Multivariable-Calculus-Linear-Algebra-Sol--William-Trench

Multiplication of a row of [A 1 Y] by a nonzero constant corresponds to multiplication of an equation in the system AX = Y by the same constant, This does not change the solutions of the

Answers to Selected Problems in Multivariable Calculus with Linear Algebra and Series

WILLIAM F TRENCH AND BERNARD KOLMAN Drexel University

= r [ a ] +r [ b ] Write (a) S = 1 + 2 + · · · + n and (b) S = n + (n-1) n n + · · · + 1 Then add (a) and (b) t o o b t a i n

T-10 B = — , C = — - For uniqueness, let

(i) A = B + C ±9 where B^ = B ± and C^ = -C ;

Then (ii) AT = B^ + C^ = B - C Adding (i) and

(ii) yields B = B; subtracting (ii) from (i) yields

AI = A n

(a) Suppose the i-th row of A consists entirely of

zeros Let C = AB; then

k > i for all terms in the last sum; hence b, = 0

for all these terms Therefore, d = 0 if i > i

(i) a Φ ±3 (ii) a = -3 (iii) a = 3

(i) a + ±1 (ii) a = -1 (iii) a = 1

T-l Multiplication of a row of [A 1 Y] by a nonzero

constant corresponds to multiplication of an

equation in the system AX = Y by the same constant, This does not change the solutions of the system Similar argument applies to the other elementary operations

2

4

6

A row in the augmented matrix of a system in

n unknowns, with zeros in the first n columns and

a 1 in the (n + 1) -st corresponds to the equation

0·χ + + 0·χ = 1, n which has no solution

For the converse, let [A|Y] be row equivalent

to [ B | Z ] , which is in row echelon form Since B has no row with a "leading" 1 in the (n + l)-st column, it follows (with the notation of Def 2.3) that j- < j? < · ·· < j,· Hence, for i = 1, 2, ,k, the i-th equation of BX = Z can be solved for x in terms of the remaining n - k unknowns, which can be specified arbitrarily

By definition of the elementary row operations, the system BX = 0 is obtained by performing on AX = 0 operations which do not change the solutions of the latter

Suppose ad - be ^ 0, b ^ O , d ^ O Then the

following matrices are row equivalent :

Suppose ad - be ^ 0 and b = 0 Then d φ 0 and

a φ 0 The following matrices are row equivalent:

T-2

T-3

T-4

similar argument disposes of the case where d = 0

If ad - bc = 0, then A is row equivalent to

T-6

only 2 x 2 matrices, other than I«, in reduced row

echelon form Since BX = 0 and CX = 0 have

nontrivial solutions, so does AX = 0, again by

J row equivalent, by Thm 2.2

Suppose A is m x n and B is p x q Since AB is

defined, (1) n = p and AB is m x q Since BA is defined,(2) m = q and BA is p x n Since AB = BA,

(3) m = p and (4) q = n The conclusion now

follows from (1), (2), (3) and (4)

T -1 T - I T T -1 T T -IT

A1 (A V = (A A)1 = I1 = I and (A V A = (AA V

T T -1 T -1 -1 T

= 1 = 1 If A = A1, then A = (A1) = (A V ; hence A is symmetric

Let E = [ e ] , E A = [ a ] , and I = [ 6 ] , where

ill 13 i j

δ = 0 if i H and δ , , = 1 i f i = j

(i) Let the row operation be multiplication of the

r-th row by c φ 0 Then e = δ if i Φ r and

e = c δ Now a = \ δ., a, = a i f i ^ r , rj rj ij / lk kj ij

n k=l and a = c \ δ a = c a Hence B = EA rj / rj kj rj

k=l (ii) Let the row operation be the interchange of

rows r and s Then e = δ if i Φ r and i φ s,

Hence B = EA (iii) Let the operation be

addition of c times the r-th row to the s-th row

(r φ s) Then e = δ + c δ δ and ij ij is rj

a = > (6., + c δ δ , )a, rj / lk is rk kj

k=l T-3

k=l k=l

6 , a rk kj

= a + e δ a ; i j i s rj hence B = EA

Since e = δ + c 6 δ and f = δ - c δ δ

l j l j IS rJ !J IJ 1 S rJ

E e " £ xi ■ R lk i s rk kj ks r j ., + c δ δ )(δ, - c 6, 6 ) δ (6, - c δ δ ) lk kj ks r j

In the notation of Definition 2.3,

l < j < j9< * » - < j = n , which implies that

j1= l , j0=2, ,j = n Hence a = 1,

i = 1, , n Since A is in reduced row echelon

form, it follows that a = 0 if i φ j; hence

A = I

T-4

T-5

T-6

Τ-7

Τ-8

Suppose a^ = 0, 1 < j < η Let B be an arbitrary

n x n matrix and C = AB Then

n n

c = \ a , b rj / rk kj 0-bkj = 0 , 1 < j < n;

k=l k=l

hence C has a row of zeros Therefore AB φ I

for any B, which means that A is singular

Let A be row equivalent to B, which is in reduced row echelon form Then B = E E A, where

