antonela ism ch08

EXERCISE SET 8.1

2. Using the usual notation, we have

T(u + v) = (2(u1 + v1) (u2 + v2) + (u3 + v3), (u2 + v2) 4(u3 + v3))

= (2u1 u2 + u3, u2 4u3) + (2v1 v2 + v3, v2 4v3)

= T(u) + T(v)

and

T(cu) = (2cu1 cu2 + cu3, cu2 4cu3) = cT(u)

3. Since T (u) = u = u = T(u) T(u) unless u = 0, the function is not linear.

5. We observe that

T(A1 + A2) = (A1 + A2)B = A1B + A2B = T(A1) + T(A2)

and

T(cA) = (cA)B = c(AB) = cT(A)

Hence, T is linear.

6. Since T(A + B) = tr(A + B) = tr(A) + tr(B) = T(A) + T(B) and T(cA) = tr(cA) = c(tr(A))= cT(A), T is linear.

375

8. (b) Since both properties fail, T is nonlinear. For instance,

10. (a) T is not linear because it sends the zero function to f(x) = 1.

(b) Since

T(f(x) + g(x)) = T((f + g)(x)) = (f + g)(x + 1)

= f(x + 1) + g(x + 1) = T(f(x)) + T(g(x))

and

T(cf(x)) = cf(x + 1) = cT(f(x))

T is linear.

12. Let v = (x1, x2) = av1 + bv2 = a(1, 1) + b(1, 0). This yields the equations x1 = a + b and x2 = a, so that v = x2v1 + (x1 x2)v2. Thus

T(v) = T(x1, x2) = x2T(v1) + (x1 x2)T(v2)

= x2(1, 2) + (x1 x2)(4, 1)

= (4x1 + 5x2, x1 3x2)

so that

T(2, 3) = (8 15, 2 + 9) = (23, 11)

14. Let v = (x1, x2, x3) = av1 + bv2 + cv3 = a(1, 1, 1) + b(1, 1, 0) + c(1, 0, 0). Thus x1 = a + b+ c, x2 = a + b, and x3 = a, so that v = x3v1 + (x2 x3)v2 + (x1 x2)v3. Therefore

T ka b

c dk a k b

kTa b

c d

= +

2 2 2 2

in general.

376 Exercise Set 8.1

T(v) = T(x1, x2, x3) = x3T(v1) + (x2 x3)T(v2) + (x1 x2)T(v3)

= x3(2, 1, 4) + (x2 x3)(3, 0, 1) + (x1 x2)(1, 5, 1)

= (x1 + 4x2 x3, 5x1 5x2 x3, x1 + 3x3)

and

T(2, 4, 1) = (2 + 16 + 1, 10 20 + 1, 2 3) = (15, 9, 1)

16. We have

T(2v1 3v2 + 4v3) = 2T(v1) 3T(v2) + 4T(v3)

= (2, 2, 4) + (0, 9, 6) + (12, 4, 8)

= (10, 7, 6)

17. (a) Since T1 is dened on all of R2, the domain of T2 T1 is R

2. We have T2 T1(x, y) =T2(T1(x, y)) = T2(2x, 3y) = (2x 3y, 2x + 3y). Since the system of equations

2x 3y = a

2x + 3y = b

can be solved for all values of a and b, the codomain is also all of R2.

(d) Since T1 is dened on all of R2, the domain of T2 T1 is R

2. We have

T2(T1(x, y)) = T2(x y, y + z, x z) = (0, 2x)

Thus the codomain of T2 T1 is the y-axis.

18. (a) Since T1 is dened on all of R2, the domain of T3 T2 T1 is R

2. We have

T3 T2 T1(x, y) = T3 T2 (2y, 3x, x 2y)

= T3(3x, x 2y, 2y)

= (3x 2y, x)

This vector can assume all possible values in R2, so the codomain is R2.

Exercise Set 8.1 377

19. (a) We have

(b) Since the range of T1 is not contained in the domain of T2, T2 T1 is not well dened.

