 2.21 Is this a subspace of P2: fa0 + a1x + a2x2¯¯ a0 + 2a1 + a2 = 4g? If so, paramatrize
its description.
X 2.22 Decide if the vector lies in the span of the set, inside of the space.
(a) Ã2
0
1!, fÃ1
0
0!;Ã0
0
1!g, in R3
(b) x ¡ x3, fx2; 2x + x2; x + x3g, in P3
(c) µ0 1
4 2¶, fµ1 0
1 1¶;µ2 0
2 3¶g, in M2£2
2.23 Which of these are members of the span [fcos2 x; sin2 xg] in the vector space
of real-valued functions of one real variable?
(a) f(x) = 1 (b) f(x) = 3 + x2 (c) f(x) = sin x (d) f(x) = cos(2x)
X 2.24 Which of these sets spans R3? That is, which of these sets has the property
that any three-tall vector can be expressed as a suitable linear combination of the
set’s elements?
(a) fÃ1
0
0!;Ã0
2
0!;Ã0
0
3!g (b) fÃ2
0
1!;Ã1
1
0!;Ã0
0
1!g (c) fÃ1
1
0!;Ã3
0
0!g
(d) fÃ1
0
1!;Ã3
1
0!;á1
0
0 !;Ã2
1
5!g (e) fÃ2
1
1!;Ã3
0
1!;Ã5
1
2!;Ã6
0
2!g
X 2.25 Paramatrize each subspace’s description. Then express each subspace as a
span.
(a) The subset f¡a b c¢¯¯ a ¡ c = 0g of the three-wide row vectors
(b) This subset of M2£2
fµa b
c d¶¯¯ a + d = 0g
(c) This subset of M2£2
fµa b
c d¶¯¯ 2a ¡ c ¡ d = 0 and a + 3b = 0g
(d) The subset fa + bx + cx3¯¯ a ¡ 2b + c = 0g of P3
(e) The subset of P2 of quadratic polynomials p such that p(7) = 0
X 2.26 Find a set to span the given subspace of the given space. (Hint. Paramatrize
each.)
(a) the xz-plane in R3
(b) fÃx
y
z!¯¯ 3x + 2y + z = 0g in R3
Section I. Definition of Vector Space 99
(c) f0B@
x
y
z
w
1CA
¯¯
2x + y + w = 0 and y + 2z = 0g in R4
(d) fa0 + a1x + a2x2 + a3x3¯¯ a0 + a1 = 0 and a2 ¡ a3 = 0g in P3
(e) The set P4 in the space P4
(f) M2£2 in M2£2
2.27 Is R2 a subspace of R3?
X 2.28 Decide if each is a subspace of the vector space of real-valued functions of one
real variable.
(a) The even functions ff : R ! R¯¯ f(¡x) = f(x) for all xg. For example, two
members of this set are f1(x) = x2 and f2(x) = cos(x).
(b) The odd functions ff : R ! R¯¯ f(¡x) = ¡f(x) for all xg. Two members are
f3(x) = x3 and f4(x) = sin(x).
2.29 Example 2.16 says that for any vector ~v that is an element of a vector space
V , the set fr ¢ ~v¯¯ r 2 Rg is a subspace of V . (This is of course, simply the span of
the singleton set f~vg.) Must any such subspace be a proper subspace, or can it be
improper?
2.30 An example following the definition of a vector space shows that the solution
set of a homogeneous linear system is a vector space. In the terminology of this
subsection, it is a subspace of Rn where the system has n variables. What about
a non-homogeneous linear system; do its solutions form a subspace (under the
inherited operations)?
2.31 Example 2.19 shows that R3 has infinitely many subspaces. Does every nontrivial
space have infinitely many subspaces?
2.32 Finish the proof of Lemma 2.9.
2.33 Show that each vector space has only one trivial subspace.
X 2.34 Show that for any subset S of a vector space, the span of the span equals the
span [[S]] = [S]. (Hint. Members of [S] are linear combinations of members of S.
Members of [[S]] are linear combinations of linear combinations of members of S.)
2.35 All of the subspaces that we’ve seen use zero in their description in some
way. For example, the subspace in Example 2.3 consists of all the vectors from R2
with a second component of zero. In contrast, the collection of vectors from R2
with a second component of one does not form a subspace (it is not closed under
scalar multiplication). Another example is Example 2.2, where the condition on
the vectors is that the three components add to zero. If the condition were that the
three components add to ong then it would not be a subspace (again, it would fail
to be closed). This exercise shows that a reliance on zero is not strictly necessary.
Consider the set
fÃx
y
z!¯¯ x + y + z = 1g
under these operations.
Ãx1
y1
z1!+ Ãx2
y2
z2!= Ãx1 + x2 ¡ 1
y1 + y2
z1 + z2 ! rÃx
y
z!= Ãrx ¡ r + 1
ry
rz !
(a) Show that it is not a subspace of R3. (Hint. See Example 2.5).
100 Chapter Two. Vector Spaces
(b) Show that it is a vector space. Note that by the prior item, Lemma 2.9 can
not apply.
(c) Show that any subspace of R3 must pass thru the origin, and so any subspace
of R3 must involve zero in its description. Does the converse hold? Does any
subset of R3 that contains the origin become a subspace when given the inherited
operations?
2.36 We can give a justification for the convention that the sum of zero-many
vectors equals the zero vector. Consider this sum of three vectors ~v1 + ~v2 +
~v3.
(a) What is the difference between this sum of three vectors and the sum of the
first two of this three?
(b) What is the difference between the prior sum and the sum of just the first
one vector?
(c) What should be the difference between the prior sum of one vector and the
sum of no vectors?
(d) So what should be the definition of the sum of no vectors?
2.37 Is a space determined by its subspaces? That is, if two vector spaces have the
same subspaces, must the two be equal?
2.38 (a) Give a set that is closed under scalar multiplication but not addition.
(b) Give a set closed under addition but not scalar multiplication.
(c) Give a set closed under neither.
2.39 Show that the span of a set of vectors does not depend on the order in which
the vectors are listed in that set.
2.40 Which trivial subspace is the span of the empty set? Is it
fÃ0
0
0!g µ R3; or f0 + 0xg µ P1;
or some other subspace?
2.41 Show that if a vector is in the span of a set then adding that vector to the set
won’t make the span any bigger. Is that also ‘only if’?
X 2.42 Subspaces are subsets and so we naturally consider how ‘is a subspace of’
interacts with the usual set operations.
(a) If A;B are subspaces of a vector space, must A\B be a subspace? Always?
Sometimes? Never?
(b) Must A [ B be a subspace?
(c) If A is a subspace, must its complement be a subspace?
(Hint. Try some test subspaces from Example 2.19.)
X 2.43 Does the span of a set depend on the enclosing space? That is, if W is a
subspace of V and S is a subset of W (and so also a subset of V ), might the span
of S in W differ from the span of S in V ?
2.44 Is the relation ‘is a subspace of’ transitive? That is, if V is a subspace of W
and W is a subspace of X, must V be a subspace of X?
X 2.45 Because ‘span of’ is an operation on sets we naturally consider how it interacts
with the usual set operations.
(a) If S µ T are subsets of a vector space, is [S] µ [T]? Always? Sometimes?
Never?
(b) If S; T are subsets of a vector space, is [S [ T] = [S] [ [T]?
(c) If S; T are subsets of a vector space, is [S \ T] = [S] \ [T]?
Section I. Definition of Vector Space 101
(d) Is the span of the complement equal to the complement of the span?
2.46 Reprove Lemma 2.15 without doing the empty set separately.
2.47 Find a structure that is closed under linear combinations, and yet is not a
vector space. (Remark. This is a bit of a trick question.)
102 Chapter Two. Vector Spaces
II Linear Independence
The prior section shows that a vector space can be understood as an unrestricted
linear combination of some of its elements—that is, as a span. For example,
the space of linear polynomials fa + bx¯¯ a; b 2 Rg is spanned by the set f1; xg.
The prior section also showed that a space can have many sets that span it.
The space of linear polynomials is also spanned by f1; 2xg and f1; x; 2xg.
At the end of that section we described some spanning sets as ‘minimal’,
but we never precisely defined that word. We could take ‘minimal’ to mean one
of two things. We could mean that a spanning set is minimal if it contains the
smallest number of members of any set with the same span. With this meaning
f1; x; 2xg is not minimal because it has one member more than the other two.
Or we could mean that a spanning set is minimal when it has no elements that
can be removed without changing the span. Under this meaning f1; x; 2xg is not
minimal because removing the 2x and getting f1; xg leaves the span unchanged.
The first sense of minimality appears to be a global requirement, in that to
check if a spanning set is minimal we seemingly must look at all the spanning sets
of a subspace and find one with the least number of elements. The second sense
of minimality is local in that we need to look only at the set under discussion
and consider the span with and without various elements. For instance, using
the second sense, we could compare the span of f1; x; 2xg with the span of f1; xg and note that the 2x is a “repeat” in that its removal doesn’t shrink the span.
In this section we will use the second sense of ‘minimal spanning set’ because
of this technical convenience. However, the most important result of this book
is that the two senses coincide; we will prove that in the section after this one.
II.1 Definition and Examples
We first characterize when a vector can be removed from a set without changing
the span of that set.
1.1 Lemma Where S is a subset of a vector space V ,
[S] = [S [ f~vg] if and only if ~v 2 [S]
for any ~v 2 V .
Proof. The left to right implication is easy. If [S] = [S [ f~vg] then, since
~v 2 [S [ f~vg], the equality of the two sets gives that ~v 2 [S].
For the right to left implication assume that ~v 2 [S] to show that [S] = [S [
f~vg] by mutual inclusion. The inclusion [S] µ [S[f~vg] is obvious. For the other
inclusion [S] ¶ [S[f~vg], write an element of [S[f~vg] as d0~v+d1~s1+¢ ¢ ¢+dm~sm
and substitute ~v’s expansion as a linear combination of members of the same set
d0(c0~t0 +¢ ¢ ¢+ck~tk)+d1~s1 +¢ ¢ ¢+dm~sm. This is a linear combination of linear
combinations and so distributing d0 results in a linear combination of vectors
from S. Hence each member of [S [ f~vg] is also a member of [S]. QED
Section II. Linear Independence 103
1.2 Example In R3, where
~v1 =0@
100
1A
~v2 =0@
010
1A
~v3 =0@
210
1A
the spans [f~v1; ~v2g] and [f~v1; ~v2; ~v3g] are equal since ~v3 is in the span [f~v1; ~v2g].
The lemma says that if we have a spanning set then we can remove a ~v to
get a new set S with the same span if and only if ~v is a linear combination of
vectors from S. Thus, under the second sense described above, a spanning set
is minimal if and only if it contains no vectors that are linear combinations of
the others in that set. We have a term for this important property.
1.3 Definition A subset of a vector space is linearly independent if none
of its elements is a linear combination of the others. Otherwise it is linearly
dependent.
Here is an important observation: although this way of writing one vector
as a combination of the others
~s0 = c1~s1 + c2~s2 + ¢ ¢ ¢ + cn~sn
visually sets ~s0 off from the other vectors, algebraically there is nothing special
in that equation about ~s0. For any ~si with a coefficient ci that is nonzero, we
can rewrite the relationship to set off ~si.
~si = (1=ci)~s0 + (¡c1=ci)~s1 + ¢ ¢ ¢ + (¡cn=ci)~sn
When we don’t want to single out any vector by writing it alone on one side of the
equation we will instead say that ~s0; ~s1; : : : ; ~sn are in a linear relationship and
write the relationship with all of the vectors on the same side. The next result
rephrases the linear independence definition in this style. It gives what is usually
the easiest way to compute whether a finite set is dependent or independent.
1.4 Lemma A subset S of a vector space is linearly independent if and only if
for any distinct ~s1; : : : ; ~sn 2 S the only linear relationship among those vectors
c1~s1 + ¢ ¢ ¢ + cn~sn = ~0 c1; : : : ; cn 2 R
is the trivial one: c1 = 0; : : : ; cn = 0.
Proof. This is a direct consequence of the observation above.
If the set S is linearly independent then no vector ~si can be written as a linear
combination of the other vectors from S so there is no linear relationship where
some of the ~s ’s have nonzero coefficients. If S is not linearly independent then
some ~si is a linear combination ~si = c1~s1+¢ ¢ ¢+ci¡1~si¡1+ci+1~si+1+¢ ¢ ¢+cn~sn of
other vectors from S, and subtracting ~si from both sides of that equation gives
a linear relationship involving a nonzero coefficient, namely the ¡1 in front of
~si. QED
104 Chapter Two. Vector Spaces
1.5 Example In the vector space of two-wide row vectors, the two-element set
f¡40 15¢; ¡¡50 25¢g is linearly independent. To check this, set
c1 ¢ ¡40 15¢+ c2 ¢ ¡¡50 25¢= ¡0 0¢ and solving the resulting system
40c1 ¡ 50c2 = 0
15c1 + 25c2 = 0
¡(15=40)½1+½2 ¡!
40c1 ¡ 50c2 = 0
(175=4)c2 = 0
shows that both c1 and c2 are zero. So the only linear relationship between the
two given row vectors is the trivial relationship.
In the same vector space, f¡40 15¢; ¡20 7:5¢g is linearly dependent since
we can satisfy
c1 ¡40 15¢+ c2 ¢ ¡20 7:5¢= ¡0 0¢ with c1 = 1 and c2 = ¡2.
1.6 Remark Recall the Statics example that began this book. We first set the
unknown-mass objects at 40 cm and 15 cm and got a balance, and then we set
the objects at ¡50 cm and 25 cm and got a balance. With those two pieces of
information we could compute values of the unknown masses. Had we instead
first set the unknown-mass objects at 40 cm and 15 cm, and then at 20 cm and
7:5 cm, we would not have been able to compute the values of the unknown
masses (try it). Intuitively, the problem is that the ¡20 7:5¢ information is a
“repeat” of the ¡40 15¢ information—that is, ¡20 7:5¢ is in the span of the
set f¡40 15¢g—and so we would be trying to solve a two-unknowns problem
with what is essentially one piece of information.
1.7 Example The set f1 + x; 1 ¡ xg is linearly independent in P2, the space
of quadratic polynomials with real coefficients, because
0 + 0x + 0x2 = c1(1 + x) + c2(1 ¡ x) = (c1 + c2) + (c1 ¡ c2)x + 0x2
gives
c1 + c2 = 0
c1 ¡ c2 = 0
¡½1+½2 ¡!
c1 + c2 = 0
2c2 = 0
since polynomials are equal only if their coefficients are equal. Thus, the only
linear relationship between these two members of P2 is the trivial one.
1.8 Example In R3, where
~v1 =0@
345
1A
~v2 =0@
292
1A
~v3 =0@
4
18
41A the set S = f~v1; ~v2; ~v3g is linearly dependent because this is a relationship
0 ¢ ~v1 + 2 ¢ ~v2 ¡ 1 ¢ ~v3 = ~0
where not all of the scalars are zero (the fact that some of the scalars are zero
doesn’t matter).
Section II. Linear Independence 105
1.9 Remark That example illustrates why, although Definition 1.3 is a clearer
statement of what independence is, Lemma 1.4 is more useful for computations.
Working straight from the definition, someone trying to compute whether S is
linearly independent would start by setting ~v1 = c2~v2 + c3~v3 and concluding
that there are no such c2 and c3. But knowing that the first vector is not
dependent on the other two is not enough. This person would have to go on to
try ~v2 = c1~v1 + c3~v3 to find the dependence c1 = 0, c3 = 1=2. Lemma 1.4 gets
the same conclusion with only one computation.
