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4.7 General Vector Spaces

In the previous six sections of this chapter, almost all the specific vector spaces appearing in our examples and problems have been vector spaces of n-tuples of real numbers. We have thus confined our attention largely to Euclidean spaces and their subspaces. In this section we discuss examples of some other types of vector spaces that play important roles in various branches of mathematics and their applications.

Example 1

Given fixed positive integers m and n, let Mmn $M_{m n}$ denote the set of all m×n $m \times n$ matrices with real number entries. Then Mmn $M_{m n}$ is a vector space with the usual operations of addition of matrices and multiplication of matrices by scalars (real numbers). That is, these operations satisfy properties (a)–(h) in the definition of a vector space (Section 4.2). In particular, the zero element in Mmn $M_{m n}$ is the m×n $m \times n$ matrix 0 matrix whose elements are all zeros, and the negative −A $- A$ of the matrix A is (as usual) the matrix whose elements are the negatives of the corresponding entries of A.

Given positive integers i and j with 1≤i≤m $1 \leq i \leq m$ and 1≤j≤n $1 \leq j \leq n$ , let Eij $E_{i j}$ denote the m×n $m \times n$ matrix whose only nonzero entry is the number 1 in the ith row and the jth column. Then it should be clear that the mn elements {Eij} ${E_{i j}}$ of Mmn $M_{m n}$ form a basis for Mmn $M_{m n}$ , and so Mmn $M_{m n}$ is a finite-dimensional vector space of dimension mn.

For instance, in the case of the vector space M22 $M_{22}$ of all 2×2 $2 \times 2$ matrices, these alleged basis elements are the four matrices

E 11 = [1000], E 12 = [0010], E 21 = [0100], and E 22 = [0001] .

$\begin{array}{c} E_{11} = [\begin{array}{l} 1 & 0 \\ 0 & 0 \end{array}], E_{12} = [\begin{array}{l} 0 & 1 \\ 0 & 0 \end{array}], \\ E_{21} = [\begin{array}{l} 0 & 0 \\ 1 & 0 \end{array}], and E_{22} = [\begin{array}{l} 0 & 0 \\ 0 & 1 \end{array}] . \end{array}$

Then any matrix A in M22 $M_{22}$ can be expressed as

A = [a c b d] = a E 11 + b E 12 + c E 21 + d E 22,

$A = [\begin{array}{c} a & b \\ c & d \end{array}] = a E_{11} + b E_{12} + c E_{21} + d E_{22},$

so the set {E11, E12, E21, E22} ${E_{11}, E_{12}, E_{21}, E_{22}}$ spans M22 $M_{22}$ . Moreover,

a E 11 + b E 12 + c E 21 + d E 22 = [a c b d] = [0000]

$a E_{11} + b E_{12} + c E_{21} + d E_{22} = [\begin{array}{c} a & b \\ c & d \end{array}] = [\begin{array}{c} 0 & 0 \\ 0 & 0 \end{array}]$

implies immediately that a=b=c=d=0 $a = b = c = d = 0$ , so the set {E11, E12, E21, E22} ${E_{11}, E_{12}, E_{21}, E_{22}}$ is also linearly independent and therefore forms a basis for M22 $M_{22}$ . Thus, we have shown that M22 $M_{22}$ is a 4-dimensional vector space.

Example 2

Let C denote the subset of M22 $M_{22}$ consisting of all 2×2 $2 \times 2$ matrices of the form

[a b - b a] .

$[\begin{array}{r} a & - b \\ b & a \end{array}] .$ (1)

Obviously the sum of any two such matrices is such a matrix, as is any scalar multiple of such a matrix. Thus C is closed under the operations of addition of matrices and multiplication by scalars and is therefore a subspace of M22 $M_{22}$ .

To determine the dimension of the subspace C, we consider the two special matrices

R e = [1001] and I m = [01 - 1 0]

$\begin{array}{l} Re = [\begin{array}{l} 1 & 0 \\ 0 & 1 \end{array}] & and & Im = [\begin{array}{l} 0 & - 1 \\ 1 & 0 \end{array}] \end{array}$ (2)

in C. Here Re is another notation for the 2×2 $2 \times 2$ identity matrix I, whereas the matrix Im is not so familiar. The matrices Re and Im are linearly independent—obviously neither is a scalar multiple of the other—and

[a b - b a] = a R e + b I m .

$[\begin{array}{r} a & - b \\ b & a \end{array}] = a Re + b Im .$ (3)

so these two matrices span C. Thus {Re, Im} ${Re, I m}$ is a basis for C, and so C is a 2-dimensional subspace of the 4-dimensional vector space M22 $M_{22}$ of 2×2 $2 \times 2$ matrices.