20. We have

(T1 T2)(p(x)) = T1(p(x + 1)) = p((x 1) + 1) = p(x)

and

(T2 T1)(p(x)) = T2(p(x 1)) = p((x + 1) 1) = p(x)

22. We have

T2 T1 (a0 + a1x + a2x2) = T2 (a0 + a1(x + 1) + a2(x + 1)

2)

= T2((a0 + a1 + a2) + (a1 + 2a2)x + a2 x2)

= (a0 + a1 + a2)x + (a1 + 2a2)x2 + a2x

3

24. (b) We have

(T3 T2) T1(v) = (T3 T2)(T1(v))

= T3(T2(T1(v)))

= T3 T2 T1(v)

25. Since (1, 0, 0) and (0, 1, 0) form an orthonormal basis for the xy-plane, we have T(x, y, z)= (x, 0, 0) + (0, y, 0) = (x, y, 0), which can also be arrived at by inspection. Then T(T(x,y, z)) = T(x, y, 0) = (x, y, 0) = T(x, y, z). This says that T leaves every point in the x-yplane unchanged.

( )( ) ( )T T A Aa c

b da dT1 2 = =

= +tr tr


26. (a) From the linearity of T and the denition of kT, we have

(kT)(u + v) = k(T(u + v)) = k(T(u) + T(v))

= kT(u) + kT(v) = (kT)(u) + (kT)(v)

and

(kT)(cu) = k(T(cu)) = kcT(u) = c(kT)(u)

Therefore, kT is linear.

28. (a) F is linear since

F((x1, y1) + (x2, y2)) = F(x1 + x2, y1 + y2)

= (a1(x1 + x2) + b1(y1 + y2), a2 (x1 + x2) + b2 (y1 + y2))

= (a1x1 + b1y1, a2x1 + b2y1) + (a1x2 + b1y2, a2x2 + b2y2)

= F(x1, y1) + F(x2, y2)

and

F(c(x, y)) = F(cx, cy)

= (a1cx + b1cy, a2cx + b2cy)

= c(a1x + b1y, a2x + b2y)

= cF(x, y)

(b) Since F(c(x, y)) = F(cx, cy) = c2F(x, y), F is not a linear operator.

30. Let v = a1v1 + + anvn be any vector in V. Then

T(v)= a1T(v1) + + anT(vn)

= a1v1 + + anvn

= v

Hence T is the identity transformation on V.


31. (b) We have

(c) We have

33. (a) True. Let c1 = c2 = 1 to establish Part (a) of the denition and let c2 = 0 to establishPart (b).

(b) False. All linear transformations have this property, and, for instance there is morethan one linear transformation from R2 to R2.

(c) True. If we let u = 0, then we have T(v) = T(v) = T(v). That is, T(v) = 0 for allvectors v in V. But there is only one linear transformation which maps every vector tothe zero vector.

(d) False. For this operator T, we have

T(v + v) = T(2v) = v0 + 2 v

But

T(v) + T(v) = 2T(v) = 2v0 + 2v

Since v0 0, these two expressions cannot be equal.

34. Since B is a basis for the vector space V, we can write any vector v in V in one and only oneway as a linear combination v = c1v1 + c2v2 + + cnvn of vectors from B. Thus for anylinear operator T, we have

T(v) = c1T(v1) + c2T(v2) + + cnT(vn)

That is, to dene T, we need only specify where it maps each of the basis vectors in B. IfT is to map B into itself, then we have n choices for each of the n vectors T(vi), and hencea total of nn possible operators T.

35. Yes. Let T Pn Pm be the given transformation, and let TR Rn+1 Rm+1 be the

corresponding linear transformation in the sense of Section 4.4. Let n Pn Rn+1 be the

function that maps a polynomial in Pn to its coordinate vector in Rn+1, and let m Pm

Rm+1 be the function that maps a polynomial in Pm to its coordinate vector in Rm+1.