1.10 Example The empty subset of a vector space is linearly independent.
There is no nontrivial linear relationship among its members as it has no members.
1.11 Example In any vector space, any subset containing the zero vector is
linearly dependent. For example, in the space P2 of quadratic polynomials,
consider the subset f1 + x; x + x2; 0g.
One way to see that this subset is linearly dependent is to use Lemma 1.4: we
have 0 ¢~v1+0 ¢~v2+1 ¢~0 = ~0, and this is a nontrivial relationship as not all of the
coefficients are zero. Another way to see that this subset is linearly dependent
is to go straight to Definition 1.3: we can express the third member of the subset
as a linear combination of the first two, namely, c1~v1 + c2~v2 = ~0 is satisfied by
taking c1 = 0 and c2 = 0 (in contrast to the lemma, the definition allows all of
the coefficients to be zero).
(There is still another way to see that this subset is dependent that is subtler.
The zero vector is equal to the trivial sum, that is, it is the sum of no vectors.
So in a set containing the zero vector, there is an element that can be written
as a combination of a collection of other vectors from the set, specifically, the
zero vector can be written as a combination of the empty collection.)
The above examples, especially Example 1.5, underline the discussion that
begins this section. The next result says that given a finite set, we can produce
a linearly independent subset by by discarding what Remark 1.6 calls “repeats”.
1.12 Theorem In a vector space, any finite subset has a linearly independent
subset with the same span.
Proof. If the set S = f~s1; : : : ; ~sng is linearly independent then S itself satisfies
the statement, so assume that it is linearly dependent.
By the definition of dependence, there is a vector ~si that is a linear combination
of the others. Discard that vector—define the set S1 = S ¡ f~sig. By
Lemma 1.1, the span does not shrink [S1] = [S].
Now, if S1 is linearly independent then we are finished. Otherwise iterate
the prior paragraph: take another vector, ~v2, this time one that is a linear
combination of other members of S1, and discard it to derive S2 = S1 ¡ f~v2g such that [S2] = [S1]. Repeat this until a linearly independent set Sj appears;
one must appear eventually because S is finite and the empty set is linearly
independent. (Formally, this argument uses induction on n, the number of
elements in the starting set. Exercise 37 asks for the details.) QED
106 Chapter Two. Vector Spaces
1.13 Example This set spans R3.
S = f0@
100
1A
;0@
020
1A
;0@
120
1A
;0@
0
¡1
11A ;0@
330
1 A
g
Looking for a linear relationship
c10@
100
1A
+ c20@
020
1A
+ c30@
120
1A
+ c40@
0
¡1
11A + c50@
330
1A
=0@
000
1A
gives a three equations/five unknowns linear system whose solution set can be
paramatrized in this way.
f
0BBBB@
c1
c2
c3
c4
c5
1CCCCA
= c3
0BBBB@
¡1
¡1
100
1CCCCA
+ c5
0BBBB@
¡3
¡3=2
001
1CCCCA
¯¯
c3; c5 2 Rg
So S is linearly dependent. Setting c3 = 0 and c5 = 1 shows that the fifth vector
is a linear combination of the first two. Thus, Lemma 1.1 says that discarding
the fifth vector
S1 = f0@
100
1A
;0@
020
1A
;0@ 120
1A
;0@
0
¡1
11 Ag
leaves the span unchanged [S1] = [S]. Now, the third vector of K1 is a linear
combination of the first two and we get
S2 = f0@
1001A ;0@
020
1A
;0@
0
¡1
11 Ag
with the same span as S1, and therefore the same span as S, but with one
difference. The set S2 is linearly independent (this is easily checked), and so
discarding any of its elements will shrink the span.
Theorem 1.12 describes producing a linearly independent set by shrinking,
that is, by taking subsets. We finish this subsection by considering how linear
independence and dependence, which are properties of sets, interact with the
subset relation between sets.
1.14 Lemma Any subset of a linearly independent set is also linearly independent.
Any superset of a linearly dependent set is also linearly dependent.
Proof. This is clear. QED
Section II. Linear Independence 107
Restated, independence is preserved by subset and dependence is preserved
by superset.
Those are two of the four possible cases of interaction that we can consider.
The third case, whether linear dependence is preserved by the subset operation,
is covered by Example 1.13, which gives a linearly dependent set S with a subset
S1 that is linearly dependent and another subset S2 that is linearly independent.
That leaves one case, whether linear independence is preserved by superset.
The next example shows what can happen.
1.15 Example In each of these three paragraphs the subset S is linearly
independent.
For the set
S = f0@
100
1 A
g
the span [S] is the x axis. Here are two supersets of S, one linearly dependent
and the other linearly independent.
dependent: f0@
100
1A
;0@
¡3
00
1 A
g
independent: f
0@
100
1A
;0@
010
1A
g
Checking the dependence or independence of these sets is easy.
For
S = f0@
100
1A
;0@
010
1 A
g
the span [S] is the xy plane. These are two supersets.
dependent: f0@
100
1A
;0@
010
1A
;0@
3
¡2
01 Ag independent: f0@
100
1A
;0@
010
1A
;0@
001
1A
g
If
S = f0@ 100
1A
;0@
0101A
;0@
001
1A
g
then [S] = R3. A linearly dependent superset is
dependent: f0@
100
1A
;0@
010
1A
;0@
001
1A
;0@
2
¡1
31Ag
but there are no linearly independent supersets of S. The reason is that for any
vector that we would add to make a superset, the linear dependence equation
0@
x
y
z1A = c10@
100
1A
+ c20@
010
1A
+ c30@
001
1A
has a solution c1 = x, c2 = y, and c3 = z.
108 Chapter Two. Vector Spaces
So, in general, a linearly independent set may have a superset that is dependent.
And, in general, a linearly independent set may have a superset that is
independent. We can characterize when the superset is one and when it is the
other.
1.16 Lemma Where S is a linearly independent subset of a vector space V ,
S [ f~vg is linearly dependent if and only if ~v 2 [S]
for any ~v 2 V with ~v 62 S.
Proof. One implication is clear: if ~v 2 [S] then ~v = c1~s1 + c2~s2 + ¢ ¢ ¢ + cn~sn
where each ~si 2 S and ci 2 R, and so ~0 = c1~s1 + c2~s2 + ¢ ¢ ¢ + cn~sn + (¡1)~v is a
nontrivial linear relationship among elements of S [ f~vg.
The other implication requires the assumption that S is linearly independent.
With S [ f~vg linearly dependent, there is a nontrivial linear relationship c0~v +
c1~s1 +c2~s2 +¢ ¢ ¢+cn~sn = ~0 and independence of S then implies that c0 6= 0, or
else that would be a nontrivial relationship among members of S. Now rewriting
this equation as ~v = ¡(c1=c0)~s1 ¡ ¢ ¢ ¢ ¡ (cn=c0)~sn shows that ~v 2 [S]. QED
(Compare this result with Lemma 1.1. Both say, roughly, that ~v is a “repeat”
if it is in the span of S. However, note the additional hypothesis here of linear
independence.)
1.17 Corollary A subset S = f~s1; : : : ; ~sng of a vector space is linearly dependent
if and only if some ~si is a linear combination of the vectors ~s1, . . . , ~si¡1
listed before it.
Proof. Consider S0 = fg, S1 = f~ s1g, S2 = f~s1; ~s2g, etc. Some index i ¸ 1 is
the first one with Si¡1 [ f~sig linearly dependent, and there ~si 2 [Si¡1]. QED
Lemma 1.16 can be restated in terms of independence instead of dependence:
if S is linearly independent and ~v 62 S then the set S [ f~vg is also linearly
independent if and only if ~v 62 [S]: Applying Lemma 1.1, we conclude that if S
is linearly independent and ~v 62 S then S [ f~vg is also linearly independent if
and only if [S [ f~vg] 6= [S]. Briefly, when passing from S to a superset S1, to
preserve linear independence we must expand the span [S1] ¾ [S].
Example 1.15 shows that some linearly independent sets are maximal—have
as many elements as possible—in that they have no supersets that are linearly
independent. By the prior paragraph, a linearly independent sets is maximal if
and only if it spans the entire space, because then no vector exists that is not
already in the span.
This table summarizes the interaction between the properties of independence
and dependence and the relations of subset and superset.
S1 ½ S S1 ¾ S
S independent S1 must be independent S1 may be either
S dependent S1 may be either S1 must be dependent
Section II. Linear Independence 109
In developing this table we’ve uncovered an intimate relationship between linear
independence and span. Complementing the fact that a spanning set is minimal
if and only if it is linearly independent, a linearly independent set is maximal if
and only if it spans the space.
In summary, we have introduced the definition of linear independence to
formalize the idea of the minimality of a spanning set. We have developed some
properties of this idea. The most important is Lemma 1.16, which tells us that
a linearly independent set is maximal when it spans the space.
Exercises
X 1.18 Decide whether each subset of R3 is linearly dependent or linearly independent.
(a) fà 1
¡3
5 !;Ã2
2
4!;Ã 4
¡4
14!g
(b) fÃ1
7
7!;Ã2
7
7!;Ã3
7
7!g
(c) fà 0
0
¡1!;Ã1
0
4!g
(d) fÃ9
9
0!;Ã2
0
1!;Ã 3
5
¡4!;Ã12
12
¡1!g
X 1.19 Which of these subsets of P3 are linearly dependent and which are independent?
(a) f3 ¡ x + 9x2; 5 ¡ 6x + 3x2; 1 + 1x ¡ 5x2g
(b) f¡x2; 1 + 4x2g
(c) f2 + x + 7x2; 3 ¡ x + 2x2; 4 ¡ 3x2g
(d) f8 + 3x + 3x2; x + 2x2; 2 + 2x + 2x2; 8 ¡ 2x + 5x2g
X 1.20 Prove that each set ff; gg is linearly independent in the vector space of all
functions from R+ to R.
(a) f(x) = x and g(x) = 1=x
(b) f(x) = cos(x) and g(x) = sin(x)
(c) f(x) = ex and g(x) = ln(x)
X 1.21 Which of these subsets of the space of real-valued functions of one real variable
is linearly dependent and which is linearly independent? (Note that we have
abbreviated some constant functions; e.g., in the first item, the ‘2’ stands for the
constant function f(x) = 2.)
(a) f2; 4 sin2(x); cos2(x)g (b) f1; sin(x); sin(2x)g (c) fx; cos(x)g
(d) f(1 + x)2; x2 + 2x; 3g (e) fcos(2x); sin2(x); cos2(x)g (f) f0; x; x2g
1.22 Does the equation sin2(x)= cos2(x) = tan2(x) show that this set of functions
fsin2(x); cos2(x); tan2(x)g is a linearly dependent subset of the set of all real-valued
functions with domain (¡¼=2::¼=2)?
1.23 Why does Lemma 1.4 say “distinct”?
X 1.24 Show that the nonzero rows of an echelon form matrix form a linearly independent
set.
110 Chapter Two. Vector Spaces
X 1.25 (a) Show that if the set f~u; ~v; ~wg linearly independent set then so is the set
f~u; ~u +~v; ~u +~v + ~wg.
(b) What is the relationship between the linear independence or dependence of
the set f~u; ~v; ~wg and the independence or dependence of f~u ¡~v; ~v ¡ ~w; ~w ¡ ~ug?
1.26 Example 1.10 shows that the empty set is linearly independent.
(a) When is a one-element set linearly independent?
(b) How about a set with two elements?
1.27 In any vector space V , the empty set is linearly independent. What about all
of V ?
1.28 Show that if f~x; ~y; ~zg is linearly independent then so are all of its proper
subsets: f~x; ~yg, f~x; ~zg, f~y; ~zg, f~xg,f~yg, f~zg, and fg. Is that ‘only if’ also?
1.29 (a) Show that this
S = fÃ1
1
0!;á1
2
0 !g
is a linearly independent subset of R3.
(b) Show that
Ã3
2
0!
is in the span of S by finding c1 and c2 giving a linear relationship.
c1 Ã1
1
0!+ c2 á1
2
0 != Ã3
2
0!
Show that the pair c1; c2 is unique.
(c) Assume that S is a subset of a vector space and that ~v is in [S], so that ~v is
a linear combination of vectors from S. Prove that if S is linearly independent
then a linear combination of vectors from S adding to ~v is unique (that is, unique
up to reordering and adding or taking away terms of the form 0 ¢ ~s). Thus S
as a spanning set is minimal in this strong sense: each vector in [S] is “hit” a
minimum number of times—only once.
(d) Prove that it can happen when S is not linearly independent that distinct
linear combinations sum to the same vector.
1.30 Prove that a polynomial gives rise to the zero function if and only if it is
the zero polynomial. (Comment. This question is not a Linear Algebra matter,
but we often use the result. A polynomial gives rise to a function in the obvious
way: x 7! cnxn + ¢ ¢ ¢ + c1x + c0.)
1.31 Return to Section 1.2 and redefine point, line, plane, and other linear surfaces
to avoid degenerate cases.
1.32 (a) Show that any set of four vectors in R2 is linearly dependent.
(b) Is this true for any set of five? Any set of three?
(c) What is the most number of elements that a linearly independent subset of
R2 can have?
X 1.33 Is there a set of four vectors in R3, any three of which form a linearly independent
set?
1.34 Must every linearly dependent set have a subset that is dependent and a
subset that is independent?
Section II. Linear Independence 111
1.35 In R4, what is the biggest linearly independent set you can find? The smallest?
The biggest linearly dependent set? The smallest? (‘Biggest’ and ‘smallest’ mean
that there are no supersets or subsets with the same property.)
X 1.36 Linear independence and linear dependence are properties of sets. We can
thus naturally ask how those properties act with respect to the familiar elementary
set relations and operations. In this body of this subsection we have covered the
subset and superset relations. We can also consider the operations of intersection,
complementation, and union.
(a) How does linear independence relate to intersection: can an intersection of
linearly independent sets be independent? Must it be?
(b) How does linear independence relate to complementation?
(c) Show that the union of two linearly independent sets need not be linearly
independent.
(d) Characterize when the union of two linearly independent sets is linearly independent,
in terms of the intersection of the span of each.
X 1.37 For Theorem 1.12,
(a) fill in the induction for the proof;
(b) give an alternate proof that starts with the empty set and builds a sequence
of linearly independent subsets of the given finite set until one appears with the
same span as the given set.
1.38 With a little calculation we can get formulas to determine whether or not a
set of vectors is linearly independent.
(a) Show that this subset of R2
fµa
c¶;µb
d¶g
is linearly independent if and only if ad ¡ bc 6= 0.
(b) Show that this subset of R3
fÃa
d
g!;Ãb
e
h!;Ãc
f
i!g
is linearly independent iff aei + bfg + cdh ¡ hfa ¡ idb ¡ gec 6= 0.
(c) When is this subset of R3
fÃa
d
g!;Ãb
e
h!g
linearly independent?
(d) This is an opinion question: for a set of four vectors from R4, must there be
a formula involving the sixteen entries that determines independence of the set?
(You needn’t produce such a formula, just decide if one exists.)
X 1.39 (a) Prove that a set of two perpendicular nonzero vectors from Rn is linearly
independent when n > 1.
(b) What if n = 1? n = 0?
(c) Generalize to more than two vectors.