Now observe that

(I m) 2 = [01 - 1 0] [01 - 1 0] = [- 1 0 0 - 1] = - R e .

${(Im)}^{2} = [\begin{array}{r} 0 & - 1 \\ 1 & 0 \end{array}] [\begin{array}{r} 0 & - 1 \\ 1 & 0 \end{array}] = [\begin{array}{r} - 1 & 0 \\ 0 & - 1 \end{array}] = - Re .$ (4)

It follows that the matrix product of any two elements aRe+bIm $a Re + b Im$ and cRe+dIm $c Re + d Im$ of C is given by

(a R e + b I m) (c R e + d I m) = = a c (R e) 2 + b c I m R e + a d R e I m + b d (I m) 2 (a c - b d) R e + (a d + b c) I m .

$\begin{array}{lcl} (a Re + b Im) (c Re + d Im) & = & a c {(Re)}^{2} + b c Im Re + a d Re Im + b d {(Im)}^{2} \\ = & (a c - b d) Re + (a d + b c) Im . \end{array}$ (5)

This explicit formula for the product of two matrices in C shows that, in addition to being closed under matrix addition and multiplication by scalars, the subspace C of M22 $M_{22}$ also is closed under matrix multiplication.

If you feel that the idea of a space of matrices being closed under matrix multiplication is a rather complex one, you are correct! Indeed, the vector space C of Example 2 can be taken as a “model” for the set of all complex numbers of the form a+bi $a + b i$ , where a and b are real numbers and i=−1−−−√ $i = \sqrt{- 1}$ denotes the “imaginary” square root of −1 $- 1$ (so that i2=−1 $i^{2} = - 1$ ). Because (Im)2=−Re ${(Im)}^{2} = - Re$ , the matrix Im plays the role of i and the matrix

[a b - b a] = a R e + b I m

$[\begin{array}{r} a & - b \\ b & a \end{array}] = a Re + b Im$

corresponds to the complex number a+bi $a + b i$ . And the matrix multiplication formula in (5) is then analogous to the formula

(a + b i) (c + d i) = (a c - b d) + (a d + b c) i

$(a + b i) (c + d i) = (a c - b d) + (a d + b c) i$

for the multiplication of complex numbers. In addition to providing a concrete interpretation of complex numbers in terms of real numbers, the matrix model C provides a convenient way to manipulate complex numbers in computer programming languages in which matrix operations are “built in” but complex operations are not.

Function Spaces

In Example 1 of Section 4.2 we introduced the vector space F $F$ of all real-valued functions defined on the real line R. If f and g are elements of F $F$ and c is a scalar, then

(f + g) (x) = f (x) + g (x)

$(f + g) (x) = f (x) + g (x)$ (6)

and

(c f) (x) = c (f (x))

$(c f) (x) = c (f (x))$

for all x in R. The zero element in F $F$ is the function 0 such that 0(x)=0 $0 (x) = 0$ for all x. We frequently refer to a function in F $F$ by simply specifying its formula. For instance, by “the function x2 $x^{2}$ ” we mean that function whose value at each x in R is x2 $x^{2}$ .

The functions f1,f2, …,fk $f_{1}, f_{2}, \dots, f_{k}$ in F $F$ are linearly dependent provided that there exist scalars c1,c2, …,ck $c_{1}, c_{2}, \dots, c_{k}$ , not all zero, such that

c 1 f 1 (x) + c 2 f 2 (x) + \dots + c k f k (x) = 0

$c_{1} f_{1} (x) + c_{2} f_{2} (x) + \dots + c_{k} f_{k} (x) = 0$

for all x in R. We often can determine whether two given functions are linearly dependent simply by observing whether one is a scalar multiple of the other.

Example 3

The functions ex $e^{x}$ and e−2x $e^{- 2 x}$ are linearly independent because either of the equations ex=ae−2x $e^{x} = {a e}^{- 2 x}$ or e−2x=bex $e^{- 2 x} = {b e}^{x}$ would imply that e3x $e^{3 x}$ is a constant, which obviously is not so. By contrast, the functions sin 2x and sin x cos x are linearly dependent, because of the trigonometric identity sin 2x=2sin xcos x $\sin 2 x = 2 \sin x \cos x$ . The three functions 1, cos2 x $\cos^{2} x$ , and sin2 x $\sin^{2} x$ are linearly dependent, because the fundamental identity cos2 x+sin2 x=1 $\cos^{2} x + \sin^{2} x = 1$ can be written in the form

(- 1) (1) + (1) (cos 2 x) + (1) (sin 2 x) = 0 .