( )( ) ( )J D e e dt ex x xx

+ = + = 3 3 10

( )(sin ) (sin ) sin( ) sin( ) sin( )J D x t dt x x = = =00

xx


By Example 7, both n and m are linear transformations. Theorem 5.4.1 implies that acoordinate map is invertible, so m

1 is also a linear transformation.

We have T=m1

TRn , so T is a composition of linear transformations. Refer to thediagram below:

Thus, by Theorem 8.1.2., T is itself a linear transformation.

T

TR

Pn

n m1Pm

Rm+1Rn+1


EXERCISE SET 8.2

1. (a) If (1, 4) is in R(T), then there must be a vector (x, y) such that T(x, y) = (2x y,8x + 4y) = (1, 4). If we equate components, we nd that 2x y = 1 or y = t and x= (1 + t)/2. Thus T maps innitely many vectors into (1, 4).

(b) Proceeding as above, we obtain the system of equations

2x y = 5

8x + 4y = 0

Since 2x y = 5 implies that 8x + 4y = 20, this system has no solution. Hence (5, 0) is not in R(T).

2. (a) The vector (5, 10) is in ker(T) since

T(5, 10) = (2(5) 10, 8(5) + 4(10)) = (0, 0)

(b) The vector (3, 2) is not in ker(T) since T(3, 2) = (4, 16) (0, 0).

3. (b) The vector (1, 3, 0) is in R(T) if and only if the following system of equations has asolution:

4x + y 2z 3w = 1

2x + y + z 4w = 3

6x 9z + 9w = 0

This system has innitely many solutions x = (3/2)(t 1), y = 10 4t, z = t, w = 1where t is arbitrary. Thus (1, 3, 0) is in R(T).

383

4. (b) Since T(0, 0, 0, 1) = (3, 4, 9) (0, 0, 0), the vector (0, 0, 0, 1) is not in ker(T).

5. (a) Since T(x2) = x3 0, the polynomial x2 is not in ker(T).

6. (a) Since T(1 + x) = x + x2, the polynomial x + x2 is in R(T).

(c) Since there doesnt exist a polynomial p(x) such that xp(x) = 3 x2, then 3 x2 is notin R(T).

7. (a) We look for conditions on x and y such that 2x y = and 8x + 4y = 0. Since theseequations are satised if and only if y = 2x, the kernel will be spanned by the vector(1, 2), which is then a basis.

(c) Since the only vector which is mapped to zero is the zero vector, the kernel is {0} andhas dimension zero so the basis is the empty set.

8. (a) Since 8x + 4y = 4(2x y) and 2x y can assume any real value, the range of T isthe set of vectors (x, 4x) or the line y = 4x. The vector (1, 4) is a basis for thisspace.

(c) The range of T is just the set of all polynomials in P3 with constant term zero. The set{x, x2, x3} is a basis for this space.

9. (a) Here n = dim(R2) = 2, rank(T) = 1 by the result of Exercise 8(a), and nullity(T) = 1by Exercise 7(a). Recall that 1 + 1 = 2.

(c) Here n = dim(P2) = 3, rank(T) = 3 by virtue of Exercise 8(c), and nullity(T) = 0 byExercise 7(c). Thus we have 3 = 3 + 0.

10. (a) We know from Example 1 that the range of T is the column space of A. Usingelementary column operations, we reduce to the matrix

Thus and as well as and form a basis for the column space

and therefore also for the range of T.

0

1

1

1

0

2

0

1

1

1

5

7

1 0 0

5 1 0

7 1 0

1 0 0

0 1 0

2 1 0

or


(b) To investigate the solution space of A, we look for conditions on x, y, and z such that

The solution of the resulting system of equations is x = 14t, y = 19t, and z = 11t

where t is arbitrary. Thus one basis for ker(T) is the vector .

(c) By Parts (a) and (b), we have rank(T) = 2 and nullity(T) = 1.

(d) Since rank(A) = 2 and A has 3 columns, nullity (A) = 3 2 = 1. This also follows fromTheorem 8.2.2 and Part (c) above.