1.40 Consider the set of functions from the open interval (¡1::1) to R.
(a) Show that this set is a vector space under the usual operations.
(b) Recall the formula for the sum of an infinite geometric series: 1+x+x2+¢ ¢ ¢ =
1=(1¡x) for all x 2 (¡1::1). Why does this not express a dependence inside of the
set fg(x) = 1=(1 ¡ x); f0(x) = 1; f1(x) = x; f2(x) = x2; : : :g (in the vector space
that we are considering)? (Hint. Review the definition of linear combination.)
112 Chapter Two. Vector Spaces
(c) Show that the set in the prior item is linearly independent.
This shows that some vector spaces exist with linearly independent subsets that
are infinite.
1.41 Show that, where S is a subspace of V , if a subset T of S is linearly independent
in S then T is also linearly independent in V . Is that ‘only if’?
Section III. Basis and Dimension 113
III Basis and Dimension
The prior section ends with the statement that a spanning set is minimal when it
is linearly independent and a linearly independent set is maximal when it spans
the space. So the notions of minimal spanning set and maximal independent
set coincide. In this section we will name this idea and study its properties.
III.1 Basis
1.1 Definition A basis for a vector space is a sequence of vectors that form
a set that is linearly independent and that spans the space.
We denote a basis with angle brackets h~¯1;~¯2; : : :i to signify that this collection
is a sequence¤—the order of the elements is significant. (The requirement
that a basis be ordered will be needed, for instance, in Definition 1.13.)
1.2 Example This is a basis for R2.
hµ24
¶;µ1
1¶i
It is linearly independent
c1 µ24
¶+ c2 µ11
¶= µ00
¶ =)
2c1 + 1c2 = 0
4c1 + 1c2 = 0 =) c1 = c2 = 0
and it spans R2.
2c1 + 1c2 = x
4c1 + 1c2 = y
=) c2 = 2x ¡ y and c1 = (y ¡ x)=2
1.3 Example This basis for R2
hµ11 ¶;µ2
4¶i
differs from the prior one because the vectors are in a different order. The
verification that it is a basis is just as in the prior example.
1.4 Example The space R2 has many bases. Another one is this.
hµ10
¶;µ0
1¶i
The verification is easy.
¤ More information on sequences is in the appendix.
114 Chapter Two. Vector Spaces
1.5 Definition For any Rn,
En = h
0BBB@
10...0
1CCCA
;0BBB@
01...0
1CCCA
; : : : ;0BBB@
00...1
1CCCA
i
is the standard (or natural ) basis. We denote these vectors by ~e1; : : : ; ~en.
(Calculus books refer to R2’s standard basis vectors ~{ and ~ instead of ~e1 and
~e2, and they refer to R3’s standard basis vectors ~{, ~, and ~k instead of ~e1, ~e2,
and ~e3.) Note that the symbol ‘~e1’ means something different in a discussion of
R3 than it means in a discussion of R2.
1.6 Example Consider the space fa ¢ cos µ + b ¢ sin µ¯¯ a; b 2 Rg of function of
the real variable µ. This is a natural basis.
h1 ¢ cos µ + 0 ¢ sin µ; 0 ¢ cos µ + 1 ¢ sin µi = hcos µ; sin µi
Another, more generic, basis is hcos µ ¡ sin µ; 2 cos µ + 3 sin µi. Verfication that
these two are bases is Exercise 22.
1.7 Example A natural basis for the vector space of cubic polynomials P3 is
h1; x; x2; x3i. Two other bases for this space are hx3; 3x2; 6x; 6i and h1; 1+x; 1+
x+x2; 1+x+x2 +x3i. Checking that these are linearly independent and span
the space is easy.
1.8 Example The trivial space f~0g has only one basis, the empty one hi.
1.9 Example The space of finite degree polynomials has a basis with infinitely
many elements h1; x; x2; : : :i.
1.10 Example We have seen bases before. In the first chapter we described
the solution set of homogeneous systems such as this one
x + y ¡ w = 0
z + w = 0
by paramatrizing.
f0BB@
¡1
100
1CCA
y +0BB@
10
¡1
1
1CCA
w¯¯ y;w 2 Rg
That is, we described the vector space of solutions as the span of a two-element
set. We can easily check that this two-vector set is also linearly independent.
Thus the solution set is a subspace of R4 with a two-element basis.
Section III. Basis and Dimension 115
1.11 Example Parameterization helps find bases for other vector spaces, not
just for solution sets of homogeneous systems. To find a basis for this subspace
of M2£2
fµa b
c 0¶¯¯ a + b ¡ 2c = 0g
we rewrite the condition as a = ¡b + 2c.
fµ¡b + 2c b
c 0¶¯¯ b; c 2 Rg = fbµ¡1 1
0 0¶+ cµ2 0
1 0¶¯¯ b; c 2 Rg
Thus, this is a natural candidate for a basis.
hµ¡1 1
0 0¶;µ2 0
1 0¶i
The above work shows that it spans the space. To show that it is linearly
independent is routine.
Consider again Example 1.2. It involves two verifications.
In the first, to check that the set is linearly independent we looked at linear
combinations of the set’s members that total to the zero vector c1~¯1+c2~¯2 = ¡0
0¢.
The resulting calculation shows that such a combination is unique, that c1 must
be 0 and c2 must be 0.
The second verification, that the set spans the space, looks at linear combinations
that total to any member of the space c1~¯1+c2~¯2 = ¡x
y¢. In Example 1.2
we noted only that the resulting calculation shows that such a combination exists,
that for each x; y there is a c1; c2. However, in fact the calculation also
shows that the combination is unique: c1 must be (y ¡ x)=2 and c2 must be
2x ¡ y.
That is, the first calculation is a special case of the second. The next result
says that this holds in general for a spanning set: the combination totaling to
the zero vector is unique if and only if the combination totaling to any vector
is unique.
1.12 Theorem In any vector space, a subset is a basis if and only if each
vector in the space can be expressed as a linear combination of elements of the
subset in a unique way.
We consider combinations to be the same if they differ only in the order of
summands or in the addition or deletion of terms of the form ‘0 ¢ ~¯ ’.
Proof. By definition, a sequence is a basis if and only if its vectors form both
a spanning set and a linearly independent set. A subset is a spanning set if
and only if each vector in the space is a linear combination of elements of that
subset in at least one way.
Thus, to finish we need only show that a subset is linearly independent if
and only if every vector in the space is a linear combination of elements from
the subset in at most one way. Consider two expressions of a vector as a linear
116 Chapter Two. Vector Spaces
combination of the members of the basis. We can rearrange the two sums, and
if necessary add some 0~¯i terms, so that the two sums combine the same ~¯’s in
the same order: ~v = c1~¯1 +c2~¯2 +¢ ¢ ¢+cn~¯n and ~v = d1~¯1 +d2~¯2 +¢ ¢ ¢+dn~¯n.
Now
c1~¯1 + c2~¯2 + ¢ ¢ ¢ + cn~¯n = d1~¯1 + d2~¯2 + ¢ ¢ ¢ + dn~¯n
holds if and only if
(c1 ¡ d1)~¯1 + ¢ ¢ ¢ + (cn ¡ dn)~¯n = ~0
holds, and so asserting that each coefficient in the lower equation is zero is the
same thing as asserting that ci = di for each i. QED
1.13 Definition In a vector space with basis B the representation of ~v with
respect to B is the column vector of the coefficients used to express ~v as a linear
combination of the basis vectors:
RepB(~v) =0BBB@
c1
c2
...
cn
1CCCA
B
where B = h~¯1; : : : ; ~¯ni and ~v = c1~¯1 + c2~¯2 + ¢ ¢ ¢ + cn~¯n. The c’s are the
coordinates of ~v with respect to B.
1.14 Example In P3, with respect to the basis B = h1; 2x; 2x2; 2x3i, the
representation of x + x2 is
RepB(x + x2) =0BB@
0
1=2
1=2
0
1CCA
B
(note that the coordinates are scalars, not vectors). With respect to a different
basis D = h1 + x; 1 ¡ x; x + x2; x + x3i, the representation
RepD(x + x2) =0BB@
0010
1CCA
D
is different.
1.15 Remark This use of column notation and the term ‘coordinates’ has
both a down side and an up side.
The down side is that representations look like vectors from Rn, which can
be confusing when the vector space we are working with is Rn, especially since
we sometimes omit the subscript base. We must then infer the intent from the
Section III. Basis and Dimension 117
context. For example, the phrase ‘in R2, where ~v = ¡3
2¢’ refers to the plane
vector that, when in canonical position, ends at (3; 2). To find the coordinates
of that vector with respect to the basis
B = hµ11
¶;µ0
2¶i
we solve
c1 µ1
1¶+ c2 µ0
2¶= µ3
2¶
to get that c1 = 3 and c2 = 1=2. Then we have this.
RepB(~v) = µ 3
¡1=2¶
Here, although we’ve ommited the subscript B from the column, the fact that
the right side is a representation is clear from the context.
The up side of the notation and the term ‘coordinates’ is that they generalize
the use that we are familiar with: in Rn and with respect to the standard
basis En, the vector starting at the origin and ending at (v1; : : : ; vn) has this
representation.
Rep
En(0B@
v1
...
vn
1CA
) =0B@
v1
...
vn
1CA
En
Our main use of representations will come in the third chapter. The definition
appears here because the fact that every vector is a linear combination
of basis vectors in a unique way is a crucial property of bases, and also to help
make two points. First, we fix an order for the elements of a basis so that
coordinates can be stated in that order. Second, for calculation of coordinates,
among other things, we shall restrict our attention to spaces with bases having
only finitely many elements. We will see that in the next subsection.
Exercises
X 1.16 Decide if each is a basis for R3.
(a) hÃ1
2
3!;Ã3
2
1!;Ã0
0
1!i (b) hÃ1
2
3!;Ã3
2
1!i (c) hà 0
2
¡1!;Ã1
1
1!;Ã2
5
0!i
(d) hà 0
2
¡1!;Ã1
1
1!;Ã1
3
0!i
X 1.17 Represent the vector with respect to the basis.
(a) µ1
2¶, B = hµ1
1¶;µ¡1
1 ¶i µ R2
(b) x2 + x3, D = h1; 1 + x; 1 + x + x2; 1 + x + x2 + x3i µ P3
(c)0B@
0
¡1
0
1
1CA
, E4 µ R4
118 Chapter Two. Vector Spaces
1.18 Find a basis for P2, the space of all quadratic polynomials. Must any such
basis contain a polynomial of each degree: degree zero, degree one, and degree two?
1.19 Find a basis for the solution set of this system.
x1 ¡ 4x2 + 3x3 ¡ x4 = 0
2x1 ¡ 8x2 + 6x3 ¡ 2x4 = 0
X 1.20 Find a basis for M2£2, the space of 2£2 matrices.
X 1.21 Find a basis for each.
(a) The subspace fa2x2 + a1x + a0¯¯ a2 ¡ 2a1 = a0g of P2
(b) The space of three-wide row vectors whose first and second components add
to zero
(c) This subspace of the 2£2 matrices
fµa b
0 c¶¯¯ c ¡ 2b = 0g
1.22 Check Example 1.6.
X 1.23 Find the span of each set and then find a basis for that span.
(a) f1 + x; 1 + 2xg in P2 (b) f2 ¡ 2x; 3 + 4x2g in P2
X 1.24 Find a basis for each of these subspaces of the space P3 of cubic polynomials.
(a) The subspace of cubic polynomials p(x) such that p(7) = 0
(b) The subspace of polynomials p(x) such that p(7) = 0 and p(5) = 0
(c) The subspace of polynomials p(x) such that p(7) = 0, p(5) = 0, and p(3) = 0
(d) The space of polynomials p(x) such that p(7) = 0, p(5) = 0, p(3) = 0,
and p(1) = 0
1.25 We’ve seen that it is possible for a basis to remain a basis when it is reordered.
Must it remain a basis?
1.26 Can a basis contain a zero vector?
X 1.27 Let h~¯1; ~¯2; ~¯3i be a basis for a vector space.
(a) Show that hc1~¯1; c2~¯2; c3~¯3i is a basis when c1; c2; c3 6= 0. What happens
when at least one ci is 0?
(b) Prove that h~®1; ~®2; ~®3i is a basis where ~®i = ~¯1 + ~¯i.
1.28 Find one vector ~v that will make each into a basis for the space.
(a) hµ1
1¶; ~vi in R2 (b) hÃ1
1
0!;Ã0
1
0!; ~vi in R3 (c) hx; 1 + x2; ~vi in P2
X 1.29 Where h~¯1; : : : ; ~¯ni is a basis, show that in this equation
c1~¯1 + ¢ ¢ ¢ + ck ~¯k = ck+1~¯k+1 + ¢ ¢ ¢ + cn~¯n
each of the ci’s is zero. Generalize.
1.30 A basis contains some of the vectors from a vector space; can it contain them
all?
1.31 Theorem 1.12 shows that, with respect to a basis, every linear combination is
unique. If a subset is not a basis, can linear combinations be not unique? If so,
must they be?
X 1.32 A square matrix is symmetric if for all indices i and j, entry i; j equals entry
j; i.
(a) Find a basis for the vector space of symmetric 2£2 matrices.
(b) Find a basis for the space of symmetric 3£3 matrices.
Section III. Basis and Dimension 119
(c) Find a basis for the space of symmetric n£n matrices.
X 1.33 We can show that every basis for R3 contains the same number of vectors.
(a) Show that no linearly independent subset of R3 contains more than three
vectors.
(b) Show that no spanning subset of R3 contains fewer than three vectors. (Hint.
Recall how to calculate the span of a set and show that this method, when applied
to two vectors, cannot yield all of R3.)
1.34 One of the exercises in the Subspaces subsection shows that the set
fÃx
y
z!¯¯ x + y + z = 1g
is a vector space under these operations.
Ãx1
y1
z1!+ Ãx2
y2
z2!= Ãx1 + x2 ¡ 1
y1 + y2
z1 + z2 ! rÃx
y
z!= Ãrx ¡ r + 1
ry
rz !
Find a basis.
III.2 Dimension
In the prior subsection we defined the basis of a vector space, and we saw that
a space can have many different bases. For example, following the definition of
a basis, we saw three different bases for R2. So we cannot talk about “the” basis
for a vector space. True, some vector spaces have bases that strike us as more
natural than others, for instance, R2’s basis E2 or R3’s basis E3 or P2’s basis
h1; x; x2i. But, for example in the space fa2x2 + a1x + a0¯¯ 2a2 ¡ a0 = a1g, no
particular basis leaps out at us as the most natural one. We cannot, in general,
associate with a space any single basis that best describes that space.
We can, however, find something about the bases that is uniquely associated
with the space. This subsection shows that any two bases for a space have the
same number of elements. So, with each space we can associate a number, the
number of vectors in any of its bases.
This brings us back to when we considered the two things that could be
meant by the term ‘minimal spanning set’. At that point we defined ‘minimal’
as linearly independent, but we noted that another reasonable interpretation of
the term is that a spanning set is ‘minimal’ when it has the fewest number of
elements of any set with the same span. At the end of this subsection, after we
have shown that all bases have the same number of elements, then we will have
shown that the two senses of ‘minimal’ are equivalent.
Before we start, we first limit our attention to spaces where at least one basis
has only finitely many members.
2.1 Definition A vector space is finite-dimensional if it has a basis with only
finitely many vectors.
