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 MmnMmn denote the set of all m×nm×n matrices with real number entries. Then MmnMmn 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 MmnMmn is the m×nm×n matrix 0 matrix whose elements are all zeros, and the negative AA 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 1im1im and 1jn1jn, let EijEij denote the m×nm×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}{Eij} of MmnMmn form a basis for MmnMmn, and so MmnMmn is a finite-dimensional vector space of dimension mn.

For instance, in the case of the vector space M22M22 of all 2×22×2 matrices, these alleged basis elements are the four matrices

E11=[1000],E12=[0100],E21=[0010],andE22=[0001].
E11=[1000],E12=[0010],E21=[0100],andE22=[0001].

Then any matrix A in M22M22 can be expressed as

A=[abcd]=aE11+bE12+cE21+dE22,
A=[acbd]=aE11+bE12+cE21+dE22,

so the set {E11, E12, E21, E22}{E11, E12, E21, E22} spans M22M22. Moreover,

aE11+bE12+cE21+dE22=[abcd]=[0000]
aE11+bE12+cE21+dE22=[acbd]=[0000]

implies immediately that a=b=c=d=0a=b=c=d=0, so the set {E11, E12, E21, E22}{E11, E12, E21, E22} is also linearly independent and therefore forms a basis for M22M22. Thus, we have shown that M22M22 is a 4-dimensional vector space.

Example 2

Let C denote the subset of M22M22 consisting of all 2×22×2 matrices of the form

[abba].
[abba].
(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 M22M22.

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

Re=[1001]andIm=[0110]
Re=[1001]andIm=[0110]
(2)

in C. Here Re is another notation for the 2×22×2 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

[abba]=aRe+bIm.
[abba]=aRe+bIm.
(3)

so these two matrices span C. Thus {Re, Im}{Re, Im} is a basis for C, and so C is a 2-dimensional subspace of the 4-dimensional vector space M22M22 of 2×22×2 matrices.

Now observe that

(Im)2=[0110][0110]=[1001]=Re.
(Im)2=[0110][0110]=[1001]=Re.
(4)

It follows that the matrix product of any two elements aRe+bImaRe+bIm and cRe+dImcRe+dIm of C is given by

(aRe+bIm)(cRe+dIm)=ac(Re)2+bcImRe+adReIm+bd(Im)2=(acbd)Re+(ad+bc)Im.
(aRe+bIm)(cRe+dIm)==ac(Re)2+bcImRe+adReIm+bd(Im)2(acbd)Re+(ad+bc)Im.
(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 M22M22 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+bia+bi, where a and b are real numbers and i=1i=1 denotes the “imaginary” square root of 11 (so that i2=1i2=1). Because (Im)2=Re(Im)2=Re, the matrix Im plays the role of i and the matrix

[abba]=aRe+bIm
[abba]=aRe+bIm

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

(a+bi)(c+di)=(acbd)+(ad+bc)i
(a+bi)(c+di)=(acbd)+(ad+bc)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 FF of all real-valued functions defined on the real line R. If f and g are elements of FF and c is a scalar, then

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

and

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

for all x in R. The zero element in FF is the function 0 such that 0(x)=00(x)=0 for all x. We frequently refer to a function in FF by simply specifying its formula. For instance, by “the function x2x2” we mean that function whose value at each x in R is x2x2.

The functions f1,f2, ,fkf1,f2, ,fk in FF are linearly dependent provided that there exist scalars c1,c2, ,ckc1,c2, ,ck, not all zero, such that

c1f1(x)+c2f2(x)++ckfk(x)=0
c1f1(x)+c2f2(x)++ckfk(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 exex and e2xe2x are linearly independent because either of the equations ex=ae2xex=ae2x or e2x=bexe2x=bex would imply that e3xe3x 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 xsin 2x=2sin xcos x. The three functions 1, cos2 xcos2 x, and sin2 xsin2 x are linearly dependent, because the fundamental identity cos2 x+sin2 x=1cos2 x+sin2 x=1 can be written in the form

(1)(1)+(1)(cos2 x)+(1)(sin2 x)=0.
(1)(1)+(1)(cos2 x)+(1)(sin2 x)=0.