12. (a) Since the range of T is the column space of A, we reduce using column operations.This yields

Thus and form a basis for the range of T.

(b) If

then x = s 4r, y = s + 2r, z = s, and w = 7r where s and r are arbitrary. Thus,

That is, the set {[1, 1, 1, 0]T, [4, 2, 0, 7]T} is one basis for ker(T).

x

y

z

w

s

=

+

1

1

1

0

4

2

0

7

r

4 1 5 2

1 2 3 0

0

0

=

x

y

z

w

0

1

1

0

1 0 0 0

0 1 0 0

14

19

11

1 1 3

5 6 4

7 4 2

x

y

z =

0

0

0


12. (c) By Parts (a) and (b), rank(T) = nullity(T) = 2.

(d) Since A has 4 columns and rank(A) = 2, we have nullity(A) = 4 2 = 2. This alsofollows from Theorem 8.2.2 and Part (c) above.

14. (a) The orthogonal projection on the xz-plane maps R3 to the entire xz-plane. Hence itsrange is the xz-plane. It maps only the y-axis to (0, 0, 0), so that its nullspace is they-axis.

16. (a) The nullity of T is 5 3 = 2.

(c) The nullity of T is 6 3 = 3.

18. (a) The dimension of the solution space of Ax = 0 is, by virtue of the Dimension Theorem,7 4 = 3.

(b) Since the range space of A has dimension 4, the range cannot be R5, which hasdimension 5.

19. By Theorem 8.2.1, the kernel of T is a subspace of R3. Since the only subspaces of R3 arethe origin, a line through the origin, a plane through the origin, or R3 itself, the resultfollows. It is clear that all of these possibilities can actually occur.

21. (a) If

then x = t, y = t, z = t. These are parametric equations for a line through the origin.

(b) Using elementary column operations, we reduce the given matrix to

1 0 0

3 5 0

2 8 0

1 3 4

3 4 7

2 2 0

0

0

0

=

x

y

z


Thus, (1, 3, 2)T and (0, 5, 8)T form a basis for the range. That range, which we caninterpret as a subspace of R3, is a plane through the origin. To nd a normal to thatplane, we compute

(1, 3, 2) (0, 5, 8) = (14, 8, 5)

Therefore, an equation for the plane is

14x 8y 5z = 0

Alternatively, but more painfully, we can use elementary row operations to reducethe matrix

to the matrix

Thus the vector (x, y, z) is in the range of T if and only if 14x 8y 5z = 0.

22. Suppose that v is any vector in V and write

(*) v = a1v1 + a2 v2 + + anvn

Now dene a transformation T V W by

T(v) = a1w1 + a2w2 + + anwn

Note that T is well-denied because, by Theorem 5.4.1, the constants a1, a2, , an in (*)are unique.

In order to complete the problem, you must show that (i) T is linear and (ii) T(vi) = wi for i = 1, 2, , n.

23. The rank of T is at most 1, since dimR = 1 and the image of T is a subspace of R. So, weknow that either rank(T) = 0 or rank(T) = 1. If rank(T) = 0, then every matrix A is in thekernel of T, so every n n matrix A has diagonal entries that sum to zero. This is clearlyfalse, so we must have that rank(T) = 1. Thus, by the Dimension Theorem (Theorem 8.2.3),dim (ker(T)) = n2 1.

1 0 1 4 3 5

0 1 1 3 5

0 0 0 14 8 5

+( )( )

x y

x y

x y z

1 3 4

3 4 7

2 2 0

x

y

z


24. (a) Suppose that {v1, v2, , vn} is a basis for V and let

(*) v = a1v1 + a2v2 + + anvn

be an arbitrary vector in V. Since T is linear, we have

(**) T(v) = a1T(v1) + + anT(vn)

The hypothesis that dim(ker(T)) = 0 implies that T(v) = 0 if and only if v = 0. Since{v1, , vn} is a linearly independent set, by (*), v = 0 if and only if

a1 = a2 = = an = 0

Thus, by (**), T(v) = 0 if and only if

a1 = a2 = = an = 0

That is, the vectors T(v1), T(v2), , T(vn) form a linearly independent set in R(T).Since they also span R(T), then dim(R(T)) = n.