120 Chapter Two. Vector Spaces
(One reason for sticking to finite-dimensional spaces is so that the representation
of a vector with respect to a basis is a finitely-tall vector, and so can be easily
written.) From now on we study only finite-dimensional vector spaces. We shall
take the term ‘vector space’ to mean ‘finite-dimensional vector space’. Other
spaces are interesting and important, but they lie outside of our scope.
To prove the main theorem we shall use a technical result.
2.2 Lemma (Exchange Lemma) Assume that B = h~¯1; : : : ;~¯ni is a basis
for a vector space, and that for the vector ~v the relationship ~v = c1~¯1 + c2~¯2 +
¢ ¢ ¢ + cn~¯n has ci 6= 0. Then exchanging ~¯i for ~v yields another basis for the
space.
Proof. Call the outcome of the exchange ˆB = h~¯1; : : : ; ~¯i¡1; ~v;~¯i+1; : : : ; ~¯ni.
We first show that ˆB is linearly independent. Any relationship d1~¯1 + ¢ ¢ ¢ +
di~v + ¢ ¢ ¢ + dn~¯n = ~0 among the members of ˆB, after substitution for ~v,
d1~¯1 + ¢ ¢ ¢ + di ¢ (c1~¯1 + ¢ ¢ ¢ + ci~¯i + ¢ ¢ ¢ + cn~¯n) + ¢ ¢ ¢ + dn~¯n = ~0 (¤)
gives a linear relationship among the members of B. The basis B is linearly
independent, so the coefficient dici of ~¯i is zero. Because ci is assumed to be
nonzero, di = 0. Using this in equation (¤) above gives that all of the other d’s
are also zero. Therefore ˆB is linearly independent.
We finish by showing that ˆB has the same span as B. Half of this argument,
that [ˆB ] µ [B], is easy; any member d1~¯1 + ¢ ¢ ¢ + di~v + ¢ ¢ ¢ + dn~¯n of [ˆB ] can
be written d1~¯1 + ¢ ¢ ¢ + di ¢ (c1~¯1 + ¢ ¢ ¢ + cn~¯n) + ¢ ¢ ¢ + dn~¯n, which is a linear
combination of linear combinations of members of B, and hence is in [B]. For
the [B] µ [ˆB] half of the argument, recall that when ~v = c1~¯1 +¢ ¢ ¢+cn~¯n with
ci 6= 0, then the equation can be rearranged to ~¯i = (¡c1=ci)~¯1+¢ ¢ ¢+(¡1=ci)~v+
¢ ¢ ¢ + (¡cn=ci)~¯n. Now, consider any member d1~¯1 + ¢ ¢ ¢ + di~¯i + ¢ ¢ ¢ + dn~¯n of
[B], substitute for ~¯i its expression as a linear combination of the members
of ˆB , and recognize (as in the first half of this argument) that the result is a
linear combination of linear combinations, of members of ˆB , and hence is in
[ˆB]. QED
2.3 Theorem In any finite-dimensional vector space, all of the bases have
the same number of elements.
Proof. Fix a vector space with at least one finite basis. Choose, from among
all of this space’s bases, one B = h~¯1; : : : ;~¯ni of minimal size. We will show
that any other basis D = h~±1;~±2; : : :i also has the same number of members, n.
Because B has minimal size, D has no fewer than n vectors. We will argue that
it cannot have more than n vectors.
The basis B spans the space and~±1 is in the space, so~±1 is a nontrivial linear
combination of elements of B. By the Exchange Lemma, ~±1 can be swapped for
a vector from B, resulting in a basis B1, where one element is ~± and all of the
n ¡ 1 other elements are ~¯ ’s.
Section III. Basis and Dimension 121
The prior paragraph forms the basis step for an induction argument. The
inductive step starts with a basis Bk (for 1 · k < n) containing k members of D
and n¡k members of B. We know that D has at least n members so there is a
~±
k+1. Represent it as a linear combination of elements of Bk. The key point: in
that representation, at least one of the nonzero scalars must be associated with
a ~¯i or else that representation would be a nontrivial linear relationship among
elements of the linearly independent set D. Exchange ~±k+1 for ~¯i to get a new
basis Bk+1 with one ~± more and one ~¯ fewer than the previous basis Bk.
Repeat the inductive step until no ~¯’s remain, so that Bn contains~±1; : : : ;~±n.
Now, D cannot have more than these n vectors because any ~±n+1 that remains
would be in the span of Bn (since it is a basis) and hence would be a linear combination
of the other ~± ’s, contradicting that D is linearly independent. QED
2.4 Definition The dimension of a vector space is the number of vectors in
any of its bases.
2.5 Example Any basis for Rn has n vectors since the standard basis En has
n vectors. Thus, this definition generalizes the most familiar use of term, that
Rn is n-dimensional.
2.6 Example The space Pn of polynomials of degree at most n has dimension
n+1. We can show this by exhibiting any basis—h1; x; : : : ; xni comes to mind—
and counting its members.
2.7 Example A trivial space is zero-dimensional since its basis is empty.
Again, although we sometimes say ‘finite-dimensional’ as a reminder, in the
rest of this book all vector spaces are assumed to be finite-dimensional. An
instance of this is that in the next result the word ‘space’ should be taken to
mean ‘finite-dimensional vector space’.
2.8 Corollary No linearly independent set can have a size greater than the
dimension of the enclosing space.
Proof. Inspection of the above proof shows that it never uses that D spans the
space, only that D is linearly independent. QED
2.9 Example Recall the subspace diagram from the prior section showing the
subspaces of R3. Each subspace shown is described with a minimal spanning
set, for which we now have the term ‘basis’. The whole space has a basis with
three members, the plane subspaces have bases with two members, the line
subspaces have bases with one member, and the trivial subspace has a basis
with zero members. When we saw that diagram we could not show that these
are the only subspaces that this space has. We can show it now. The prior
corollary proves that the only subspaces of R3 are either three-, two-, one-, or
zero-dimensional. Therefore, the diagram indicates all of the subspaces. There
are no subspaces somehow, say, between lines and planes.
2.10 Corollary Any linearly independent set can be expanded to make a basis.
122 Chapter Two. Vector Spaces
Proof. If a linearly independent set is not already a basis then it must not
span the space. Adding to it a vector that is not in the span preserves linear
independence. Keep adding, until the resulting set does span the space, which
the prior corollary shows will happen after only a finite number of steps. QED
2.11 Corollary Any spanning set can be shrunk to a basis.
Proof. Call the spanning set S. If S is empty then it is already a basis (the
space must be a trivial space). If S = f~0g then it can be shrunk to the empty
basis, thereby making it linearly independent, without changing its span.
Otherwise, S contains a vector ~s1 with ~s1 6= ~0 and we can form a basis
B1 = h~s1i. If [B1] = [S] then we are done.
If not then there is a ~s2 2 [S] such that ~s2 62 [B1]. Let B2 = h~s1; ~ s2i; if
[B2] = [S] then we are done.
We can repeat this process until the spans are equal, which must happen in
at most finitely many steps. QED
2.12 Corollary In an n-dimensional space, a set of n vectors is linearly independent
if and only if it spans the space.
Proof. First we will show that a subset with n vectors is linearly independent
if and only if it is a basis. ‘If’ is trivially true—bases are linearly independent.
‘Only if’ holds because a linearly independent set can be expanded to a basis,
but a basis has n elements, so that this expansion is actually the set we began
with.
To finish, we will show that any subset with n vectors spans the space if and
only if it is a basis. Again, ‘if’ is trivial. ‘Only if’ holds because any spanning
set can be shrunk to a basis, but a basis has n elements and so this shrunken
set is just the one we started with. QED
The main result of this subsection, that all of the bases in a finite-dimensional
vector space have the same number of elements, is the single most important
result in this book because, as Example 2.9 shows, it describes what vector
spaces and subspaces there can be. We will see more in the next chapter.
2.13 Remark The case of infinite-dimensional vector spaces is somewhat controversial.
The statement ‘any infinite-dimensional vector space has a basis’
is known to be equivalent to a statement called the Axiom of Choice (see
[Blass 1984]). Mathematicians differ philosophically on whether to accept or
reject this statement as an axiom on which to base mathematics (although, the
great majority seem to accept it). Consequently the question about infinitedimensional
vector spaces is still somewhat up in the air. (A discussion of the
Axiom of Choice can be found in the Frequently Asked Questions list for the
Usenet group sci.math. Another accessible reference is [Rucker].)
Exercises
Assume that all spaces are finite-dimensional unless otherwise stated.
X 2.14 Find a basis for, and the dimension of, P2.
Section III. Basis and Dimension 123
2.15 Find a basis for, and the dimension of, the solution set of this system.
x1 ¡ 4x2 + 3x3 ¡ x4 = 0
2x1 ¡ 8x2 + 6x3 ¡ 2x4 = 0
X 2.16 Find a basis for, and the dimension of,M2£2, the vector space of 2£2 matrices.
2.17 Find the dimension of the vector space of matrices
µa b
c d¶
subject to each condition.
(a) a; b; c; d 2 R
(b) a ¡ b + 2c = 0 and d 2 R
(c) a + b + c = 0, a + b ¡ c = 0, and d 2 R
X 2.18 Find the dimension of each.
(a) The space of cubic polynomials p(x) such that p(7) = 0
(b) The space of cubic polynomials p(x) such that p(7) = 0 and p(5) = 0
(c) The space of cubic polynomials p(x) such that p(7) = 0, p(5) = 0, and p(3) =
0
(d) The space of cubic polynomials p(x) such that p(7) = 0, p(5) = 0, p(3) = 0,
and p(1) = 0
2.19 What is the dimension of the span of the set fcos2 µ; sin2 µ; cos 2µ; sin 2µg? This
span is a subspace of the space of all real-valued functions of one real variable.
2.20 Find the dimension of C47, the vector space of 47-tuples of complex numbers.
2.21 What is the dimension of the vector space M3£5 of 3£5 matrices?
X 2.22 Show that this is a basis for R4.
h0B@
1
0
0
0
1CA
;0B@
1
1
0
0
1CA
;0B@
1
1
1
0
1CA
;0B@
1
1
1
1
1CA
i
(The results of this subsection can be used to simplify this job.)
2.23 Refer to Example 2.9.
(a) Sketch a similar subspace diagram for P2.
(b) Sketch one for M2£2.
X 2.24 Observe that, where S is a set, the functions f : S ! R form a vector space
under the natural operations: f +g (s) = f(s)+g(s) and r ¢ f (s) = r ¢ f(s). What
is the dimension of the space resulting for each domain?
(a) S = f1g (b) S = f1; 2g (c) S = f1; : : : ; ng
2.25 (See Exercise 24.) Prove that this is an infinite-dimensional space: the set of
all functions f : R ! R under the natural operations.
2.26 (See Exercise 24.) What is the dimension of the vector space of functions
f : S ! R, under the natural operations, where the domain S is the empty set?
2.27 Show that any set of four vectors in R2 is linearly dependent.
2.28 Show that the set h~®1; ~®2; ~®3i ½ R3 is a basis if and only if there is no plane
through the origin containing all three vectors.
2.29 (a) Prove that any subspace of a finite dimensional space has a basis.
(b) Prove that any subspace of a finite dimensional space is finite dimensional.
2.30 Where is the finiteness of B used in Theorem 2.3?
124 Chapter Two. Vector Spaces
X 2.31 Prove that if U and W are both three-dimensional subspaces of R5 then U \W
is non-trivial. Generalize.
2.32 Because a basis for a space is a subset of that space, we are naturally led to
how the property ‘is a basis’ interacts with set operations.
(a) Consider first how bases might be related by ‘subset’. Assume that U;W are
subspaces of some vector space and that U µ W. Can there exist bases BU for
U and BW for W such that BU µ BW? Must such bases exist?
For any basis BU for U, must there be a basis BW forW such that BU µ BW?
For any basis BW forW, must there be a basis BU for U such that BU µ BW?
For any bases BU;BW for U and W, must BU be a subset of BW?
(b) Is the intersection of bases a basis? For what space?
(c) Is the union of bases a basis? For what space?
(d) What about complement?
(Hint. Test any conjectures against some subspaces of R3.)
X 2.33 Consider how ‘dimension’ interacts with ‘subset’. Assume U and W are both
subspaces of some vector space, and that U µ W.
(a) Prove that dim(U) · dim(W).
(b) Prove that equality of dimension holds if and only if U = W.
(c) Show that the prior item does not hold if they are infinite-dimensional.
? 2.34 For any vector ~v in Rn and any permutation ¾ of the numbers 1, 2, . . . , n
(that is, ¾ is a rearrangement of those numbers into a new order), define ¾(~v)
to be the vector whose components are v¾(1), v¾(2), . . . , and v¾(n) (where ¾(1) is
the first number in the rearrangement, etc.). Now fix ~v and let V be the span of
f¾(~v)¯¯ ¾ permutes 1, . . . , ng. What are the possibilities for the dimension of V ?
[Wohascum no. 47]
III.3 Vector Spaces and Linear Systems
We will now reconsider linear systems and Gauss’ method, aided by the tools
and terms of this chapter. We will make three points.
For the first point, recall the first chapter’s Linear Combination Lemma and
its corollary: if two matrices are related by row operations A ¡! ¢ ¢ ¢ ¡! B then
each row of B is a linear combination of the rows of A. That is, Gauss’ method
works by taking linear combinations of rows. Therefore, the right setting in
which to study row operations in general, and Gauss’ method in particular, is
the following vector space.
3.1 Definition The row space of a matrix is the span of the set of its rows. The
row rank is the dimension of the row space, the number of linearly independent
rows.
3.2 Example If
A = µ2 3
4 6¶
Section III. Basis and Dimension 125
then Rowspace(A) is this subspace of the space of two-component row vectors.
fc1 ¢ ¡2 3¢+ c2 ¢ ¡4 6¢¯¯ c1; c2 2 Rg
The linear dependence of the second on the first is obvious and so we can simplify
this description to fc ¢ ¡2 3¢¯¯ c 2 Rg.
3.3 Lemma If the matrices A and B are related by a row operation
A
½i$½j ¡! B or A
k½i ¡! B or A
k½i+½j ¡! B
(for i 6= j and k 6= 0) then their row spaces are equal. Hence, row-equivalent
matrices have the same row space, and hence also, the same row rank.
Proof. By the Linear Combination Lemma’s corollary, each row of B is in the
row space of A. Further, Rowspace(B) µ Rowspace(A) because a member of
the set Rowspace(B) is a linear combination of the rows of B, which means it
is a combination of a combination of the rows of A, and hence, by the Linear
Combination Lemma, is also a member of Rowspace(A).
For the other containment, recall that row operations are reversible: A ¡! B
if and only if B ¡! A. With that, Rowspace(A) µ Rowspace(B) also follows
from the prior paragraph, and so the two sets are equal. QED
Thus, row operations leave the row space unchanged. But of course, Gauss’
method performs the row operations systematically, with a specific goal in mind,
echelon form.
3.4 Lemma The nonzero rows of an echelon form matrix make up a linearly
independent set.
Proof. A result in the first chapter, Lemma III.2.5, states that in an echelon
form matrix, no nonzero row is a linear combination of the other rows. This is
a restatement of that result into new terminology. QED
Thus, in the language of this chapter, Gaussian reduction works by eliminating
linear dependences among rows, leaving the span unchanged, until no
nontrivial linear relationships remain (among the nonzero rows). That is, Gauss’
method produces a basis for the row space.