$(- 1) (1) + (1) (\cos^{2} x) + (1) (\sin^{2} x) = 0 .$

A subspace of F $F$ is called a function space. An example of a function space is the vector space P $P$ of all polynomials in F $F$ (Example 4 in Section 4.4). Recall that a function p(x) in F $F$ is called a polynomial of degree n≧0 $n ≧ 0$ if it can be expressed in the form

p (x) = a 0 + a 1 x + a 2 x 2 + \dots + a n x n

$p (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n}$ (7)

with an≠0 $a_{n} \neq 0$ . Polynomials are added and multiplied by scalars in the usual manner—by collecting coefficients of like powers of x. Clearly any linear combination of polynomials is a polynomial, so P $P$ is, indeed, a subspace of F $F$ . (Recall the subspace criterion of Theorem 1 in Section 4.2.)

Example 4

Given n≥0 $n \geq 0$ , denote by Pn $P_{n}$ the set of all polynomials of degree at most n. That is, Pn $P_{n}$ consists of all polynomials of degrees 0, 1, 2, …, n. Any linear combination of two polynomials of degree at most n is again a polynomial of degree at most n, so Pn $P_{n}$ is a subspace of P $P$ (and of F $F$ ). The formula in (7) shows that the n+1 $n + 1$ polynomials

1, x, x 2, x 3, \dots, x n

$1, x, x^{2}, x^{3}, \dots, x^{n}$ (8)

span Pn $P_{n}$ . To show that these “monomials” are linearly independent, suppose that

c 0 + c 1 x + c 2 x 2 + \dots + c n x n = 0

$c_{0} + c_{1} x + c_{2} x^{2} + \dots + c_{n} x^{n} = 0$ (9)

for all x in R. If a0, a1, …, an $a_{0}, a_{1}, \dots, a_{n}$ are fixed distinct real numbers, then each ai $a_{i}$ satisfies Eq. (9).

In matrix notation this means that

⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ 111 ⋮ 1 a 0 a 1 a 2 ⋮ a n a 20 a 21 a 22 ⋮ a 2 n \dots \dots \dots ⋱ \dots a n 0 a n 1 a n 2 ⋮ a n n ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ c 0 c 1 c 2 ⋮ c n ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ 000 ⋮ 0 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ .

$[\begin{array}{c} 1 & a_{0} & a_{0}^{2} & \dots & a_{0}^{n} \\ 1 & a_{1} & a_{1}^{2} & \dots & a_{1}^{n} \\ 1 & a_{2} & a_{2}^{2} & \dots & a_{2}^{n} \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ 1 & a_{n} & a_{n}^{2} & \dots & a_{n}^{n} \end{array}] [\begin{array}{c} c_{0} \\ c_{1} \\ c_{2} \\ ⋮ \\ c_{n} \end{array}] = [\begin{array}{c} 0 \\ 0 \\ 0 \\ ⋮ \\ 0 \end{array}] .$ (10)

The coefficient matrix in (10) is a Vandermonde matrix V, and Problems 61–63 in Section 3.6 imply that any Vandermonde matrix is nonsingular. Therefore the system Vc=0 $V c = 0$ has only the trivial solution c=0 $c = 0$ , so it follows from (10) that c0=c1=c2=⋯=cn=0 $c_{0} = c_{1} = c_{2} = \dots = c_{n} = 0$ . Thus we have proved that the n+1 $n + 1$ monomials 1, x, x2, …, xn $1, x, x^{2}, \dots, x^{n}$ are linearly independent and hence constitute a basis for Pn $P_{n}$ . Consequently Pn $P_{n}$ is an (n+1) $(n + 1)$ -dimensional vector space.

The fact that the monomials in (8) are linearly independent implies that the coefficients in a polynomial are unique. That is, if

a 0 + a 1 x + \dots + a n x n = b 0 + b 1 x + \dots + b n x n

$a_{0} + a_{1} x + \dots + a_{n} x^{n} = b_{0} + b_{1} x + \dots + b_{n} x^{n}$

for all x, then a0=b0, a1=b1, … $a_{0} = b_{0}, a_{1} = b_{1}, \dots$ , and an=bn $a_{n} = b_{n}$ . (See the remark that follows the definition of linear independence in Section 4.3.) This identity principle for polynomials is often used without proof in elementary algebra. A typical application is the method of partial-fraction decomposition illustrated in Example 5.