26. If J(p)= = -2b, then the kernel of J is the set of all

p(x) = ax + b for which b = 0 except the y-axis.

27. If f(x) is in the kernel of D D, then f (x) = 0 or f(x) = ax + b. Since these are the onlyeligible functions f(x) for which f (x) = 0 (Why?), the kernel of D D is the set of allfunctions f(x) = ax + b, or all straight lines in the plane. Similarly, the kernel of D D Dis the set of all functions f(x) = ax2 + bx + c, or all straight lines except the y-axis andcertain parabolas in the plane.

28. (a) kernel, range

(b) (1, 1, 1)

(c) the dimension of V

(d) 3

29. (a) Since the range of T has dimension 3 minus the nullity of T, then the range of T hasdimension 2. Therefore it is a plane through the origin.

( ) ( / )ax b dx ax bx+ = +

211 11

2


(b) As in Part (a), if the range of T has dimension 2, then the kernel must have dimension1. Hence, it is a line through the origin.

30. (a) If T(f(x)) = f (x), then T(ax3 + bx2 + cx + d) = 0 for all polynomials in P3. Moreover,these are the only functions which are mapped to the zero vector, that is, the functionwhich is identically zero.

(b) If T transforms f(x) to its (n + 1)st derivative, then kernel (T) = Pn.


EXERCISE SET 8.3

1. (a) Clearly ker(T) = {(0, 0)}, so T is one-to-one.

(c) Since T(x, y) = (0, 0) if and only if x = y and x = y, the kernel is {0, 0} and T is one-to-one.

(e) Here T(x, y) = (0, 0, 0) if and only if x and y satisfy the equations x y = 0, x + y= 0, and 2x 2y = 0. That is, (x, y) is in ker(T) if and only if x = y, so the kernel ofT is this line and T is not one-to-one.

2. (a) Since det(A) = 1 0, then T has an inverse. By direct calculation

and so

3. (a) Since det(A) = 0, or equivalently, rank(A) < 3, T has no inverse.

Tx

xA

x

x

x x

x

=

=

+1 1

2

1 1

2

1 2

1

2

2 55 2x

A =

1 1 2

2 5

391

3. (c) Since A is invertible, we have

4. (a) Since the kernel of this transformation is the line x = 2y, the transformation is not one-to-one.

(c) Since the kernel of the transformation is {0, 0}, the transformation is one-to-one.

5. (a) The kernel of T is the line y = x since all points on this line (and only those points)map to the origin.

(b) Since the kernel is not {0, 0}, the transformation is not one-to-one.

7. (b) Since nullity(T) = n rank(T) = 1, T is not one-to-one.

(c) Here T cannot be one-to-one since rank(T) n < m, so nullity(T) 1.

8. (b) If p(x) = a0 + a1x + a2x2, then

T(p(x)) = a0 + a1 (x + 1) + a2(x + 1)2

= (a0 + a1 + a2) + (a1 + 2a2)x + a2x2

Thus T(p(x)) = 0 if and only if a2 = a1 + 2 a2 = a0 + a1 + a2 = 0, or a2 = a1 = a0 = 0.That is, the nullity of T is zero so T is one-to-one.

T

x

x

x

A

x

x

x

=

=1

1

2

3

11

2

3

1

2

1

2

1

21

2

1

2

1

21

2

1

2

1

2

x

x

x

1

2

3

=

+1

2 1 2x x x33

1 2 3

1 2 3

1

21

2

( ) + +( )

+ ( )

x x x

x x x


10. (a) Since innitely many vectors (0, 0, , 0, xn) map to the zero vector, T is not one-to-one.

(c) Here T is one-to-one and T1 (x1, x2, , xn) = (xn, x1, , xn1).