3.5 Example From any matrix, we can produce a basis for the row space by
performing Gauss’ method and taking the nonzero rows of the resulting echelon
form matrix. For instance,
0@1
3
1
1 4 1
2 0 51A ¡½1+½¡2¡½!1+½23
6½2+½3 ¡! 0@
1 3 1
0 1 0
0 0 31A
produces the basis h¡1 3 1¢; ¡0 1 0¢; ¡0 0 3¢i for the row space. This
is a basis for the row space of both the starting and ending matrices, since the
two row spaces are equal.
126 Chapter Two. Vector Spaces
Using this technique, we can also find bases for spans not directly involving
row vectors.
3.6 Definition The column space of a matrix is the span of the set of its
columns. The column rank is the dimension of the column space, the number
of linearly independent columns.
Our interest in column spaces stems from our study of linear systems. An
example is that this system
c1 + 3c2 + 7c3 = d1
2c1 + 3c2 + 8c3 = d2
c2 + 2c3 = d3
4c1 + 4c3 = d4
has a solution if and only if the vector of d’s is a linear combination of the other
column vectors,
c10BB@
1204
1CCA
+ c20BB@
3310
1CCA
+ c30BB@
7824
1CCA
=0BB@
d1
d2
d3
d4
1CCA
meaning that the vector of d’s is in the column space of the matrix of coefficients.
3.7 Example Given this matrix,
0BB@
1 3 7
2 3 8
0 1 2
4 0 4
1CCA
to get a basis for the column space, temporarily turn the columns into rows and
reduce.
0@
1 2 0 4
3 3 1 0
7 8 2 41A ¡3½1+½¡7¡½!1+½23
¡2½2+½3 ¡! 0@
1 2 0 4
0 ¡3 1 ¡12
0 0 0 0 1A
Now turn the rows back to columns.
h0BB@
1204
1CCA
;0BB@
0
¡3
1
¡12
1CCA
i
The result is a basis for the column space of the given matrix.
3.8 Definition The transpose of a matrix is the result of interchanging the
rows and columns of that matrix. That is, column j of the matrix A is row j
of Atrans, and vice versa.
Section III. Basis and Dimension 127
So the instructions for the prior example are “transpose, reduce, and transpose
back”.
We can even, at the price of tolerating the as-yet-vague idea of vector spaces
being “the same”, use Gauss’ method to find bases for spans in other types of
vector spaces.
3.9 Example To get a basis for the span of fx2 + x4; 2x2 + 3x4;¡x2 ¡ 3x4g in the space P4, think of these three polynomials as “the same” as the row
vectors ¡0 0 1 0 1¢, ¡0 0 2 0 3¢, and ¡0 0 ¡1 0 ¡3¢, apply
Gauss’ method
0@
0 0 1 0 1
0 0 2 0 3
0 0 ¡1 0 ¡31A ¡2½1+½½¡1+!½3 2
2½2+½3 ¡! 0@
0 0 1 0 1
0 0 0 0 1
0 0 0 0 01A
and translate back to get the basis hx2 +x4; x4i. (As mentioned earlier, we will
make the phrase “the same” precise at the start of the next chapter.)
Thus, our first point in this subsection is that the tools of this chapter give
us a more conceptual understanding of Gaussian reduction.
For the second point of this subsection, consider the effect on the column
space of this row reduction.
µ1 2
2 4¶ ¡2½1+½2 ¡! µ1 2
0 0¶
The column space of the left-hand matrix contains vectors with a second component
that is nonzero. But the column space of the right-hand matrix is different
because it contains only vectors whose second component is zero. It is this
knowledge that row operations can change the column space that makes next
result surprising.
3.10 Lemma Row operations do not change the column rank.
Proof. Restated, if A reduces to B then the column rank of B equals the
column rank of A.
We will be done if we can show that row operations do not affect linear relationships
among columns (e.g., if the fifth column is twice the second plus the
fourth before a row operation then that relationship still holds afterwards), because
the column rank is just the size of the largest set of unrelated columns. But
this is exactly the first theorem of this book: in a relationship among columns,
c1 ¢
0BBB@
a1;1
a2;1
...
am;1
1CCCA
+ ¢ ¢ ¢ + cn ¢
0BBB@
a1;n
a2;n
...
am;n
1CCCA
= 0BBB@
00...0
1CCCA
row operations leave unchanged the set of solutions (c1; : : : ; cn). QED
128 Chapter Two. Vector Spaces
Another way, besides the prior result, to state that Gauss’ method has something
to say about the column space as well as about the row space is to consider
again Gauss-Jordan reduction. Recall that it ends with the reduced echelon form
of a matrix, as here.
0@
1 3 1 6
2 6 3 16
1 3 1 61A ¡! ¢ ¢ ¢ ¡! 0@
1 3 0 2
0 0 1 4
0 0 0 01A Consider the row space and the column space of this result. Our first point
made above says that a basis for the row space is easy to get: simply collect
together all of the rows with leading entries. However, because this is a reduced
echelon form matrix, a basis for the column space is just as easy: take the
columns containing the leading entries, that is, h~e1; ~e2i. (Linear independence
is obvious. The other columns are in the span of this set, since they all have a
third component of zero.) Thus, for a reduced echelon form matrix, bases for
the row and column spaces can be found in essentially the same way—by taking
the parts of the matrix, the rows or columns, containing the leading entries.
3.11 Theorem The row rank and column rank of a matrix are equal.
Proof. First bring the matrix to reduced echelon form. At that point, the
row rank equals the number of leading entries since each equals the number
of nonzero rows. Also at that point, the number of leading entries equals the
column rank because the set of columns containing leading entries consists of
some of the ~ei’s from a standard basis, and that set is linearly independent and
spans the set of columns. Hence, in the reduced echelon form matrix, the row
rank equals the column rank, because each equals the number of leading entries.
But Lemma 3.3 and Lemma 3.10 show that the row rank and column rank
are not changed by using row operations to get to reduced echelon form. Thus
the row rank and the column rank of the original matrix are also equal. QED
3.12 Definition The rank of a matrix is its row rank or column rank.
So our second point in this subsection is that the column space and row
space of a matrix have the same dimension. Our third and final point is that
the concepts that we’ve seen arising naturally in the study of vector spaces are
exactly the ones that we have studied with linear systems.
3.13 Theorem For linear systems with n unknowns and with matrix of coefficients
A, the statements
(1) the rank of A is r
(2) the space of solutions of the associated homogeneous system has dimension
n ¡ r
are equivalent.
So if the system has at least one particular solution then for the set of solutions,
the number of parameters equals n¡r, the number of variables minus the rank
of the matrix of coefficients.
Section III. Basis and Dimension 129
Proof. The rank of A is r if and only if Gaussian reduction on A ends with r
nonzero rows. That’s true if and only if echelon form matrices row equivalent
to A have r-many leading variables. That in turn holds if and only if there are
n ¡ r free variables. QED
3.14 Remark [Munkres] Sometimes that result is mistakenly remembered to
say that the general solution of an n unknown system of m equations uses n¡m
parameters. The number of equations is not the relevant figure, rather, what
matters is the number of independent equations (the number of equations in
a maximal independent set). Where there are r independent equations, the
general solution involves n ¡ r parameters.
3.15 Corollary Where the matrix A is n£n, the statements
(1) the rank of A is n
(2) A is nonsingular
(3) the rows of A form a linearly independent set
(4) the columns of A form a linearly independent set
(5) any linear system whose matrix of coefficients is A has one and only one
solution
are equivalent.
Proof. Clearly (1) () (2) () (3) () (4). The last, (4) () (5), holds
because a set of n column vectors is linearly independent if and only if it is a
basis for Rn, but the system
c10BBB@
a1;1
a2;1
...
am;1
1CCCA
+ ¢ ¢ ¢ + cn0BBB@
a1;n
a2;n
...
am;n
1CCCA
=0BBB@
d1
d2
...
dn
1CCCA
has a unique solution for all choices of d1; : : : ; dn 2 R if and only if the vectors
of a’s form a basis. QED
Exercises
3.16 Transpose each.
(a) µ2 1
3 1¶ (b) µ2 1
1 3¶ (c) µ1 4 3
6 7 8¶ (d) Ã0
0
0!
(e) ¡¡1 ¡2¢ X 3.17 Decide if the vector is in the row space of the matrix.
(a) µ2 1
3 1¶, ¡1 0¢ (b) Ã 0 1 3
¡1 0 1
¡1 2 7!, ¡1 1 1¢
X 3.18 Decide if the vector is in the column space.
130 Chapter Two. Vector Spaces
(a) µ1 1
1 1¶, µ1
3¶ (b) Ã1 3 1
2 0 4
1 ¡3 ¡3!, Ã1
0
0!
X 3.19 Find a basis for the row space of this matrix.
0B@
2 0 3 4
0 1 1 ¡1
3 1 0 2
1 0 ¡4 ¡1
1CA
X 3.20 Find the rank of each matrix.
(a) Ã2 1 3
1 ¡1 2
1 0 3! (b) Ã 1 ¡1 2
3 ¡3 6
¡2 2 ¡4! (c) Ã1 3 2
5 1 1
6 4 3!
(d) Ã0 0 0
0 0 0
0 0 0!
X 3.21 Find a basis for the span of each set.
(a) f¡1 3¢; ¡¡1 3¢; ¡1 4¢; ¡2 1¢g µM1£2
(b) fÃ1
2
1!;Ã 3
1
¡1!;Ã 1
¡3
¡3!g µ R3
(c) f1 + x; 1 ¡ x2; 3 + 2x ¡ x2g µ P3
(d) fµ1 0 1
3 1 ¡1¶;µ1 0 3
2 1 4¶;µ¡1 0 ¡5
¡1 ¡1 ¡9¶g µM2£3
3.22 Which matrices have rank zero? Rank one?
X 3.23 Given a; b; c 2 R, what choice of d will cause this matrix to have the rank of
one? µa b
c d¶
3.24 Find the column rank of this matrix. µ1 3 ¡1 5 0 4
2 0 1 0 4 1¶
3.25 Show that a linear system with at least one solution has at most one solution
if and only if the matrix of coefficients has rank equal to the number of its columns.
X 3.26 If a matrix is 5£9, which set must be dependent, its set of rows or its set of
columns?
3.27 Give an example to show that, despite that they have the same dimension,
the row space and column space of a matrix need not be equal. Are they ever
equal?
3.28 Show that the set f(1;¡1; 2;¡3); (1; 1; 2; 0); (3;¡1; 6;¡6)g does not have the
same span as f(1; 0; 1; 0); (0; 2; 0; 3)g. What, by the way, is the vector space?
X 3.29 Show that this set of column vectors (Ãd1
d2
d3!¯¯ there are x, y, and z such that
3x + 2y + 4z = d1
x ¡ z = d2
2x + 2y + 5z = d3)
is a subspace of R3. Find a basis.
3.30 Show that the transpose operation is linear:
(rA + sB)trans = rAtrans + sBtrans
for r; s 2 R and A;B 2Mm£n,
Section III. Basis and Dimension 131
X 3.31 In this subsection we have shown that Gaussian reduction finds a basis for
the row space.
(a) Show that this basis is not unique—different reductions may yield different
bases.
(b) Produce matrices with equal row spaces but unequal numbers of rows.
(c) Prove that two matrices have equal row spaces if and only if after Gauss-
Jordan reduction they have the same nonzero rows.
3.32 Why is there not a problem with Remark 3.14 in the case that r is bigger
than n?
3.33 Show that the row rank of an m£n matrix is at most m. Is there a better
bound?
X 3.34 Show that the rank of a matrix equals the rank of its transpose.
3.35 True or false: the column space of a matrix equals the row space of its transpose.
X 3.36 We have seen that a row operation may change the column space. Must it?
3.37 Prove that a linear system has a solution if and only if that system’s matrix
of coefficients has the same rank as its augmented matrix.
3.38 An m£n matrix has full row rank if its row rank is m, and it has full column
rank if its column rank is n.
(a) Show that a matrix can have both full row rank and full column rank only
if it is square.
(b) Prove that the linear system with matrix of coefficients A has a solution for
any d1, . . . , dn’s on the right side if and only if A has full row rank.
(c) Prove that a homogeneous system has a unique solution if and only if its
matrix of coefficients A has full column rank.
(d) Prove that the statement “if a system with matrix of coefficients A has any
solution then it has a unique solution” holds if and only if A has full column
rank.
3.39 How would the conclusion of Lemma 3.3 change if Gauss’ method is changed
to allow multiplying a row by zero?
X 3.40 What is the relationship between rank(A) and rank(¡A)? Between rank(A)
and rank(kA)? What, if any, is the relationship between rank(A), rank(B), and
rank(A + B)?
III.4 Combining Subspaces
This subsection is optional. It is required only for the last sections of Chapter
Three and Chapter Five and for occasional exercises, and can be passed over
without loss of continuity.
This chapter opened with the definition of a vector space, and the middle
consisted of a first analysis of the idea. This subsection closes the chapter
by finishing the analysis, in the sense that ‘analysis’ means “method of determining
the . . . essential features of something by separating it into parts”
[Macmillan Dictionary].
132 Chapter Two. Vector Spaces
A common way to understand things is to see how they can be built from
component parts. For instance, we think of R3 as put together, in some way,
from the x-axis, the y-axis, and z-axis. In this subsection we will make this
precise; we will describe how to decompose a vector space into a combination of
some of its subspaces. In developing this idea of subspace combination, we will
keep the R3 example in mind as a benchmark model.
Subspaces are subsets and sets combine via union. But taking the combination
operation for subspaces to be the simple union operation isn’t what we
want. For one thing, the union of the x-axis, the y-axis, and z-axis is not all of
R3, so the benchmark model would be left out. Besides, union is all wrong for
this reason: a union of subspaces need not be a subspace (it need not be closed;
for instance, this R3 vector
0@
100
1A
+0@
010
1A
+0@
001
1A
=0@
111
1A
is in none of the three axes and hence is not in the union). In addition to
the members of the subspaces, we must at least also include all of the linear
combinations.
4.1 Definition Where W1; : : : ;Wk are subspaces of a vector space, their sum
is the span of their union W1 +W2 + ¢ ¢ ¢ +Wk = [W1 [W2 [ : : :Wk].
(The notation, writing the ‘+’ between sets in addition to using it between
vectors, fits with the practice of using this symbol for any natural accumulation
operation.)
4.2 Example The R3 model fits with this operation. Any vector ~w 2 R3 can
be written as a linear combination c1~v1 + c2~v2 + c3~v3 where ~v1 is a member of
the x-axis, etc., in this way
0@
w1
w2
w3
1A
= 1 ¢0@
w1
00 1A
+ 1 ¢0@
0
w2
01A + 1 ¢0@
00
w3
1A
and so R3 = x-axis + y-axis + z-axis.
4.3 Example A sum of subspaces can be less than the entire space. Inside of
P4, let L be the subspace of linear polynomials fa + bx¯¯ a; b 2 Rg and let C be
the subspace of purely-cubic polynomials fcx3¯¯ c 2 Rg. Then L + C is not all
of P4. Instead, it is the subspace L + C = fa + bx + cx3¯¯ a; b; c 2 Rg.
4.4 Example A space can be described as a combination of subspaces in more
than one way. Besides the decomposition R3 = x-axis + y-axis + z-axis, we can
also write R3 = xy-plane + yz-plane. To check this, we simply note that any
~w 2 R3 can be written
0@
w1
w2
w3
1A
= 1 ¢0@
w1
w2
01A + 1 ¢0@
00
w3
1A
Section III. Basis and Dimension 133
as a linear combination of a member of the xy-plane and a member of the
yz-plane.