Example 5

Find constants A, B, and C such that

6 x ( x - 1 ) ( x + 1 ) ( x + 2 ) = A x - 1 + B x + 1 + C x + 2

$\frac{6 x}{(x - 1) (x + 1) (x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{x + 2}$ (11)

for all x (other than x=1, −1 $x = 1, - 1$ , or −2 $- 2$ ).

Solution

Multiplication of each side of the equation in (11) by the denominator on the left-hand side yields

6 x = A (x + 1) (x + 2) + B (x - 1) (x + 2) + C (x - 1) (x + 1); 6 x = (2 A - 2 B - C) + (3 A + B) x + (A + B + C) x 2 .

$\begin{array}{l} 6 x = A (x + 1) (x + 2) + B (x - 1) (x + 2) + C (x - 1) (x + 1); \\ 6 x = (2 A - 2 B - C) + (3 A + B) x + (A + B + C) x^{2} . \end{array}$

Then the identity principle for polynomials yields the linear equations

2 A 3 A A - + + 2 B B B - + C C = = = 06 0,

$\begin{array}{rcrcrcl} 2 A & - & 2 B & - & C & = & 0 \\ 3 A & + & B & = & 6 \\ A & + & B & + & C & = & 0, \end{array}$

which we readily solve for A=1, B=3 $A = 1, B = 3$ , and C=−4 $C = - 4$ . Therefore

6 x ( x - 1 ) ( x + 1 ) ( x + 2 ) = 1 x - 1 + 3 x + 1 - 4 x + 2

$\frac{6 x}{(x - 1) (x + 1) (x + 2)} = \frac{1}{x - 1} + \frac{3}{x + 1} - \frac{4}{x + 2}$

if x≠1, −1, −2 $x \neq 1, - 1, - 2$ .

Example 6

Show that the four polynomials

1, x, 3 x 2 - 1, and 5 x 3 - 3 x

$\begin{array}{l} 1, & x, & 3 x^{2} - 1, & and & 5 x^{3} - 3 x \end{array}$ (12)

form a basis for P3 $P_{3}$ .

Solution

In order to show simultaneously that these four polynomials span P3 $P_{3}$ and are linearly independent, it suffices to see that every polynomial

p (x) = b 0 + b 1 x + b 2 x 2 + b 3 x 3

$p (x) = b_{0} + b_{1} x + b_{2} x^{2} + b_{3} x^{3}$ (13)

in P3 $P_{3}$ can be expressed uniquely as a linear combination

c 0 (1) + = c 1 (x) + c 2 (3 x 2 - 1) + c 3 (5 x 3 - 3 x) (c 0 - c 2) + (c 1 - 3 c 3) x + (3 c 2) x 2 + (5 c 3) x 3

$\begin{array}{lcl} c_{0} (1) & + & c_{1} (x) + c_{2} (3 x^{2} - 1) + c_{3} (5 x^{3} - 3 x) \\ = & (c_{0} - c_{2}) + (c_{1} - 3 c_{3}) x + (3 c_{2}) x^{2} + (5 c_{3}) x^{3} \end{array}$ (14)

of the polynomials in (12). But upon comparing the coefficients in (13) and (14), we see that we need only observe that the linear system

⎡ ⎣ ⎢ ⎢ ⎢ ⎢ 10000100 - 1 030 0 - 3 05 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ c 0 c 1 c 2 c 3 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ b 0 b 1 b 2 b 3 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥

$[\begin{array}{r} 1 & 0 & - 1 & 0 \\ 0 & 1 & 0 & - 3 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 5 \end{array}] [\begin{array}{c} c_{0} \\ c_{1} \\ c_{2} \\ c_{3} \end{array}] = [\begin{array}{c} b_{0} \\ b_{1} \\ b_{2} \\ b_{3} \end{array}]$

obviously has a unique solution for c0, c1, c2, c3 $c_{0}, c_{1}, c_{2}, c_{3}$ in terms of b0, b1, b2, b3 $b_{0}, b_{1}, b_{2}, b_{3}$ .

Because the vector space P $P$ of all polynomials contains n+1 $n + 1$ linearly independent functions for every integer n≥0 $n \geq 0$ , it follows that P $P$ is an infinite-dimensional vector space. So too is any function space, such as F $F$ itself, that contains P $P$ . Another important function space is the set C(0) $C^{(0)}$ of all continuous functions on R; C(0) $C^{(0)}$ is a subspace of F $F$ because every linear combination of continuous functions is again a continuous function. Every polynomial is a continuous function, so C(0) $C^{(0)}$ contains P $P$ and is therefore an infinite-dimensional vector space. Similarly, the set C(k) $C^{(k)}$ of all functions in F $F$ that have continuous kth-order derivatives is an infinite-dimensional function space.