A subspace of FF is called a function space. An example of a function space is the vector space PP of all polynomials in FF (Example 4 in Section 4.4). Recall that a function p(x) in FF is called a polynomial of degree n0n0 if it can be expressed in the form

p(x)=a0+a1x+a2x2++anxn
p(x)=a0+a1x+a2x2++anxn
(7)

with an0an0. 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 PP is, indeed, a subspace of FF. (Recall the subspace criterion of Theorem 1 in Section 4.2.)

Example 4

Given n0n0, denote by PnPn the set of all polynomials of degree at most n. That is, PnPn 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 PnPn is a subspace of PP (and of FF). The formula in (7) shows that the n+1n+1 polynomials

1, x, x2, x3, , xn
1, x, x2, x3, , xn
(8)

span PnPn. To show that these “monomials” are linearly independent, suppose that

c0+c1x+c2x2++cnxn=0
c0+c1x+c2x2++cnxn=0
(9)

for all x in R. If a0, a1, , ana0, a1, , an are fixed distinct real numbers, then each aiai satisfies Eq. (9).

In matrix notation this means that

[1a0a20an01a1a21an11a2a22an21ana2nann][c0c1c2cn]=[0000].
1111a0a1a2ana20a21a22a2nan0an1an2annc0c1c2cn=0000.
(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=0Vc=0 has only the trivial solution c=0c=0, so it follows from (10) that c0=c1=c2==cn=0c0=c1=c2==cn=0. Thus we have proved that the n+1n+1 monomials 1, x, x2, , xn1, x, x2, , xn are linearly independent and hence constitute a basis for PnPn. Consequently PnPn 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

a0+a1x++anxn=b0+b1x++bnxn
a0+a1x++anxn=b0+b1x++bnxn

for all x, then a0=b0, a1=b1, a0=b0, a1=b1, , and an=bnan=bn. (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

6x(x1)(x+1)(x+2)=Ax1+Bx+1+Cx+2
6x(x1)(x+1)(x+2)=Ax1+Bx+1+Cx+2
(11)

for all x (other than x=1, 1x=1, 1, or 22).

Solution

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

6x=A(x+1)(x+2)+B(x1)(x+2)+C(x1)(x+1);6x=(2A2BC)+(3A+B)x+(A+B+C)x2.
6x=A(x+1)(x+2)+B(x1)(x+2)+C(x1)(x+1);6x=(2A2BC)+(3A+B)x+(A+B+C)x2.

Then the identity principle for polynomials yields the linear equations

2A2BC=03A+B=6A+B+C=0,
2A3AA++2BBB+CC===060,

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

6x(x1)(x+1)(x+2)=1x1+3x+14x+2
6x(x1)(x+1)(x+2)=1x1+3x+14x+2

if x1, 1, 2x1, 1, 2.

Example 6

Show that the four polynomials

1,x,3x21,and5x33x
1,x,3x21,and5x33x
(12)

form a basis for P3P3.

Solution

In order to show simultaneously that these four polynomials span P3P3 and are linearly independent, it suffices to see that every polynomial

p(x)=b0+b1x+b2x2+b3x3
p(x)=b0+b1x+b2x2+b3x3
(13)

in P3P3 can be expressed uniquely as a linear combination

c0(1)+c1(x)+c2(3x21)+c3(5x33x)=(c0c2)+(c13c3)x+(3c2)x2+(5c3)x3
c0(1)+=c1(x)+c2(3x21)+c3(5x33x)(c0c2)+(c13c3)x+(3c2)x2+(5c3)x3
(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

[1010010300300005][c0c1c2c3]=[b0b1b2b3]
1000010010300305c0c1c2c3=b0b1b2b3

obviously has a unique solution for c0, c1, c2, c3c0, c1, c2, c3 in terms of b0, b1, b2, b3b0, b1, b2, b3.

Because the vector space PP of all polynomials contains n+1n+1 linearly independent functions for every integer n0n0, it follows that PP is an infinite-dimensional vector space. So too is any function space, such as FF itself, that contains PP. 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 FF because every linear combination of continuous functions is again a continuous function. Every polynomial is a continuous function, so C(0)C(0) contains PP and is therefore an infinite-dimensional vector space. Similarly, the set C(k)C(k) of all functions in FF 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=ky(k constant)
y'=ky(k constant)
(15)

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

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

ay+by+cy=0
ay''+by'+cy=0
(16)

(with constant coefficients a, b, and c) is a 2-dimensional function space SS, 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=1a=1 and b=c=0b=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)=0y''(x)=0, then two integrations give first

y(x)=y(x) dx=(0) dx=A
y'(x)=y''(x) dx=(0) dx=A

and then

y(x)=y(x) dx=Adx=Ax+B,
y(x)=y'(x) dx=Adx=Ax+B,

where A and B are arbitrary constants of integration. Thus every solution of (17) is of the form y(x)=Ax+By(x)=Ax+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 P1P1 of linear polynomials generated by the basis {1,x}{1,x}.