The above definition gives one way in which a space can be thought of as a
combination of some of its parts. However, the prior example shows that there is
at least one interesting property of our benchmark model that is not captured by
the definition of the sum of subspaces. In the familiar decomposition of R3, we
often speak of a vector’s ‘x part’ or ‘y part’ or ‘z part’. That is, in this model,
each vector has a unique decomposition into parts that come from the parts
making up the whole space. But in the decomposition used in Example 4.4, we
cannot refer to the “xy part” of a vector—these three sums
0@
123
1A
=0@
120
1A
+0@
003
1A
=0@
100
1A
+0@
023
1A
=0@
110
1A
+0@
013
1A
all describe the vector as comprised of something from the first plane plus something
from the second plane, but the “xy part” is different in each.
That is, when we consider how R3 is put together from the three axes “in
some way”, we might mean “in such a way that every vector has at least one
decomposition”, and that leads to the definition above. But if we take it to
mean “in such a way that every vector has one and only one decomposition”
then we need another condition on combinations. To see what this condition
is, recall that vectors are uniquely represented in terms of a basis. We can use
this to break a space into a sum of subspaces such that any vector in the space
breaks uniquely into a sum of members of those subspaces.
4.5 Example The benchmark is R3 with its standard basis E3 = h~e1; ~e2; ~e3i.
The subspace with the basis B1 = h~e1i is the x-axis. The subspace with the
basis B2 = h~e2i is the y-axis. The subspace with the basis B3 = h~e3i is the
z-axis. The fact that any member of R3 is expressible as a sum of vectors from
these subspaces 0@
x
y
z1A =0@ x00
1A
+0@
0
y
01A +0@
00
z1A is a reflection of the fact that E3 spans the space—this equation
0@
x
y
z1A = c10@
100
1A
+ c20@
010
1A
+ c30@
001
1A
has a solution for any x; y; z 2 R. And, the fact that each such expression is
unique reflects that fact that E3 is linearly independent—any equation like the
one above has a unique solution.
4.6 Example We don’t have to take the basis vectors one at a time, the same
idea works if we conglomerate them into larger sequences. Consider again the
space R3 and the vectors from the standard basis E3. The subspace with the
134 Chapter Two. Vector Spaces
basis B1 = h~e1; ~e3i is the xz-plane. The subspace with the basis B2 = h~e2i is
the y-axis. As in the prior example, the fact that any member of the space is a
sum of members of the two subspaces in one and only one way
0@
x
y
z1A =0@
x0
z1A +0@
0
y
01A
is a reflection of the fact that these vectors form a basis—this system
0@
x
y
z1A = (c10@
100
1A
+ c30@
001
1A
) + c20@
010
1A
has one and only one solution for any x; y; z 2 R.
These examples illustrate a natural way to decompose a space into a sum
of subspaces in such a way that each vector decomposes uniquely into a sum of
vectors from the parts. The next result says that this way is the only way.
4.7 Definition The concatenation of the sequences B1 = h~¯1;1; : : : ;~¯1;n1 i,
. . . , Bk = h~¯k;1; : : : ;~¯k;nk i is their adjoinment.
B1
_
B2
_
¢ ¢ ¢
_
Bk = h~¯1;1; : : : ;~¯1;n1 ;~¯2;1; : : : ;~¯k;nk i
4.8 Lemma Let V be a vector space that is the sum of some of its subspaces
V = W1 + ¢ ¢ ¢ + Wk. Let B1, . . . , Bk be any bases for these subspaces. Then
the following are equivalent.
(1) For every ~v 2 V , the expression ~v = ~w1 + ¢ ¢ ¢ + ~wk (with ~wi 2 Wi) is
unique.
(2) The concatenation B1
_
¢ ¢ ¢
_
Bk is a basis for V .
(3) The nonzero members of f~w1; : : : ; ~wkg (with ~wi 2 Wi) form a linearly
independent set—among nonzero vectors from different Wi’s, every linear
relationship is trivial.
Proof. We will show that (1) =) (2), that (2) =) (3), and finally that
(3) =) (1). For these arguments, observe that we can pass from a combination
of ~w’s to a combination of ~¯’s
d1 ~w1 + ¢ ¢ ¢ + dk ~wk
= d1(c1;1~¯1;1 + ¢ ¢ ¢ + c1;n1
~¯1;n1 ) + ¢ ¢ ¢ + dk(ck;1~¯k;1 + ¢ ¢ ¢ + ck;nk
~¯k;nk )
= d1c1;1 ¢ ~¯1;1 + ¢ ¢ ¢ + dkck;nk ¢ ~¯k;nk (¤)
and vice versa.
For (1) =) (2), assume that all decompositions are unique. We will show
that B1
_
¢ ¢ ¢
_
Bk spans the space and is linearly independent. It spans the
space because the assumption that V = W1 + ¢ ¢ ¢ + Wk means that every ~v
Section III. Basis and Dimension 135
can be expressed as ~v = ~w1 + ¢ ¢ ¢ + ~wk, which translates by equation (¤) to an
expression of ~v as a linear combination of the ~¯’s from the concatenation. For
linear independence, consider this linear relationship.
~0 = c1;1~¯1;1 + ¢ ¢ ¢ + ck;nk
~¯k;nk
Regroup as in (¤) (that is, take d1, . . . , dk to be 1 and move from bottom to
top) to get the decomposition ~0 = ~w1 + ¢ ¢ ¢ + ~wk. Because of the assumption
that decompositions are unique, and because the zero vector obviously has the
decomposition ~0 = ~0+¢ ¢ ¢+~0, we now have that each ~wi is the zero vector. This
means that ci;1~¯i;1 +¢ ¢ ¢+ci;ni
~¯i;ni = ~0. Thus, since each Bi is a basis, we have
the desired conclusion that all of the c’s are zero.
For (2) =) (3), assume that B1
_
¢ ¢ ¢
_
Bk is a basis for the space. Consider
a linear relationship among nonzero vectors from different Wi’s,
~0 = ¢ ¢ ¢ + di ~wi + ¢ ¢ ¢
in order to show that it is trivial. (The relationship is written in this way
because we are considering a combination of nonzero vectors from only some of
the Wi’s; for instance, there might not be a ~w1 in this combination.) As in (¤),
~0 = ¢ ¢ ¢+di(ci;1~¯i;1+¢ ¢ ¢+ci;ni
~¯i;ni )+¢ ¢ ¢ = ¢ ¢ ¢+dici;1¢~¯i;1+¢ ¢ ¢+dici;ni ¢~¯i;ni+¢ ¢ ¢ and the linear independence of B1
_
¢ ¢ ¢
_
Bk gives that each coefficient dici;j is
zero. Now, ~wi is a nonzero vector, so at least one of the ci;j ’s is not zero, and
thus di is zero. This holds for each di, and therefore the linear relationship is
trivial.
Finally, for (3) =) (1), assume that, among nonzero vectors from different
Wi’s, any linear relationship is trivial. Consider two decompositions of a vector
~v = ~w1 + ¢ ¢ ¢ + ~wk and ~v = ~u1 + ¢ ¢ ¢ + ~uk in order to show that the two are the
same. We have
~0 = ( ~w1 + ¢ ¢ ¢ + ~wk) ¡ (~u1 + ¢ ¢ ¢ + ~uk) = ( ~w1 ¡ ~u1) + ¢ ¢ ¢ + ( ~wk ¡ ~uk)
which violates the assumption unless each ~wi ¡ ~ui is the zero vector. Hence,
decompositions are unique. QED
4.9 Definition A collection of subspaces fW1; : : : ;Wkg is independent if no
nonzero vector from any Wi is a linear combination of vectors from the other
subspaces W1; : : : ;Wi¡1;Wi+1; : : : ;Wk.
4.10 Definition A vector space V is the direct sum (or internal direct sum)
of its subspaces W1; : : : ;Wk if V = W1 + W2 + ¢ ¢ ¢ + Wk and the collection
fW1; : : : ;Wkg is independent. We write V = W1 ©W2 © : : : ©Wk.
4.11 Example The benchmark model fits: R3 = x-axis © y-axis © z-axis.
4.12 Example The space of 2£2 matrices is this direct sum.
fµa 0
0 d¶¯¯ a; d 2 Rg © fµ0 b
0 0¶¯¯ b 2 Rg © fµ0 0
c 0¶¯¯ c 2 Rg
136 Chapter Two. Vector Spaces
It is the direct sum of subspaces in many other ways as well; direct sum decompositions
are not unique.
4.13 Corollary The dimension of a direct sum is the sum of the dimensions
of its summands.
Proof. In Lemma 4.8, the number of basis vectors in the concatenation equals
the sum of the number of vectors in the subbases that make up the concatenation.
QED
The special case of two subspaces is worth mentioning separately.
4.14 Definition When a vector space is the direct sum of two of its subspaces,
then they are said to be complements.
4.15 Lemma A vector space V is the direct sum of two of its subspaces W1
and W2 if and only if it is the sum of the two V = W1+W2 and their intersection
is trivial W1 \W2 = f~0 g.
Proof. Suppose first that V = W1 © W2. By definition, V is the sum of the
two. To show that the two have a trivial intersection, let ~v be a vector from
W1 \ W2 and consider the equation ~v = ~v. On the left side of that equation
is a member of W1, and on the right side is a linear combination of members
(actually, of only one member) of W2. But the independence of the spaces then
implies that ~v = ~0, as desired.
For the other direction, suppose that V is the sum of two spaces with a trivial
intersection. To show that V is a direct sum of the two, we need only show
that the spaces are independent—no nonzero member of the first is expressible
as a linear combination of members of the second, and vice versa. This is
true because any relationship ~w1 = c1 ~w2;1 + ¢ ¢ ¢ + dk ~w2;k (with ~w1 2 W1 and
~w2;j 2 W2 for all j) shows that the vector on the left is also in W2, since the
right side is a combination of members of W2. The intersection of these two
spaces is trivial, so ~w1 = ~0. The same argument works for any ~w2. QED
4.16 Example In the space R2, the x-axis and the y-axis are complements, that
is, R2 = x-axis©y-axis. A space can have more than one pair of complementary
subspaces; another pair here are the subspaces consisting of the lines y = x and
y = 2x.
4.17 Example In the space F = fa cos µ + b sin µ¯¯ a; b 2 Rg, the subspaces
W1 = fa cos µ¯¯ a 2 Rg and W2 = fb sin µ¯¯ b 2 Rg are complements. In addition
to the fact that a space like F can have more than one pair of complementary
subspaces, inside of the space a single subspace like W1 can have more than one
complement—another complement of W1 is W3 = fb sin µ + b cos µ¯¯ b 2 Rg.
4.18 Example In R3, the xy-plane and the yz-planes are not complements,
which is the point of the discussion following Example 4.4. One complement of
the xy-plane is the z-axis. A complement of the yz-plane is the line through
(1; 1; 1).
Section III. Basis and Dimension 137
4.19 Example Following Lemma 4.15, here is a natural question: is the simple
sum V = W1 + ¢ ¢ ¢ +Wk also a direct sum if and only if the intersection of the
subspaces is trivial? The answer is that if there are more than two subspaces
then having a trivial intersection is not enough to guarantee unique decomposition
(i.e., is not enough to ensure that the spaces are independent). In R3, let
W1 be the x-axis, let W2 be the y-axis, and let W3 be this.
W3 = f0@
q
q
r1A
¯¯
q; r 2 Rg
The check that R3 = W1 +W2 +W3 is easy. The intersection W1 \W2 \W3 is
trivial, but decompositions aren’t unique.
0@
x
y
z1A =0@
000
1A
+0@
0
y ¡ x
01A +0@
x
x
z1A =0@
x ¡ y
00
1A
+0@
000
1A
+0@
y
y
z1A
(This example also shows that this requirement is also not enough: that all
pairwise intersections of the subspaces be trivial. See Exercise 30.)
In this subsection we have seen two ways to regard a space as built up from
component parts. Both are useful; in particular, in this book the direct sum
definition is needed to do the Jordan Form construction in the fifth chapter.
Exercises
X 4.20 Decide if R2 is the direct sum of each W1 and W2.
(a) W1 = fµx
0¶¯¯ x 2 Rg, W2 = fµx
x¶¯¯ x 2 Rg
(b) W1 = fµs
s¶¯¯ s 2 Rg, W2 = fµ s
1:1s¶¯¯ s 2 Rg
(c) W1 = R2, W2 = f~0g
(d) W1 = W2 = fµt
t¶¯¯ t 2 Rg
(e) W1 = fµ1
0¶+ µx
0¶¯¯ x 2 Rg, W2 = fµ¡1
0 ¶+ µ0
y¶¯¯ y 2 Rg
X 4.21 Show that R3 is the direct sum of the xy-plane with each of these.
(a) the z-axis
(b) the line
fÃz
z
z!¯¯ z 2 Rg
4.22 Is P2 the direct sum of fa + bx2¯¯ a; b 2 Rg and fcx¯¯ c 2 Rg?
X 4.23 In Pn, the even polynomials are the members of this set
E = fp 2 Pn¯¯ p(¡x) = p(x) for all xg
and the odd polynomials are the members of this set.
O = fp 2 Pn¯¯ p(¡x) = ¡p(x) for all xg
Show that these are complementary subspaces.
138 Chapter Two. Vector Spaces
4.24 Which of these subspaces of R3
W1: the x-axis, W2: the y-axis, W3: the z-axis,
W4: the plane x + y + z = 0, W5: the yz-plane
can be combined to
(a) sum to R3? (b) direct sum to R3?
X 4.25 Show that Pn = fa0¯¯ a0 2 Rg © : : : © fanxn¯¯ an 2 Rg.
4.26 What is W1 +W2 if W1 µ W2?
4.27 Does Example 4.5 generalize? That is, is this true or false: if a vector space V
has a basis h~¯1; : : : ; ~¯ni then it is the direct sum of the spans of the one-dimensional
subspaces V = [f~¯1g] © : : : © [f~¯ng]?
4.28 Can R4 be decomposed as a direct sum in two different ways? Can R1?
4.29 This exercise makes the notation of writing ‘+’ between sets more natural.
Prove that, where W1; : : : ;Wk are subspaces of a vector space,
W1 + ¢ ¢ ¢ +Wk = f~w1 + ~w2 + ¢ ¢ ¢ + ~wk¯¯ ~w1 2 W1; : : : ; ~wk 2 Wkg;
and so the sum of subspaces is the subspace of all sums.
4.30 (Refer to Example 4.19. This exercise shows that the requirement that pariwise
intersections be trivial is genuinely stronger than the requirement only that
the intersection of all of the subspaces be trivial.) Give a vector space and three
subspaces W1, W2, and W3 such that the space is the sum of the subspaces, the
intersection of all three subspaces W1 \W2 \W3 is trivial, but the pairwise intersections
W1 \W2, W1 \W3, and W2 \W3 are nontrivial.
X 4.31 Prove that if V = W1 ©: : :©Wk then Wi \Wj is trivial whenever i 6= j. This
shows that the first half of the proof of Lemma 4.15 extends to the case of more
than two subspaces. (Example 4.19 shows that this implication does not reverse;
the other half does not extend.)
4.32 Recall that no linearly independent set contains the zero vector. Can an
independent set of subspaces contain the trivial subspace?
X 4.33 Does every subspace have a complement?
X 4.34 Let W1;W2 be subspaces of a vector space.