11. (a) We know that T will have an inverse if and only if its kernel is the zero vector, whichmeans if and only if none of the numbers ai = 0.

12. Note that T1 and T2 can be represented by the matrices

that is,

(a) Since det(AT1) = 2 0 and det(AT2) = 5 0, it follows that both T1 and T2 are one-to-one.

(b) The transformations T11 and T2

1 are represented by AT11 and AT2

1, respectively, where

Then T2 T1 can be represented by

and (T2 T1)1 can be represented by

A AT T2 11 1

10

3 1

1 3( ) =

A AT T2 13 1

1 3=

A AT T1 21 11

2

1 1

1 1

1

5

2 1

1 2

=

=

and

Tx

yA

x

yT

x

yA

x

yT T1 21 2

=

=and

A AT T1 21 1

1 1

2 1

1 2=

=

and


that is

12. (c) It is easy to verify that

AT11 AT2

1 = (AT2 AT1)1

Alternative Solution: If you want to do this the hard way, notice that if T1(x, y) =

(x + y, x y), then T11(x, y) = and if T2(x, y) = (2x + y, x 2y)

then T21(x, y) = . Also T2 T1(x, y) = T2(x + y, x y) =

(3x + y, x + 3y) and (T1 T2)1(x, y) = .

13. (a) By inspection, T11(p(x)) = p(x)/x, where p(x) must, of course, be in the range of T1

and hence have constant term zero. Similarly T21(p(x)) = p(x 1), where, again, p(x)

must be in the range of T2. Therefore (T2 T1)1(p(x)) = p(x 1)/x for appropriate

polynomials p(x).

16. If there were such a transformation T, it would have nullity zero (because it would be one-to-one) and rank equal to the dimension of W (because its range would be W). But theDimension Theorem guarantees that rank(T) + nullity(T) = dim(V), which is, byhypothesis, greater than dim(V). Thus there can be no such transformation.

17. (a) Since T sends the nonzero matrix to the zero matrix, it is not one-to-one.

(c) Since T sends only the zero matrix to the zero matrix, it is one-to-one. By inspection,T1(A) = T(A).

Alternative Solution: T can be represented by the matrix

0 1

0 0

3

10

3

10

x y x y +

,

2

5

2

5

x y x y+

,

x y x y+

2 2,

( )T Tx

y

x

y2 11 1

10

3 1

1 3

=

=

+

1

10

3

3

x y

x y


By direct calculation, TB = (TB)1, so T = T1.

18. If T(x, y) = (x + ky, y) = (0, 0), then x = y = 0. Hence ker(T) = {(0, 0)} and therefore Tis one-to-one. Since T(T(x, y)) = T(x + ky, y) = (x + ky + k(y), (y)) = (x, y), we haveT1 = T.

19. Suppose that w1 and w2 are in R(T). We must show that

T1(w1 + w2) = T1(w1) + T

1(w2)

and

T1(kw1) = kT1(w1)

Because T is one-to-one, the above equalities will hold if and only if the results of applyingT to both sides are indeed valid equalities. This follows immediately from the linearity of thetransformation T.

21. It is easy to show that T is linear. However, T is not one-to-one, since, for instance, it sendsthe function f(x) = x 5 to the zero vector.

24. (a) False. Since T is not one-to-one, it has no inverse. For T1 to exist, T must map eachpoint in its domain to a unique point in its range.

(b) True. If T1(v1) = T1(v2) where v1 v2, then T2 T1(v1) = T2 T1(v2) where v1 v2.

(c) True, since such rotations and reections are one-to-one linear transformations.

25. Yes. The transformation is linear and only (0, 0, 0) maps to the zero polynomial. Clearlydistinct triples in R3 map to distinct polynomials in P2.

TB =

0 0 0 1

0 1 0 0

0 0 1 0

1 0 0 0


26. It does. T is a linear operator on M22. If EA = O for any matrix A, then A = E1O = O. Thus

T is one-to-one. Alternatively, if EA1 = EA2, then E(A1 A2) = O, so A1 = A2 = E1O = O.