E , , E, are elementary matrices Consequently,

B is nonsingular if and only if A is nonsingular From Exercise T-5, it now follows that A is

nonsingular if and only if B = I

The conclusion follows from Exercise T-4,

Section 1.2, and Corollary 3.2

T-9

1

ad-bc

d -c

- b |

a The result follows from Theorems 3.4 and 3.6

28 (a)

1 2 1

3" - 3" 3

1 1 2(b) - 3 - 3 3

I I I

3 3-"6

30 27

Interchange of adjacent elements j and j

increases the number of inversions by 1, if

1 < j ,_, or decreases it by 1 if j > j ,, ·

Now suppose s - r > 2 An interchange of j and

j can be effected by successively interchanging

j with j , , j (s - r interchanges of adjacent elements) to obtain

an odd integer, the conclusion follows

Let D be the determinant obtained by inter­

changing two columns of D Let D and D be

obtained by interchanging the rows and columns of

D and D, respectively Then D is obtained by interchanging two rows of D; hence D = -D

(Thm 4.2) However, by Thm 4 1

D = D and D = D

Therefore D = -D

Use Thm 4.4

Similar to proof given for (a) and (c)

The right side of the equation

T-7 Let the elementary operation be:

(a) interchange of two rows Then det E = -1 and det EB = -det B; hence det EB = (det E) (det B)

(b) multiplication of a row by a constant c φ 0

Then det E = c and det EB = c det B; hence det EB = (det E)(det B)

(c) addition of a multiple of a row to another row Then det E = 1 and det EB = det B; hence

det EB = (det E)(det B)

T-8 det E E B = (det Εχ) det(E2 E B)

= (det E )(det E2) det(E3 EkB)

= (det E )(det E ) (det E ) det B

= (det E E )(det E ) (det E ) det B

= det (E E )det B

T-9 From Eq (23), adj A = DA~" ; the result follows

from Exercise T-3

T-10 If det A φ 0, then A is nonsingular (Thm 4.10);

hence AX = 0 has only the trivial solution

(Cor 3.3, Sect 1.3) If det A = 0, then A is singular (Thm 4.10); hence A = E E B,

where B is in reduced row echelon form, and has a row of zeros (Cor 3.2, Sect 1.3) The system

BX = 0 has the same solutions as the system

B-X = 0, where B- is obtained by omitting the last (zero) row of B Since B-X = 0 is a homogeneous system of n equations in n - 1 unknowns, it has a

solution X φ 0 (Thm 2.6, Section, 1.2) Since

T-18 If A = A then AA = I; hence (det A ) 2 = 1,

- y )

(χ-1

-z) z) -z)

(y-z)(y-x)

~(z + x) (y-z)(y-x)

1

J2L

(z-x)(z-y) -(x + y) (z-x)(z-y)

1 (x-y)(x-z) (y-z)(y-x) (z-x)(z-y)

T-20 Apply Definition 4.3 to A - ti

T-21 Let A be upper triangular; then a = 0 if j < i

T-l aO = a(0 + 0) = aO + aO; now add -(aO) to the first

and last member to obtain 0 = aO

T-2 0 = 0U = [1 + (-l)U] = 1-U + (-l)U = U + (-l)U;

hence (-l)U = -U

T-3 Add -U to both sides and use the associative law T-4 Write (a - b)U = 0 and use (vi) of Thm 1.1

T-7 Suppose T is a subspace of S, and U is in S Then,

by (b) of Def 1.2, (-l)U is in T, and then, by (a) of Def 1.2, U + (-l)U = 0 is in T The

remaining properties of Def 1.1 hold in T, since they hold in S Conversely, if T is a vector space, then (a) and (b) hold, by Def 1.2

T-8 Use the following properties of continuous

functions: (a) cf is continuous if c is constant and f is continuous (b) f + g is continuous if

f and g are

T-9 In T-8, replace "continuous" by "n-times

differentiable."