(a) Assume that the set S1 spans W1, and that the set S2 spans W2. Can S1[S2
span W1 +W2? Must it?
(b) Assume that S1 is a linearly independent subset of W1 and that S2 is a
linearly independent subset of W2. Can S1 [S2 be a linearly independent subset
of W1 +W2? Must it?
4.35 When a vector space is decomposed as a direct sum, the dimensions of the
subspaces add to the dimension of the space. The situation with a space that is
given as the sum of its subspaces is not as simple. This exercise considers the
two-subspace special case.
(a) For these subspaces of M2£2 find W1 \ W2, dim(W1 \ W2), W1 + W2, and
dim(W1 +W2).
W1 = fµ0 0
c d¶¯¯ c; d 2 Rg W2 = fµ0 b
c 0¶¯¯ b; c 2 Rg
(b) Suppose that U and W are subspaces of a vector space. Suppose that the
sequence h~¯1; : : : ; ~¯ki is a basis for U \ W. Finally, suppose that the prior
sequence has been expanded to give a sequence h~¹1; : : : ; ~¹j ; ~¯1; : : : ; ~¯ki that is a
basis for U, and a sequence h~¯1; : : : ; ~¯k; ~!1; : : : ; ~!pi that is a basis for W. Prove
that this sequence
h~¹1; : : : ; ~¹j ; ~¯1; : : : ; ~¯k; ~!1; : : : ; ~!pi
Section III. Basis and Dimension 139
is a basis for for the sum U +W.
(c) Conclude that dim(U +W) = dim(U) + dim(W) ¡ dim(U \W).
(d) Let W1 and W2 be eight-dimensional subspaces of a ten-dimensional space.
List all values possible for dim(W1 \W2).
4.36 Let V = W1 © : : : © Wk and for each index i suppose that Si is a linearly
independent subset of Wi. Prove that the union of the Si’s is linearly independent.
4.37 A matrix is symmetric if for each pair of indices i and j, the i; j entry equals
the j; i entry. A matrix is antisymmetric if each i; j entry is the negative of the j; i
entry.
(a) Give a symmetric 2£2 matrix and an antisymmetric 2£2 matrix. (Remark.
For the second one, be careful about the entries on the diagional.)
(b) What is the relationship between a square symmetric matrix and its transpose?
Between a square antisymmetric matrix and its transpose?
(c) Show that Mn£n is the direct sum of the space of symmetric matrices and
the space of antisymmetric matrices.
4.38 Let W1;W2;W3 be subspaces of a vector space. Prove that (W1\W2)+(W1\
W3) µ W1 \ (W2 +W3). Does the inclusion reverse?
4.39 The example of the x-axis and the y-axis in R2 shows that W1©W2 = V does
not imply that W1 [W2 = V . Can W1 ©W2 = V and W1 [W2 = V happen?
X 4.40 Our model for complementary subspaces, the x-axis and the y-axis in R2,
has one property not used here. Where U is a subspace of Rn we define the
orthocomplement of U to be
U? = f~v 2 Rn¯¯ ~v ~u = 0 for all ~u 2 Ug
(read “U perp”).
(a) Find the orthocomplement of the x-axis in R2.
(b) Find the orthocomplement of the x-axis in R3.
(c) Find the orthocomplement of the xy-plane in R3.
(d) Show that the orthocomplement of a subspace is a subspace.
(e) Show that if W is the orthocomplement of U then U is the orthocomplement
of W.
(f) Prove that a subspace and its orthocomplement have a trivial intersection.
(g) Conclude that for any n and subspace U µ Rn we have that Rn = U © U?.
(h) Show that dim(U) + dim(U?) equals the dimension of the enclosing space.
X 4.41 Consider Corollary 4.13. Does it work both ways—that is, supposing that
V = W1 + ¢ ¢ ¢ + Wk, is V = W1 © : : : © Wk if and only if dim(V ) = dim(W1) +
¢ ¢ ¢ + dim(Wk)?
4.42 We know that if V = W1 © W2 then there is a basis for V that splits into a
basis for W1 and a basis for W2. Can we make the stronger statement that every
basis for V splits into a basis for W1 and a basis for W2?
4.43 We can ask about the algebra of the ‘+’ operation.
(a) Is it commutative; is W1 +W2 = W2 +W1?
(b) Is it associative; is (W1 +W2) +W3 = W1 + (W2 +W3)?
(c) Let W be a subspace of some vector space. Show that W +W = W.
(d) Must there be an identity element, a subspace I such that I +W = W +I =
W for all subspaces W?
(e) Does left-cancelation hold: if W1 + W2 = W1 + W3 then W2 = W3? Right
cancelation?
4.44 Consider the algebraic properties of the direct sum operation.
(a) Does direct sum commute: does V = W1 ©W2 imply that V = W2 ©W1?
140 Chapter Two. Vector Spaces
(b) Prove that direct sum is associative: (W1 ©W2) ©W3 = W1 © (W2 ©W3).
(c) Show that R3 is the direct sum of the three axes (the relevance here is that by
the previous item, we needn’t specify which two of the threee axes are combined
first).
(d) Does the direct sum operation left-cancel: does W1 ©W2 = W1 ©W3 imply
W2 = W3? Does it right-cancel?
(e) There is an identity element with respect to this operation. Find it.
(f) Do some, or all, subspaces have inverses with respect to this operation: is
there a subspace W of some vector space such that there is a subspace U with
the property that U ©W equals the identity element from the prior item?
Topic: Fields 141
Topic: Fields
Linear combinations involving only fractions or only integers are much easier
for computations than combinations involving real numbers, because computing
with irrational numbers is awkward. Could other number systems, like the
rationals or the integers, work in the place of R in the definition of a vector
space?
Yes and no. If we take “work” to mean that the results of this chapter
remain true then an analysis of which properties of the reals we have used in
this chapter gives the following list of conditions an algebraic system needs in
order to “work” in the place of R.
Definition. A field is a set F with two operations ‘+’ and ‘¢’ such that
(1) for any a; b 2 F the result of a + b is in F and
² a + b = b + a
² if c 2 F then a + (b + c) = (a + b) + c
(2) for any a; b 2 F the result of a ¢ b is in F and
² a ¢ b = b ¢ a
² if c 2 F then a ¢ (b ¢ c) = (a ¢ b) ¢ c
(3) if a; b; c 2 F then a ¢ (b + c) = a ¢ b + a ¢ c
(4) there is an element 0 2 F such that
² if a 2 F then a + 0 = a
² for each a 2 F there is an element ¡a 2 F such that (¡a) + a = 0
(5) there is an element 1 2 F such that
² if a 2 F then a ¢ 1 = a
² for each element a 6= 0 of F there is an element a¡1 2 F such that
a¡1 ¢ a = 1.
The number system consisting of the set of real numbers along with the usual
addition and multiplication operation is a field, naturally. Another field is the
set of rational numbers with its usual addition and multiplication operations.
An example of an algebraic structure that is not a field is the integer number
system—it fails the final condition.
Some examples are surprising. The set f0; 1g under these operations:
+ 0 1
0 0 1
1 1 0
¢ 0 1
0 0 0
1 0 1
is a field (see Exercise 4).
142 Chapter Two. Vector Spaces
We could develop Linear Algebra as the theory of vector spaces with scalars
from an arbitrary field, instead of sticking to taking the scalars only from R. In
that case, almost all of the statements in this book would carry over by replacing
‘R’ with ‘F’, and thus by taking coefficients, vector entries, and matrix entries
to be elements of F (“almost” because statements involving distances or angles
are exceptions). Here are some examples; each applies to a vector space V over
a field F.
¤ For any ~v 2 V and a 2 F, (i) 0 ¢ ~v = ~0, and (ii) ¡1 ¢ ~v + ~v = ~0, and
(iii) a ¢ ~0 = ~0.
¤ The span (the set of linear combinations) of a subset of V is a subspace
of V .
¤ Any subset of a linearly independent set is also linearly independent.
¤ In a finite-dimensional vector space, any two bases have the same number
of elements.
(Even statements that don’t explicitly mention F use field properties in their
proof.)
We won’t develop vector spaces in this more general setting because the
additional abstraction can be a distraction. The ideas we want to bring out
already appear when we stick to the reals.
The only exception is in Chapter Five. In that chapter we must factor
polynomials, so we will switch to considering vector spaces over the field of
complex numbers. We will discuss this more, including a brief review of complex
arithmetic, when we get there.
Exercises
1 Show that the real numbers form a field.
2 Prove that these are fields.
(a) The rational numbers Q (b) The complex numbers C
3 Give an example that shows that the integer number system is not a field.
4 Consider the set f0; 1g subject to the operations given above. Show that it is a
field.
5 Give suitable operations to make the set f0; 1; 2g a field.
Topic: Crystals 143
Topic: Crystals
Everyone has noticed that table salt comes in little cubes.
Remarkably, the explanation for the cubical external shape is the simplest one
possible: the internal shape, the way the atoms lie, is also cubical. The internal
structure is pictured below. Salt is sodium cloride, and the small spheres shown
are sodium while the big ones are cloride. (To simplify the view, only the
sodiums and clorides on the front, top, and right are shown.)
The specks of salt that we see when we spread a little out on the table consist of
many repetitions of this fundamental unit. That is, these cubes of atoms stack
up to make the larger cubical structure that we see. A solid, such as table salt,
with a regular internal structure is a crystal.
We can restrict our attention to the front face. There, we have this pattern
repeated many times.
The distance between the corners of this cell is about 3:34 °Angstroms (an
°Angstrom is 10¡10 meters). Obviously that unit is unwieldly for describing
points in the crystal lattice. Instead, the thing to do is to take as a unit the
length of each side of the square. That is, we naturally adopt this basis.
hµ3:34
0 ¶;µ 0
3:34¶i
Then we can describe, say, the corner in the upper right of the picture above as
3~¯1 + 2~¯2.
144 Chapter Two. Vector Spaces
Another crystal from everyday experience is pencil lead. It is graphite,
formed from carbon atoms arranged in this shape.
This is a single plane of graphite. A piece of graphite consists of many of these
planes layered in a stack. (The chemical bonds between the planes are much
weaker than the bonds inside the planes, which explains why graphite writes—
it can be sheared so that the planes slide off and are left on the paper.) A
convienent unit of length can be made by decomposing the hexagonal ring into
three regions that are rotations of this unit cell.
A natural basis then would consist of the vectors that form the sides of that
unit cell. The distance along the bottom and slant is 1:42 °Angstroms, so this
hµ1:42
0 ¶;µ1:23
:71 ¶i
is a good basis.
The selection of convienent bases extends to three dimensions. Another
familiar crystal formed from carbon is diamond. Like table salt, it is built from
cubes, but the structure inside each cube is more complicated than salt’s. In
addition to carbons at each corner,
there are carbons in the middle of each face.
Topic: Crystals 145
(To show the added face carbons clearly, the corner carbons have been reduced
to dots.) There are also four more carbons inside the cube, two that are a
quarter of the way up from the bottom and two that are a quarter of the way
down from the top.
(As before, carbons shown earlier have been reduced here to dots.) The distance
along any edge of the cube is 2:18 °Angstroms. Thus, a natural basis for
describing the locations of the carbons, and the bonds between them, is this.
h0@
2:18
00
1A
;0@
0
2:18
01A ;0@
00
2:181Ai
Even the few examples given here show that the structures of crystals is complicated
enough that some organized system to give the locations of the atoms,
and how they are chemically bound, is needed. One tool for that organization
is a convienent basis. This application of bases is simple, but it shows a context
where the idea arises naturally. The work in this chapter just takes this simple
idea and develops it.
Exercises
1 How many fundamental regions are there in one face of a speck of salt? (With a
ruler, we can estimate that face is a square that is 0:1 cm on a side.)
2 In the graphite picture, imagine that we are interested in a point 5:67 °Angstroms
up and 3:14 °Angstroms over from the origin.
(a) Express that point in terms of the basis given for graphite.
(b) How many hexagonal shapes away is this point from the origin?
(c) Express that point in terms of a second basis, where the first basis vector is
the same, but the second is perpendicular to the first (going up the plane) and
of the same length.
3 Give the locations of the atoms in the diamond cube both in terms of the basis,
and in °Angstroms.
4 This illustrates how the dimensions of a unit cell could be computed from the
shape in which a substance crystalizes ([Ebbing], p. 462).
(a) Recall that there are 6:022£1023 atoms in a mole (this is Avagadro’s number).
From that, and the fact that platinum has a mass of 195:08 grams per mole,
calculate the mass of each atom.
(b) Platinum crystalizes in a face-centered cubic lattice with atoms at each lattice
point, that is, it looks like the middle picture given above for the diamond crystal.
Find the number of platinums per unit cell (hint: sum the fractions of platinums
that are inside of a single cell).
(c) From that, find the mass of a unit cell.
(d) Platinum crystal has a density of 21:45 grams per cubic centimeter. From
this, and the mass of a unit cell, calculate the volume of a unit cell.
146 Chapter Two. Vector Spaces
(e) Find the length of each edge.
(f) Describe a natural three-dimensional basis.
Topic: Dimensional Analysis 147
Topic: Dimensional Analysis
“You can’t add apples and oranges,” the old saying goes. It reflects our experience
that in applications the quantities have units and keeping track of those
units is worthwhile. Everyone has done calculations such as this one that use
the units as a check.
60
sec
min ¢ 60
min
hr ¢ 24
hr
day ¢ 365
day
year
= 31 536 000
sec
year
However, the idea of including the units can be taken beyond bookkeeping. It
can be used to draw conclusions about what relationships are possible among
the physical quantities.
To start, consider the physics equation: distance = 16 ¢ (time)2. If the
distance is in feet and the time is in seconds then this is a true statement about
falling bodies. However it is not correct in other unit systems; for instance, it
is not correct in the meter-second system. We can fix that by making the 16 a
dimensional constant.
dist = 16
ft
sec2 ¢ (time)2
For instance, the above equation holds in the yard-second system.
distance in yards = 16
(1=3) yd
sec2 ¢ (time in sec)2 =
16
3
yd
sec2 ¢ (time in sec)2
So our first point is that by “including the units” we mean that we are restricting
our attention to equations that use dimensional constants.
By using dimensional constants, we can be vague about units and say only
that all quantities are measured in combinations of some units of length L,
mass M, and time T. We shall refer to these three as dimensions (these are the
only three dimensions that we shall need in this Topic). For instance, velocity
could be measured in feet=second or fathoms=hour, but in all events it involves
some unit of length divided by some unit of time so the dimensional formula
of velocity is L=T. Similarly, the dimensional formula of density is M=L3. We
shall prefer using negative exponents over the fraction bars and we shall include
the dimensions with a zero exponent, that is, we shall write the dimensional
formula of velocity as L1M0T¡1 and that of density as L¡3M1T0.
In this context, “You can’t add apples to oranges” becomes the advice to
check that all of an equation’s terms have the same dimensional formula. An example
is this version of the falling body equation: d¡gt2 = 0. The dimensional
formula of the d term is L1M0T0. For the other term, the dimensional formula
of g is L1M0T¡2 (g is the dimensional constant given above as 16 ft=sec2)
and the dimensional formula of t is L0M0T1, so that of the entire gt2 term is
L1M0T¡2(L0M0T1)2 = L1M0T0. Thus the two terms have the same dimensional
formula. An equation with this property is dimensionally homogeneous.