That is, A1 = A2.

27. No. T is a linear operator by Theorem 3.4.2. However, it is not one-to-one since T(a) = a a = 0 = T(0). That is, T maps a to the zero vector, so if T is one-to-one, a must be thezero vector. But then T would be the zero transformation, which is certainly not one-to-one.


EXERCISE SET 8.4

2. (a) Let B = {1, x, x2} = {u1, u2, u3} and B = {1, x} = {v1, v2}. Observe that

T(u1) = T(1) = 1 = 1v1

T(u2) = T(x) = 1 2x = 1v1 2v2

T(u3) = T(x2) = 3x = 3v2

Hence the matrix of T with respect to B and B is

(b) We have

and

[T(c0 + c1x + c2x2)]B = [(c0 + c1) (2c1 + 3c2)x]B

=

+

c c

c c0 1

1 22 3

[ ] [ ]',T

c

c

cB B Bx =

=

1 1 0

0 2 3

0

1

2

c c

c c0 1

1 22 3

+

1 1 0

0 2 3

397

4. (a) Since

and

we have

(b) Since

and

Formula (5a) does, indeed, hold.

6. (a) Observe that

T(v1) = (1, 1, 0) = v1 v2

T

T

vv vv vv vv

vv

2 1 2 3

3

1 1 13

2

1

2

1

2

0 0 1

( ) = ( ) = + +( ) = (

, ,

, , )) = + 12

1

2

1

21 2 3vv vv vv

Tx x

x xx x x

BB

x( )[ ] = +

= +( ) +1 21 2

1 2 1 2 22uu uu

[ ] [ ]Tx

xB BB

x =

=

2 1

2 0

2 1

1

2

22 0

2 1

2

1 2 2 1 2

( ) + +( )

=

x xB

u u u

00 22

2 1

1 2

2

=

+

xx x

x x

x

[ ]T B =

2 1

2 0

T( )u u2 11

1=

=

T( )uu uu uu1 1 20

22 2=

= +


Thus the matrix of T with respect to B is

(b) If we solve for (x1, x2, x3)B = a1v1 + a2v2 + a3v3, we nd that

Thus we have

and

[T(x)]B = [(x1 x2, x2 x1, x1 x3)]B

=

+

+ +

( ) ( ) ( )

( ) ( )

x x x x x x

x x x x

1 2 2 1 1 3

1 2 2 1

2(( )

( ) ( ) ( )

x x

x x x x x x

1 3

1 2 2 1 1 3

2

2

+

=

+

+

3 2

22

2

2

1 2 3

1 2 3

1 3

x x x

x x x

x x

[ ] [ ]T B Bx =

13

2

1

2

11

2

1

2

01

2

1

2

+

+ +

+

x x x

x x x

x x

1 2 3

1 2 3

1 2

2

2

=

+

+

x

x x x

x x

3

1 2 3

1

2

3 2

22 22 3

1 3

2

2

+

+

x

x x

x = + + + +

B

x x x x x x x x x1 2 3 1 2 3 1 2 32 2 2

T

13

2

1

2

11

2

1

2

01

2

1

2


6. (c) No. The transformation T is not invertible because x1 x2 = (x2 x1) and thusrank(T) = 2 3. We can see this because the reduced row-echelon form of the matrix[T]B of the transformation is

8. (a) Since

T(1) = x

T(x) = x(x 3) = 3x + x2

T(x2) = x(x 3)2 = 9x 6x2 + x3

we have

(b) Since we are working with standard bases, we have

so that T(1 + x x2) = 11x + 7x2 x3.

(c) We have T(1 + x x2) = x(1 + (x 3) (x 3)2) = 11x + 7x2 x3.