Example 8

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

y2y=0.
y''2y'=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)=2v(x).
v'(x)=2v(x).

But this is a familiar first-order differential equation with general solution v(x)=Ce2xv(x)=Ce2x. Then integration gives

y(x)=y(x) dx=v(x) dx=Ce2x dx=12Ce2x+A.
y(x)=y'(x) dx=v(x) dx=Ce2x dx=12Ce2x+A.

Thus every solution of (18) is of the form y(x)=A+Be2xy(x)=A+Be2x (where B=12CB=12C), 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,e2x}.

Example 9

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

yy=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=dvdx=dvdydydx=vdvdy=y;
y''=dvdx=dvdydydx=vdvdy=y;

so v dv=y dyv dv=y dy. Integration therefore gives

12v2=12y2+C,sov2=y2±a2,
12v2=12y2+C,sov2=y2±a2,
(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/dxv=dy/dx, it then follows from (20) that

x=±dyy2a2.
x=±dyy2a2.
(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=auy=au and the standard integral

duu21=cosh1 u+C
duu21=cosh1 u+C

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

x=cosh1ya+b,
x=cosh1ya+b,

so finally,

y=a cosh(xb)=a(cosh xcosh bsinh xsinh b),
y=a cosh(xb)=a(cosh xcosh bsinh xsinh 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 bA=a cosh b and B=a sinh bB=a sinh b. Thus every solution of the differential equation yy=0y''y=0 in (19) is of the form y(x)=A cosh x+B sinh xy(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 SS 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=ex+ex2,sinh x=exex2
cosh x=ex+ex2,sinh x=exex2

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

y(x)=Cex+Dex.
y(x)=Cex+Dex.

Consequently, {ex, ex}{ex, ex} is a second basis for the 2-dimensional solution space SS.

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

4.7 Problems

In Problems 1–4, determine whether or not the indicated set of 3×33×3 matrices is a subspace of M33M33.

  1. The set of all diagonal 3×33×3 matrices.

  2. The set of all symmetric 3×33×3 matrices (that is, matrices A=[aij]A=[aij] such that aij=ajiaij=aji for 1i3, 1j31i3, 1j3).

  3. The set of all nonsingular 3×33×3 matrices.

  4. The set of all singular 3×33×3 matrices.

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

  1. The set of all f such that f(0)=0f(0)=0.

  2. The set of all f such that f(x)0f(x)0 for all x.

  3. The set of all f such that f(0)=0f(0)=0 and f(1)=1f(1)=1.

  4. 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+a3x3a0+a1x+a2x2+a3x3 is given. Determine whether or not the set of all such polynomials satisfying this condition is a subspace of the space PP of all polynomials.

  1. a30a30

     

  2. a0=a1=0a0=a1=0

     

  3. a0+a1+a2+a3=0a0+a1+a2+a3=0

     

  4. a0, a1, a2a0, a1, a2, and a3a3 are all integers

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

  1. sin x and cos x

  2. exex and xexxex

     

  3. 1+x, 1x1+x, 1x, and 1x21x2

     

  4. 1+x, x+x21+x, x+x2, and 1x21x2

     

  5. cos 2x, sin2 xcos 2x, sin2 x, and cos2 xcos2 x

     

  6. 2 cos x+3 sin x2 cos x+3 sin x and 4 cos x+5 sin x4 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.

  1. x5(x2)(x3)=Ax2+Bx3x5(x2)(x3)=Ax2+Bx3

     

  2. 2x(x21)=Ax+Bx1+Cx+12x(x21)=Ax+Bx1+Cx+1

     

  3. 8x(x2+4)=Ax+Bx+Cx2+48x(x2+4)=Ax+Bx+Cx2+4

     

  4. 2x(x+1)(x+2)(x+3)=Ax+1+Bx+2+Cx+32x(x+1)(x+2)(x+3)=Ax+1+Bx+2+Cx+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.)