Solution Spaces of Differential Equations

In Section 1.5 we saw that the (now familiar) first-order differential equation

y' = k y (k constant)

$\begin{array}{l} y ′ = k y & (k constant) \end{array}$ (15)

has the general solution y(x)=Cekx $y (x) = {C e}^{k x}$ . Thus the “solution space” of (15) is the set of all constant multiples of the single function ekx $e^{k x}$ , and it is therefore the 1-dimensional function space with basis {ekx} ${e^{k x}}$ .

In Section 5.1 we will see that the set of all solutions y(x) of a linear second-order differential equation of the form

a y'' + b y' + c y = 0

$a y ″ + b y ′ + c y = 0$ (16)

(with constant coefficients a, b, and c) is a 2-dimensional function space S $S$ , called the solution space of the differential equation. Here we will illustrate this general fact with three simple examples of second-order equations that can be solved in an elementary manner.

Example 7

With a=1 $a = 1$ and b=c=0 $b = c = 0$ in (16), we get the differential equation

y'' = 0 .

$y ″ = 0 .$ (17)

If y(x) is any solution of this equation, so y''(x)=0 $y ″ (x) = 0$ , then two integrations give first

y' (x) = \int y'' (x) d x = \int (0) d x = A

$y ′ (x) = \int y ″ (x) d x = \int (0) d x = A$

and then

y (x) = \int y' (x) d x = \int A d x = A x + B,

$y (x) = \int y ′ (x) d x = \int A d x = A x + B,$

where A and B are arbitrary constants of integration. Thus every solution of (17) is of the form y(x)=Ax+B $y (x) = A x + B$ , and clearly every such function is a solution of the differential equation. Thus the solution space of (17) is simply the 2-dimensional space P1 $P_{1}$ of linear polynomials generated by the basis {1,x} ${1, x}$ .

Example 8

With a=1, b=−2 $a = 1, b = - 2$ , and c=0 $c = 0$ in (16), we get the differential equation

y'' - 2 y' = 0 .

$y ″ - 2 y ′ = 0 .$ (18)

If y(x) is any solution of this equation and if v(x)=y'(x) $v (x) = y ′ (x)$ , then (18) says that

v' (x) = 2 v (x) .

$v ′ (x) = 2 v (x) .$

But this is a familiar first-order differential equation with general solution v(x)=Ce2x $v (x) = {C e}^{2 x}$ . Then integration gives

y (x) = \int y' (x) d x = \int v (x) d x = \int C e 2 x d x = 1 2 C e 2 x + A .

$y (x) = \int y^{′} (x) d x = \int v (x) d x = \int C e^{2 x} d x = \frac{1}{2} C e^{2 x} + A .$

Thus every solution of (18) is of the form y(x)=A+Be2x $y (x) = A + {B e}^{2 x}$ (where B=12C $B = \frac{1}{2} C$ ), and you should show by substitution in (18) that every function of this form is a solution of the differential equation. Therefore, the solution space of (18) is the 2-dimensional function space that is generated by the basis {1,e2x} ${1, e^{2 x}}$ .

Example 9

With a=1, b=0 $a = 1, b = 0$ , and c=−1 $c = - 1$ in (16), we get the differential equation

y'' - y = 0 .

$y ″ - y = 0 .$ (19)

If y(x) is any solution of this equation and if v(x)=y'(x) $v (x) = y ′ (x)$ , then (19) says that

y'' = d v d x = d v d y d y d x = v d v d y = y;

$y ″ = \frac{d v}{d x} = \frac{d v}{d y} \frac{d y}{d x} = v \frac{d v}{d y} = y;$

so v dv=y dy $v d v = y d y$ . Integration therefore gives

1 2 v 2 = 1 2 y 2 + C, so v 2 = y 2 \pm a 2,

$\begin{array}{l} \frac{1}{2} v^{2} = \frac{1}{2} y^{2} + C, & so & v^{2} = y^{2} \pm a^{2}, \end{array}$ (20)

where the final sign depends on the sign of C. The final result is independent of this choice (Problem 27); we’ll take the negative sign for the sake of illustration. Since v=dy/dx $v = d y / d x$ , it then follows from (20) that

x = \pm \int d y y 2 - a 2 - - - - - - \sqrt .