°-As to spanning S[X-, X~ , X~] , it suffices to show that

each vector is a linear combination of the other two:

Χ Λ "~ X-j T" X« ; X_ — X~ — X~ ? Λ „ — X_ X_

T-l

T-2

T-S If T = {Ul ,· · · , Un}

then one of the U 's

1

others Remove this

T-3 Let T = {Xl' ' Xm} be a linearly independent set

If X is in 5 and X is not a linear combination ofXl'··.' Xm, then T' = {Xl'···' Xm, X} is linearlyindependent, which is a contradiction, since T' hasmore then m elements Thus T spans 5; since T islinearly independent, it is a basis for 5

T-4 Let dim 5 = dim T = n Let {Xl' ' Xn} be a basisfor T Then {Xl' ' Xn} is also a basis for 5, by

Ex T-3 Hence 5 T

is not linearly independent,

is a linear combination of thevector from T to obtain T' Then T' also spans 5 But this is a contradiction,since it implies that dim 5 < n

T-6 By Tbm 2.6, there is a basis which contains

Ul , , Un· Since dim S = n, this basis cannotcontain any other vectors

T-7 Without loss of generality, suppose that Ul ,···, Urn

(m 2 n) are linearly independent, and if m < n,

Um+l , , Un can be written as linear combinations

of Ul , , Urn· Then {Ul ,···, Urn} spans

S[Ul ,···, Un]; hence dim 5[Ul ,···, Un] =m

T-B Recall that det AT = det A

T-9 Let 51 = {Ul ,·.·, Uk} and

52 = {Ul ,···, Uk' Uk+l ,···, Un}·

0, where not all of the

a are zero, then a^U- + l 1 1 + a k \

+ 0·1λ _, + ···+ 0·ϋ is a nontrivial linear k+1 n

combination of vectors in S~ which vanishes

(b) If S is linearly dependent, then so is S~ , by (a)

1

-5

5 -2

with respect to the natural basis;

with respect to B and C; L

2

3

= L· -J

5

- 1 j

13

~ J

L(0) = L(0 + 0) = L(0) + L(0); hence L(0) = 0 Suppose V and V are in range L; then L(U ) = V and L(U?) = V , where U and U are in S: Then

LCU-L + U2) = LC^) + L(U2) = νχ + V2; hence V ± + V2

is in range L If a is a constant, then

L(aU ) = aV ; hence aV is in range L Thus range

L is a subspace of T, by Def 1.2, Sect 2.1

T-4 Every X in S can be w r i t t e n as X = a-lL + · · · + a U ; J l i n n

hence every vector in range L can be written as

Y = L(a,U, + · · · + a U ) = a,L(U1 1 n n l l n n n) + · · · + a L(U )

Therefore {L(U ) , , L(U )} spans range L

T-5 Let U = 01 1 n n 1 1 n n Ίϋ- + ··· + 3 U and V = ynU, + ··· + γ U ,

and suppose L(U) = L(V) Then 3 = y (1 < i < n) ,

and therefore U = V; hence L is one-to-one If

X = is an arbitrary vector in R ; then

L(3-,U, + ··· + 3 U ) = X; hence L is onto 1 1 n n

T-6 If (X) = X for every X in R , then

T-8 Let {υ1 n Ί, , U } be a basis for S If

ajLO^) + + a L(U ) = 0, then n n

alUl + '*

+ a U ) = 0 (Thm 3.1) n n Thus + a U = 0 , because ker L = {0}, and n n

a = =1 n a =0, since IL , , , , U are linearly I n J

independent Therefore {L(IL ) , , L(U ) } i s 1 n

linearly independent, and, from Exercise T-4, spans

range L; hence it is a basis for range L

T-9 Let A = [a ] be m x n Define

L(A) = [a LV a12, , a ^ , a ^ , a^, , a^,.,

l T

ml mn

T-10 If L is one-to-one, then dim(ker L) = 0 (Thm 3.3);

then by Thm 3.5, dim(range L) = n = dim T Hence

L is onto Conversely, if L is onto, then

dim(range L) = n and dim(ker L) = 0 (Thm 3.5);

hence L is one-to-one (Thm 3.3)

T-ll Let U , , U be linearly independent in S If

L(U ) , , L(U ) are linearly dependent, then there

are scalars a-, , a, , not all zero, such that

a1L(U1) + + akL(iy = Θ

Let U = a U + ··· + a U (which is nonzero); then

L(U) = Θ, so L is not one-to-one (Thm 3.3) For

the converse, let U + 0; then L(U) φ Θ, by

hypothesis; hence L is one-to-one (Thm 3.3)

T-12 Use Ex T-ll and the fact L:S -> Rn, defined by

L(U) = (U)_,, is one-to-one

D

T-13 In Thm 3.8, let T=s, and let L(U) = U for every

U(L = identity) The result follows from Eq (15)

10 3 12 3 14 {X , Χ2> 16 yes

-1 Let A be an m x n matrix, and suppose A contains

columns j , , j from A, where 1 < j < ··· < j l K — 1 k and 1 < k < n If the rows of A are linearly

dependent, there are constants c , , c , not all

I m zero, such that

Cl[all' a12'···' aln] + C2[a21' a2 2 " · " a2n]