  1. y=0y'''=0

     

  2. y(4)=0y(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.

  1. y5y=0y''5y'=0

     

  2. y+10y=0y''+10y'=0

     

  3. Take the positive sign in Eq. (20), and then use the standard integral

    duu2+1=sinh1 u+C
    duu2+1=sinh1 u+C

    to derive the same general solution y(x)=A cosh x+B sinh xy(x)=A cosh x+B sinh x given in Eq. (22).

  4. Use the method of Example 9 and the standard integral

    du1u2=sinh1 u+C
    du1u2=sinh1 u+C

    to derive the general solution y(x)=A cos x+B sin xy(x)=A cos x+B sin x of the second-order differential equation y+y=0y''+y=0. Thus its solution space has basis {cos x,sin x}{cos x,sin x}.

  5. Let V be the set of all infinite sequences {xn}={x1, x2, x3, }{xn}={x1, x2, x3, } of real numbers. Let addition of elements of V and multiplication by scalars be defined as follows:

    {xn}+{yn}={xn+yn}
    {xn}+{yn}={xn+yn}

    and

    c{xn}={cxn}.
    c{xn}={cxn}.
    1. Show that V is a vector space with these operations.

    2. Prove that V is infinite dimensional.

  6. Let V be the vector space of Problem 29 and let the subset W consist of those elements {xn}{xn} of V such that xn=xn1+xn2xn=xn1+xn2 for n2n2. Thus a typical element of W is the Fibonacci sequence

    {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, }.
    {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, }.
    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.

  1. Motivated by Example 2, let us define a function or transformation T from the set of all complex numbers to the space of all 2×22×2 matrices of the form in (1) as follows. Given a complex number z=a+biz=a+bi, let

    T(z)=[abba].
    T(z)=[abba].
    1. Suppose that z1z1 and z2z2 are complex numbers and c1c1 and c2c2 are real numbers. Show that

      T(c1z1+c2z2)=c1T(z1)+c2T(z2).
      T(c1z1+c2z2)=c1T(z1)+c2T(z2).
    2. Show that

      T(z1z2)=T(z1)T(z2)
      T(z1z2)=T(z1)T(z2)

      for all complex numbers z1z1 and z2z2. 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(z1)={T(z)}1,
      T(z1)={T(z)}1,

      where z1=1/zz1=1/z and {T(z)}1{T(z)}1 is the inverse of the matrix T(z).

  2. Let A and B be 4×44×4 (real) matrices partitioned into 2×22×2 submatrices or “blocks”:

    A=[A11A12A21A22],B=[B11B12B21B22].
    A=[A11A21A12A22],B=[B11B21B12B22].

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

    AB=[A11B11+A12B21A11B12+A12B22A21B11+A22B21A21B12+A22B22].
    AB=[A11B11+A12B21A21B11+A22B21A11B12+A12B22A21B12+A22B22].
  3. Given a 2×22×2 matrix M=[zij]M=[zij] whose entries are complex numbers, let T(M) denote the 4×44×4 matrix of real numbers given in block form by

    T(M)=[T(z11)T(z12)T(z21)T(z22)].
    T(M)=[T(z11)T(z21)T(z12)T(z22)].

    The transformation T on the right-hand side is the one we defined in Problem 31. Suppose that M and N are 2×22×2 complex matrices. Use Problems 31(b) and 32 to show that

    T(MN)=T(M)T(N).
    T(MN)=T(M)T(N).

    Hence one can find the product of the 2×22×2 complex matrices M and N by calculating the product of the 4×44×4 real matrices T(M) and T(N). For instance, if

    M=[1+i2+i1+2i1+3i]
    M=[11++i2i21++i3i]

    and

    N=[3i12i2i32i],
    N=[3i2i12i32i],

    then

    T(M)T(N)=[1121111212132131]×[3112132121321223]=[921122921110101471010714].
    T(M)T(N)==1112112121131231×3121131212322123921010291010112147211714.

    Therefore,

    MN=[9+2i112i10+10i14+7i].
    MN=[910++2i10i1114+2i7i].

    If M is a nonsingular 2×2 complex matrix, it can be shown that T(M1)={T(M)}1. The n×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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