$x = \pm \int \frac{d y}{\sqrt{y^{2} - a^{2}}} .$ (21)

Here again the final result is independent of the choice of sign; this time we’ll take the positive sign. Then the substitution y=au $y = a u$ and the standard integral

\int d u u 2 - 1 - - - - - \sqrt = cosh - 1 u + C

$\int \frac{d u}{\sqrt{u^{2} - 1}} = \cosh^{- 1} u + C$

(see Section 7.6 of Edwards and Penney, Calculus: Early Transcendentals, 7th edition, Hoboken, NJ: Pearson, 2008) yield

x = cosh - 1 y a + b,

$x = \cosh^{- 1} \frac{y}{a} + b,$

so finally,

y = a cosh (x - b) = a (cosh x cosh b - sinh x sinh b),

$y = a \cosh (x - b) = a (\cosh x \cosh b - \sinh x \sinh b),$

which can be written as

y (x) = A cosh x + B sinh x,

$y (x) = A \cosh x + B \sinh x,$ (22)

with A=a cosh b $A = a \cosh b$ and B=−a sinh b $B = - a \sinh b$ . Thus every solution of the differential equation y''−y=0 $y ″ - y = 0$ in (19) is of the form y(x)=A cosh x+B sinh x $y (x) = A \cosh x + B \sinh x$ , and you should show by substitution in (19) that every function of this form is a solution of the differential equation. Therefore, the solution space S $S$ of (19) is the 2-dimensional function space that is generated by the basis {cosh x,sinh x} ${\cosh x, \sinh x}$ . If we substitute into (22) the relations

cosh x = e x + e - x 2, sinh x = e x - e - x 2

$\begin{array}{l} \cosh x = \frac{e^{x} + e^{- x}}{2}, & \sinh x = \frac{e^{x} - e^{- x}}{2} \end{array}$

that define the hyperbolic cosine and sine functions in terms of exponentials, we see that every solution of y''−y=0 $y ″ - y = 0$ can also be written in the form of the linear combination

y (x) = C e x + D e - x .

$y (x) = C e^{x} + D e^{- x} .$

Consequently, {ex, e−x} ${e^{x}, e^{- x}}$ is a second basis for the 2-dimensional solution space S $S$ .

In Section 5.3 we will learn a much quicker way to solve linear second-order differential equations like the equation y''−y=0 $y ″ - y = 0$ of Example 9 and the equation y''+y=0 $y ″ + y = 0$ of Problem 28.

4.7 Problems

In Problems 1–4, determine whether or not the indicated set of 3×3 $3 \times 3$ matrices is a subspace of M33 $M_{33}$ .

The set of all diagonal 3×3 $3 \times 3$ matrices.
The set of all symmetric 3×3 $3 \times 3$ matrices (that is, matrices A=[aij] $A = [a_{i j}]$ such that aij=aji $a_{i j} = a_{j i}$ for 1≤i≤3, 1≤j≤3 $1 \leq i \leq 3, 1 \leq j \leq 3$ ).
The set of all nonsingular 3×3 $3 \times 3$ matrices.
The set of all singular 3×3 $3 \times 3$ matrices.

In Problems 5–8, determine whether or not each indicated set of functions is a subspace of the space F $F$ of all real-valued functions on R.

The set of all f such that f(0)=0 $f (0) = 0$ .
The set of all f such that f(x)≠0 $f (x) \neq 0$ for all x.
The set of all f such that f(0)=0 $f (0) = 0$ and f(1)=1 $f (1) = 1$ .
The set of all f such that f(−x)=−f(x) $f (- x) = - f (x)$ for all x.

In Problems 9–12, a condition on the coefficients of a polynomial a0+a1x+a2x2+a3x3 $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3}$ is given. Determine whether or not the set of all such polynomials satisfying this condition is a subspace of the space P $P$ of all polynomials.

a3≠0 $a_{3} \neq 0$
a0=a1=0 $a_{0} = a_{1} = 0$
a0+a1+a2+a3=0 $a_{0} + a_{1} + a_{2} + a_{3} = 0$
a0, a1, a2 $a_{0}, a_{1}, a_{2}$ , and a3 $a_{3}$ are all integers

In Problems 13–18, determine whether the given functions are linearly independent.

sin x and cos x
ex $e^{x}$ and xex ${x e}^{x}$
1+x, 1−x $1 + x, 1 - x$ , and 1−x2 $1 - x^{2}$
1+x, x+x2 $1 + x, x + x^{2}$ , and 1−x2 $1 - x^{2}$
cos 2x, sin2 x $\cos 2 x, \sin^{2} x$ , and cos2 x $\cos^{2} x$
2 cos x+3 sin x $2 \cos x + 3 \sin x$ and 4 cos x+5 sin x $4 \cos x + 5 \sin x$