+ ··· + c [ a , a , a ] = [0, 0, , 0]

m ml mz mn Then

-3 Let L:Rn -> Rm be defined by L(X) = AX By Theorem

3.5, dim(range L) + dim(ker L) = n Since

R(A) = dim(range L) and N(A) = dim(ker L) , the

conclusion follows

-4 The same sequence of elementary row operations that

leads from A to B automatically leads from A., to

V

Suppose columns p , , p of A form a basis for the column space of A; then they are linearly

independent If A is the submatrix consisting of columns p , , p of A, then A is of rank r and consequently has a nonzero subdeterminant of order

r (Theorem 4.3) Let B be the submatrix consisting

of columns p-, , p of B Since B is row 1

equivalent to A , and elementary row operations

* preserve nonvanishing of determinants, B has a nonzero determinant of order r The proof of the converse is similar

Let B be as defined in Thm 4.6, with p = 1,

p = 2, , p = k Then det B φ 0 if and only if

b,, = b11 22 kk 0 0 = ··· = b = 1 Now the conclusion follows from Thm 4.6

Let A be an m x n matrix with k nonzero rows, which

we denote by R , , R, ; thus

R = [rl il i2 in #1 , r.0, , r ] With j , , j as introduced in Def 2.3, Sect 1.2, r = 0 if i < j and r., = 1 If

c JL R 1 + c 2 R 2 + · · · + c ^ = [0, 0 , , 0 ] ,

examination of the j -th component on the left yields c = 0, then examination of the j9-th

column yields c = 0, and so forth Therefore

R , , R^ are linearly independent

T-5

T-6

T-7

T-8 AX = Y has a solution if and only if Y is a linear

combination of the columns of A

T-9 Follows from Cor 2.1, Sect 2.2

T-10 From Ex T-9,det A = 0 if and only if R(A) < n; now

X (d) = - X j/30 / 6

X-Y = a , ( X - Y ) + a1 1 I I m m 0 ( X -Y) + · · · + c (X -Y) = 0 ,

(a) In particular, take X = Y, which yields

(b) Χ·(Υ - Z) = 0 for all X in R Apply (a) to

in the same direction

T-8 |X + Y|2 = (X + Υ)·(Χ + Y) = |x|2 + |Y|2 + 2(Χ·Υ), from which the conclusion follows

T-9 Follows directly from Def 5.2

T-10 Follows directly from Def 5.2, and the definition of matrix multiplication

T-11 Take X = U - V in part (d) of Thm 5.1 to obtain (a) and (b) For (c), simply observe that |-x| = |x| For (d), let X = U - W and Y = W - V in Thm 5.3

L 2t

fio

1

ilo

T-l Suppose AXI = AXI and AX2 = AX2; then

A(XI + X2) = AXI + AX2 = A(XI + X2) and, if c is ascalar, then A(cXl ) = cAXl = cAXl = A(cXl ) Theconclusion follows

T-2 See Exercise T-20, Sect 1.4

T-3 Since Xl' ' X l' X are linearly dependent, thereJ- Jare constants a l , , aj _l , b, not all zero, suchthat

T-4 The remainder of the proof follows directly from

Eq (18): if λ φ 0, then (18) determines a

uniquely; if λ = 0, then (18) cannot be satisfied

Q ^ Q , then C = (Q λ ? λ ) A(PQ) = (PQ) λ A(PQ) ,

T-6 Let X 1" , X be the columns of P; then n

T-8 In the equation preceding Thm 6.12, observe that

P AP = diag [λ 1 " ν·

T-9 Let B = Ρ" AP, and note that I = P^IP Then

B - λΐ = P_1AP - λΡ_1ΙΡ = P_1(A - λΙ)Ρ; hence

det(B - λΐ) = (det P"1) det(A - λΐ) det P = det A -λΐ

Similar matrices have the same eigenvalues

T-10 If A is triangular, then A - XI is also triangular;

T-12 The constant term in ρ(λ) = det(A - XI) is, on the

one hand, equal to the product of the roots of ρ(λ)

(which are the eigenvalues of A), and > on the other

hand, equal to p(0) = det A

T-13 If I = ATA, then det I = det ATA = det AT det A

T Since det 1 = 1 and det A = det A, it follows that

2 (det A) = 1 ; hence det A = ±1

T - 1 4 I f A T A = B T B = I , t h e n (AB) T (AB) = B T A T AB = B T B = I ;

h e n c e AB i s o r t h o g o n a l

- I T T T T T T T T - 1

T - 1 5 (P AP) = ( P A P ) = P A (P ) = P A P = P AP

T-16 Let X-, , X be orthonormal eigenvectors of A, and 1 n

let P be the orthogonal matrix which has X , , X

as columns Then A = PDP , where D is diagonal

Now the result follows from Exercise T-15