Quantities with dimensional formula L0M0T0 are dimensionless. For example,
we measure an angle by taking the ratio of the subtended arc to the
radius
148 Chapter Two. Vector Spaces
r
arc
which is the ratio of a length to a length L1M0T0=L1M0T0 and thus angles
have the dimensional formula L0M0T0.
The classic example of using the units for more than bookkeeping, using
them to draw conclusions, considers the formula for the period of a pendulum.
p = –some expression involving the length of the string, etc.–
The period is in units of time L0M0T1. So the quantities on the other side of
the equation must have dimensional formulas that combine in such a way that
their L’s and M’s cancel and only a single T remains. The table on page 149 has
the quantities that an experienced investigator would consider possibly relevant.
The only dimensional formulas involving L are for the length of the string and
the acceleration due to gravity. For the L’s of these two to cancel, when they
appear in the equation they must be in ratio, e.g., as (`=g)2, or as cos(`=g), or
as (`=g)¡1. Therefore the period is a function of `=g.
This is a remarkable result: with a pencil and paper analysis, before we ever
took out the pendulum and made measurements, we have determined something
about the relationship among the quantities.
To do dimensional analysis systematically, we need to know two things (arguments
for these are in [Bridgman], Chapter II and IV). The first is that each
equation relating physical quantities that we shall see involves a sum of terms,
where each term has the form
mp1
1 mp2
2 ¢ ¢ ¢mpk
k
for numbers m1, . . . , mk that measure the quantities.
For the second, observe that an easy way to construct a dimensionally homogeneous
expression is by taking a product of dimensionless quantities or
by adding such dimensionless terms. Buckingham’s Theorem states that any
complete relationship among quantities with dimensional formulas can be algebraically
manipulated into a form where there is some function f such that
f(¦1; : : : ;¦n) = 0
for a complete set f¦1; : : : ;¦ng of dimensionless products. (The first example
below describes what makes a set of dimensionless products ‘complete’.) We
usually want to express one of the quantities, m1 for instance, in terms of the
others, and for that we will assume that the above equality can be rewritten
m1 = m¡p2
2 ¢ ¢ ¢m¡pk
k ¢ ˆ f(¦2; : : : ;¦n)
where ¦1 = m1mp2
2 ¢ ¢ ¢mpk
k is dimensionless and the products ¦2, . . . , ¦n don’t
involve m1 (as with f, here ˆ f is just some function, this time of n¡1 arguments).
Thus, to do dimensional analysis we should find which dimensionless products
are possible.
For example, consider again the formula for a pendulum’s period.
Topic: Dimensional Analysis 149
quantity
dimensional
formula
period p L0M0T1
length of string ` L1M0T0
mass of bob m L0M1T0
acceleration due to gravity g L1M0T¡2
arc of swing µ L0M0T0
By the first fact cited above, we expect the formula to have (possibly sums
of terms of) the form pp1`p2mp3gp4µp5 . To use the second fact, to find which
combinations of the powers p1, . . . , p5 yield dimensionless products, consider
this equation.
(L0M0T1)p1 (L1M0T0)p2 (L0M1T0)p3 (L1M0T¡2)p4 (L0M0T0)p5 = L0M0T0
It gives three conditions on the powers.
p2 + p4 = 0
p3 = 0
p1 ¡ 2p4 = 0
Note that p3 is 0 and so the mass of the bob does not affect the period. Gaussian
reduction and parametrization of that system gives this
f
0BBBB@
p1
p2
p3
p4
p5
1CCCCA
=0BBBB@
1
¡1=2
0
1=2
0
1CCCCA
p1 +0BBBB@
00001
1CCCCA
p5¯¯ p1; p5 2 Rg
(we’ve taken p1 as one of the parameters in order to express the period in terms
of the other quantities).
Here is the linear algebra. The set of dimensionless products contains all
terms pp1`p2mp3ap4µp5 subject to the conditions above. This set forms a vector
space under the ‘+’ operation of multiplying two such products and the ‘¢’
operation of raising such a product to the power of the scalar (see Exercise 5).
The term ‘complete set of dimensionless products’ in Buckingham’s Theorem
means a basis for this vector space.
We can get a basis by first taking p1 = 1, p5 = 0 and then p1 = 0, p5 = 1. The
associated dimensionless products are ¦1 = p`¡1=2g1=2 and ¦2 = µ. Because
the set f¦1;¦2g is complete, Buckingham’s Theorem says that
p = `1=2g¡1=2 ¢ ˆ f(µ) = p`=g ¢ ˆ f(µ)
where ˆ f is a function that we cannot determine from this analysis (a first year
physics text will show by other means that for small angles it is approximately
the constant function ˆ f(µ) = 2¼).
150 Chapter Two. Vector Spaces
Thus, analysis of the relationships that are possible between the quantities
with the given dimensional formulas has produced a fair amount of information:
a pendulum’s period does not depend on the mass of the bob, and it rises
with the square root of the length of the string.
For the next example we try to determine the period of revolution of two
bodies in space orbiting each other under mutual gravitational attraction. An
experienced investigator could expect that these are the relevant quantities.
quantity
dimensional
formula
period p L0M0T1
mean separation r L1M0T0
first mass m1 L0M1T0
second mass m2 L0M1T0
grav. constant G L3M¡1T¡2
To get the complete set of dimensionless products we consider the equation
(L0M0T1)p1 (L1M0T0)p2 (L0M1T0)p3 (L0M1T0)p4 (L3M¡1T¡2)p5 = L0M0T0
which results in a system
p2 + 3p5 = 0
p3 + p4 ¡ p5 = 0
p1 ¡ 2p5 = 0
with this solution.
f
0BBBB@
1
¡3=2
1=2
0
1=2
1CCCCA
p1 +0BBBB@
00 ¡
1
10 1CCCCA
p4¯¯ p1; p4 2 Rg
As earlier, the linear algebra here is that the set of dimensionless products
of these quantities forms a vector space, and we want to produce a basis
for that space, a ‘complete’ set of dimensionless products. One such set, gotten
from setting p1 = 1 and p4 = 0, and also setting p1 = 0 and p4 = 1
is f¦1 = pr¡3=2m1=2
1 G1=2; ¦2 = m¡1
1 m2g. With that, Buckingham’s Theorem
says that any complete relationship among these quantities is stateable this
form.
p = r3=2m¡1=2
1 G¡1=2 ¢ ˆ f(m¡1
1 m2) = r3=2
pGm1 ¢ ˆ f(m2=m1)
Remark. An important application of the prior formula is when m1 is the
mass of the sun and m2 is the mass of a planet. Because m1 is very much greater
than m2, the argument to ˆ f is approximately 0, and we can wonder whether
this part of the formula remains approximately constant as m2 varies. One way
to see that it does is this. The sun is so much larger than the planet that the
Topic: Dimensional Analysis 151
mutual rotation is approximately about the sun’s center. If we vary the planet’s
mass m2 by a factor of x (e.g., Venus’s mass is x = 0:815 times Earth’s mass),
then the force of attraction is multiplied by x, and x times the force acting on
x times the mass gives, since F = ma, the same acceleration, about the same
center (approximately). Hence, the orbit will be the same and so its period
will be the same, and thus the right side of the above equation also remains
unchanged (approximately). Therefore, ˆ f(m2=m1) is approximately constant
as m2 varies. This is Kepler’s Third Law: the square of the period of a planet
is proportional to the cube of the mean radius of its orbit about the sun.
The final example was one of the first explicit applications of dimensional
analysis. Lord Raleigh considered the speed of a wave in deep water and suggested
these as the relevant quantities.
quantity
dimensional
formula
velocity of the wave v L1M0T¡1
density of the water d L¡3M1T0
acceleration due to gravity g L1M0T¡2
wavelength ¸ L1M0T0
The equation
(L1M0T¡1)p1 (L¡3M1T0)p2 (L1M0T¡2)p3 (L1M0T0)p4 = L0M0T0
gives this system
p1 ¡ 3p2 + p3 + p4 = 0
p2 = 0
¡p1 ¡ 2p3 = 0
with this solution space
f0BB@
10
¡1=2
¡1=2
1CCA
p1¯¯ p1 2 Rg
(as in the pendulum example, one of the quantities d turns out not to be involved
in the relationship). There is one dimensionless product, ¦1 = vg¡1=2¸¡1=2, and
so v is p¸g times a constant ( ˆ f is constant since it is a function of no arguments).
As the three examples above show, dimensional analysis can bring us far
toward expressing the relationship among the quantities. For further reading,
the classic reference is [Bridgman]—this brief book is delightful. Another source
is [Giordano, Wells, Wilde]. A description of dimensional analysis’s place in
modeling is in [Giordano, Jaye, Weir].
Exercises
1 Consider a projectile, launched with initial velocity v0, at an angle µ. An investigation
of this motion might start with the guess that these are the relevant
152 Chapter Two. Vector Spaces
quantities. [de Mestre]
quantity
dimensional
formula
horizontal position x L1M0T0
vertical position y L1M0T0
initial speed v0 L1M0T¡1
angle of launch µ L0M0T0
acceleration due to gravity g L1M0T¡2
time t L0M0T1
(a) Show that fgt=v0; gx=v2
0; gy=v2
0; µg is a complete set of dimensionless products.
(Hint. This can be done by finding the appropriate free variables in the
linear system that arises, but there is a shortcut that uses the properties of a
basis.)
(b) These two equations of motion for projectiles are familiar: x = v0 cos(µ)t and
y = v0 sin(µ)t ¡ (g=2)t2. Manipulate each to rewrite it as a relationship among
the dimensionless products of the prior item.
2 [Einstein] conjectured that the infrared characteristic frequencies of a solid may
be determined by the same forces between atoms as determine the solid’s ordanary
elastic behavior. The relevant quantities are
quantity
dimensional
formula
characteristic frequency º L0M0T¡1
compressibility k L1M¡1T2
number of atoms per cubic cm N L¡3M0T0
mass of an atom m L0M1T0
Show that there is one dimensionless product. Conclude that, in any complete
relationship among quantities with these dimensional formulas, k is a constant
times º¡2N¡1=3m¡1. This conclusion played an important role in the early study
of quantum phenomena.
3 The torque produced by an engine has dimensional formula L2M1T¡2. We may
first guess that it depends on the engine’s rotation rate (with dimensional formula
L0M0T¡1), and the volume of air displaced (with dimensional formula L3M0T0).
[Giordano, Wells, Wilde]
(a) Try to find a complete set of dimensionless products. What goes wrong?
(b) Adjust the guess by adding the density of the air (with dimensional formula
L¡3M1T0). Now find a complete set of dimensionless products.
4 Dominoes falling make a wave. We may conjecture that the wave speed v depends
on the the spacing d between the dominoes, the height h of each domino, and the
acceleration due to gravity g. [Tilley]
(a) Find the dimensional formula for each of the four quantities.
(b) Show that f¦1 = h=d;¦2 = dg=v2g is a complete set of dimensionless products.
(c) Show that if h=d is fixed then the propagation speed is proportional to the
square root of d.
5 Prove that the dimensionless products form a vector space under the ~+ operation
of multiplying two such products and the ~¢ operation of raising such the product
to the power of the scalar. (The vector arrows are a precaution against confusion.)
That is, prove that, for any particular homogeneous system, this set of products
Topic: Dimensional Analysis 153
of powers of m1, . . . , mk
fmp1
1 : : :mpk
k¯¯ p1, . . . , pk satisfy the systemg
is a vector space under:
mp1
1 : : :mpk
k
~+
m
q
1
1 : : :mqk
k = mp1+q1
1 : : :mpk+qk
k
and
r~¢(mp1
1 : : :mpk
k ) = mrp1
1 : : :mrpk
k
(assume that all variables represent real numbers).
6 The advice about apples and oranges is not right. Consider the familiar equations
for a circle C = 2¼r and A = ¼r2.
(a) Check that C and A have different dimensional formulas.
(b) Produce an equation that is not dimensionally homogeneous (i.e., it adds
apples and oranges) but is nonetheless true of any circle.
(c) The prior item asks for an equation that is complete but not dimensionally
homogeneous. Produce an equation that is dimensionally homogeneous but not
complete.
(Just because the old saying isn’t strictly right, doesn’t keep it from being a useful
strategy. Dimensional homogeneity is often used as a check on the plausibility
of equations used in models. For an argument that any complete equation can
easily be made dimensionally homogeneous, see [Bridgman], Chapter I, especially
page 15.)
Index
accuracy
of Gauss’ method, 68–71
rounding error, 69
angle, 42
augmented matrix, 14
back-substitution, 5
C language, 68
canonical form
for row equivalence, 58
Cauchy-Schwartz Inequality, 41
Chemistry problem, 1, 9
chemistry problem, 22
circuits
parallel, 73
series, 73
series-parallel, 74
column, 13
vector, 15
component, 15
computer algebra systems, 62–63
conditioning number, 71
direction vector, 35
dot product, 40
double precision, 69
echelon form, 5
free variable, 12
leading variable, 5
reduced, 47
elementary reduction operations, 4
pivoting, 4
rescaling, 4
swapping, 4
elementary row operations, 4
entry, 13
equivalence relation
row equivalence, 50
flat, 36
form, 56
free variable, 12
Gauss’ method, 2
accuracy, 68–71
back-substitution, 5
elementary operations, 4
Gauss-Jordan, 47
Gauss-Jordan, 47
homogeneous equation, 21
ill-conditioned, 69
induction, 23
inner product, 40
Input-Output Analysis, 64–67
Kirchhoff’s Laws, 73
leading variable, 5
length, 39
Leontief, W., 64
linear combination, 52
Linear Combination Lemma, 53
linear equation, 2
coefficients, 2
constant, 2
homogeneous, 21
satisfied by a vector, 15
solution of, 2
Gauss’ method, 3
Gauss-Jordan, 47
system of, 2
linear surface, 36
LINPACK, 62
Maple, 62
Mathematica, 62
mathematical induction, 23
MATLAB, 62
matrix, 13
augmented, 14
column, 13
conditioning number, 71
entry, 13
nonsingular, 27
row, 13
row equivalence, 50
singular, 27
transpose, 19
matrix:form, 56
mean
arithmetic, 44
geometric, 44
MuPAD, 62
networks, 72–77
Kirchhoff’s Laws, 73
nonsingular
matrix, 27
Octave, 62
orthogonal, 42
parallelogram rule, 34
parameter, 13
partial pivoting, 70
partition
row equivalence classes, 51
perpendicular, 42
Physics problem, 1
pivoting
full, 70
partial
scaled, 70
pivoting on rows, 4
potential, 72
proof techniques
induction, 23
reduced echelon form, 47
representative
for row equivalence classes, 58
rescaling rows, 4
resistance, 72
resistance:equivalent, 76
resistor, 72
row, 13
vector, 15
row equivalence, 50
scalar multiple
vector, 15, 34
scalar product, 40
scaled partial pivoting, 70
Schwartz Inequality, 41
SciLab, 62
single precision, 68
singular
matrix, 27
Statics problem, 5
sum
vector, 15, 34
swapping rows, 4
system of linear equations, 2
Gauss’ method, 2
solving, 2
transpose, 19
Triangle Inequality, 40
vector, 15, 33
angle, 42
canonical position, 34
column, 15
component, 15
direction, 35
dot product, 40
free, 33
length, 39
orthogonal, 42
row, 15
satisfies an equation, 15
scalar multiple, 15, 34
sum, 15, 34
unit, 44
zero, 22
voltage drop, 73
Wheatstone bridge, 74
zero vector, 22