In Problems 19–22, use the method of Example 5 to find the constants A, B, and C in the indicated partial-fraction decompositions.

x−5(x−2)(x−3)=Ax−2+Bx−3 $\frac{x - 5}{(x - 2) (x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}$
2x(x2−1)=Ax+Bx−1+Cx+1 $\frac{2}{x (x^{2} - 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}$
8x(x2+4)=Ax+Bx+Cx2+4 $\frac{8}{x (x^{2} + 4)} = \frac{A}{x} + \frac{B x + C}{x^{2} + 4}$
2x(x+1)(x+2)(x+3)=Ax+1+Bx+2+Cx+3 $\frac{2 x}{(x + 1) (x + 2) (x + 3)} = \frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3}$

In Problems 23 and 24, use the method of Example 7 to find a basis for the solution space of the given differential equation. (It’s 3-dimensional in Problem 23 and 4-dimensional in Problem 24.)

y'''=0 $y ‴ = 0$
y(4)=0 $y^{(4)} = 0$

In Problems 25 and 26, use the method of Example 8 to find a basis for the 2-dimensional solution space of the given differential equation.

y''−5y'=0 $y ″ - 5 y ′ = 0$
y''+10y'=0 $y ″ + 10 y ′ = 0$
Take the positive sign in Eq. (20), and then use the standard integral

$\int d u u 2 + 1 - - - - - \sqrt = sinh - 1 u + C$ $\int \frac{d u}{\sqrt{u^{2} + 1}} = \sinh^{- 1} u + C$

to derive the same general solution y(x)=A cosh x+B sinh x $y (x) = A \cosh x + B \sinh x$ given in Eq. (22).
Use the method of Example 9 and the standard integral

$\int d u 1 - u 2 - - - - - \sqrt = sinh - 1 u + C$ $\int \frac{d u}{\sqrt{1 - u^{2}}} = \sinh^{- 1} u + C$

to derive the general solution y(x)=A cos x+B sin x $y (x) = A \cos x + B \sin x$ of the second-order differential equation y''+y=0 $y ″ + y = 0$ . Thus its solution space has basis {cos x,sin x} ${\cos x, \sin x}$ .
Let V be the set of all infinite sequences {xn}={x1, x2, x3, …} ${x_{n}} = {x_{1}, x_{2}, x_{3}, \dots}$ of real numbers. Let addition of elements of V and multiplication by scalars be defined as follows:

${x n} + {y n} = {x n + y n}$ ${x_{n}} + {y_{n}} = {x_{n} + y_{n}}$

and

$c {x n} = {c x n} .$ $c {x_{n}} = {c x_{n}} .$
1. Show that V is a vector space with these operations.
2. Prove that V is infinite dimensional.
Let V be the vector space of Problem 29 and let the subset W consist of those elements {xn} ${x_{n}}$ of V such that xn=xn−1+xn−2 $x_{n} = x_{n - 1} + x_{n - 2}$ for n≥2 $n \geq 2$ . Thus a typical element of W is the Fibonacci sequence

${1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \dots} .$ ${1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \dots} .$
1. Show that W is a subspace of V.
2. Prove that W is 2-dimensional.

Problems 31 through 33 develop a method of multiplying matrices whose entries are complex numbers.

Motivated by Example 2, let us define a function or transformation T from the set of all complex numbers to the space of all 2×2 $2 \times 2$ matrices of the form in (1) as follows. Given a complex number z=a+bi $z = a + b i$ , let

$T (z) = [a b - b a] .$ $T (z) = [\begin{array}{r} a & - b \\ b & a \end{array}] .$
1. Suppose that z1 $z_{1}$ and z2 $z_{2}$ are complex numbers and c1 $c_{1}$ and c2 $c_{2}$ are real numbers. Show that
  
  $T (c 1 z 1 + c 2 z 2) = c 1 T (z 1) + c 2 T (z 2) .$ $T (c_{1} z_{1} + c_{2} z_{2}) = c_{1} T (z_{1}) + c_{2} T (z_{2}) .$
2. Show that
  
  $T (z 1 z 2) = T (z 1) T (z 2)$ $T (z_{1} z_{2}) = T (z_{1}) T (z_{2})$
  
  for all complex numbers z1 $z_{1}$ and z2 $z_{2}$ . Note the complex multiplication on the left in contrast with the matrix multiplication on the right.
3. Prove that if z is an arbitrary nonzero complex number, then
  
  $T (z - 1) = {T (z)} - 1,$ $T (z^{- 1}) = {T (z)}^{- 1},$
  
  where z−1=1/z $z^{- 1} = 1 / z$ and {T(z)}−1 ${T (z)}^{- 1}$ is the inverse of the matrix T(z).
Let A and B be 4×4 $4 \times 4$ (real) matrices partitioned into 2×2 $2 \times 2$ submatrices or “blocks”:

$A = [A 11 A 21 A 12 A 22], B = [B 11 B 21 B 12 B 22] .$ $\begin{array}{l} A = [\begin{array}{l} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array}], & B = [\begin{array}{l} B_{11} & B_{12} \\ B_{21} & B_{22} \end{array}] . \end{array}$

Then verify that AB can be calculated in “blockwise” fashion:

$A B = [A 11 B 11 + A 12 B 21 A 21 B 11 + A 22 B 21 A 11 B 12 + A 12 B 22 A 21 B 12 + A 22 B 22] .$ $A B = [\begin{array}{l} A_{11} B_{11} + A_{12} B_{21} & A_{11} B_{12} + A_{12} B_{22} \\ A_{21} B_{11} + A_{22} B_{21} & A_{21} B_{12} + A_{22} B_{22} \end{array}] .$
Given a 2×2 $2 \times 2$ matrix M=[zij] $M = [z_{i j}]$ whose entries are complex numbers, let T(M) denote the 4×4 $4 \times 4$ matrix of real numbers given in block form by

$T (M) = [T (z 11) T (z 21) T (z 12) T (z 22)] .$ $T (M) = [\begin{array}{l} T (z_{11}) & T (z_{12}) \\ T (z_{21}) & T (z_{22}) \end{array}] .$

The transformation T on the right-hand side is the one we defined in Problem 31. Suppose that M and N are 2×2 $2 \times 2$ complex matrices. Use Problems 31(b) and 32 to show that

$T (M N) = T (M) T (N) .$ $T (M N) = T (M) T (N) .$

Hence one can find the product of the 2×2 $2 \times 2$ complex matrices M and N by calculating the product of the 4×4 $4 \times 4$ real matrices T(M) and T(N). For instance, if

$M = [11 + + i 2 i 21 + + i 3 i]$ $M = [\begin{array}{lcrlcr} 1 & + & i & 2 & + & i \\ 1 & + & 2 i & 1 & + & 3 i \end{array}]$

and

$N = [3 - i 2 - i 1 - 2 i 3 - 2 i],$ $N = [\begin{array}{l} 3 - i & 1 - 2 i \\ 2 - i & 3 - 2 i \end{array}],$

then

$T (M) T (N) = = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ 1112 - 1 1 - 2 1 2113 - 1 2 - 3 1 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ \times ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ 3 - 1 2 - 1 1312 1 - 2 3 - 2 2123 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ 921010 - 2 9 - 10 10 11 - 2 147 211 - 7 14 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ .$ $\begin{array}{l} T (M) T (N) \\ \begin{array}{l} = & [\begin{array}{r} 1 & - 1 & 2 & - 1 \\ 1 & 1 & 1 & 2 \\ 1 & - 2 & 1 & - 3 \\ 2 & 1 & 3 & 1 \end{array}] \times [\begin{array}{r} 3 & 1 & 1 & 2 \\ - 1 & 3 & - 2 & 1 \\ 2 & 1 & 3 & 2 \\ - 1 & 2 & - 2 & 3 \end{array}] \\ = & [\begin{array}{r} 9 & - 2 & 11 & 2 \\ 2 & 9 & - 2 & 11 \\ 10 & - 10 & 14 & - 7 \\ 10 & 10 & 7 & 14 \end{array}] . \end{array} \end{array}$

Therefore,

$M N = [910 + + 2 i 10 i 1114 - + 2 i 7 i] .$ $M N = [\begin{array}{r} 9 & + & 2 i & 11 & - & 2 i \\ 10 & + & 10 i & 14 & + & 7 i \end{array}] .$

If M is a nonsingular $2 \times 2$ complex matrix, it can be shown that $T (M^{- 1}) = {T (M)}^{- 1}$ . The $n \times n$ versions of these results are sometimes used to carry out complex matrix operations in computer languages that support only real arithmetic.

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Table of Contents for 4.7 General Vector Spaces

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Table of Contents for
4.7 General Vector Spaces