9. (a) Since A is the matrix of T with respect to B, then we know that the rst and secondcolumns of A must be [T(v1)]B and [T(v2)]B, respectively. That is

T

T

B

B

( )

( )

vv

vv

1

2

1

2

3

5

=

=

[ ] [ ],T B B B =

x

0 0 0

1 3 9

0 1 6

0 0 1

1

1

1

0

11

7

1

=

[ ] ,T B B =

0 0 0

1 3 9

0 1 6

0 0 1

1 0 1

0 1 1

0 0 0

.I


Alternatively, since v1 = 1v1 + 0v2 and v2 = 0v1 + 1v2, we have

and

(b) From Part (a),

and

(c) Since we already know T(v1) and T(v2), all we have to do is express [x1 x2]T in terms

of v1 and v2. If

then

x1 = a b

x2 = 3a + 4b

or

a = (4x1 + x2)/7

b = (3x1 + x2)/7

Thus

x

x

x x1

2

1 24

7

3

5

3

=

+

+ xx x

x x

x x

1 2

1 2

1 2

7

2

29

18

7107 24

7

+

=

+

+

x

xa b a b1

21 2

1

3

1

4

= + =

+

vv vv

T( )vv vv vv2 1 23 52

29= + =

T( )vv vv vv1 1 223

5= =

( )T AB

vv20

1

3

5 =

=

T AB

( )vv1 =

=

1

0

1

2


9. (d) By the above formula,

11. (a) The columns of A, by denition, are [T(v1)]B, [T(v2)]B, and [T(v3)]B, respectively.

(b) From Part (a),

T(v1) = v1 + 2v2 + 6v3 = 16 + 51x + 19x2

T(v2) = 3v1 2v3 = 6 5x + 5x2

T(v3) = v1 + 5v2 + 4v3 = 7 + 40x + 15x2

(c) Let a0 + a1x + a2x2 = b0v1 + b1v2 + b2v3. Then

a0 = b1 + 3b2

a1 = 3b0 + 3b1 + 7b2

a2 = 3b0 + 2b1 + 2b2

This system of equations has the solution

b0 = (a0 a1 + 2a2)/3

b1 = (5a0 + 3a1 3a2)/8

b2 = (a0 + a1 a2)/8

Thus

T(a0 + a1x + a2x2) = b0T(v1) + b1T(v2) + b2T(v3)

=

+

+ +

+

239 161 247

24

201 111 247

8

0 1 2

0 1 2

a a a

a a ax

61 31 107

120 1 2 2a a a x

+

T1

1

19 7

83 7

=


(d) By the above formula,

T(1 + x2) = 2 + 56x + 14x2

13. (a) Since

T1(1) = 2 and T1(x) = 3x2

T2(1) = 3x T2(x) = 3x2 and T2(x2) = 3x

3

T2 T1(1) = 6x and T2 T1(x) = 9x3

we have

and

(b) We observe that here

[T2 T1]B,B = [T2]B,B [T1]B,B

T TB B2 1

0 0

6 0

0 0

0 9

,

=

[ ] [ ], ,T TB B B B1 2

2 0

0 0

0 3

0 0 0

3 0

=

=

00

0 3 0

0 0 3


15. If T is a contraction or a dilation of V, then T maps any basis B = {v1, , vn} of V to{kv1, , kvn} where k is a nonzero constant. Therefore the matrix of T with respect to B is

17. The standard matrix for T is just the m n matrix whose columns are the transforms of thestandard basis vectors. But since B is indeed the standard basis for Rn, the matrices are thesame. Moreover, since B is the standard basis for Rm, the resulting transformation will yieldvector components relative to the standard basis, rather than to some other basis.

18. (a) Since D(1) = 0, D(x) = 1, and D(x2) = 2x, then

is the matrix of D with respect to B.

(b) Since D(2) = 0, D(2 3x) = 3 = 32

p1 and D(2 3x + 8x2) = 3 + 16x = 23

6p1

166

p2,

the matrix of D with respect to B is

03

2

23

6

0 016

3

0 0 0

0 1 0

0 0 2

0 0 0

k

k

k

k

0 0 0

0 0 0

0 0 0

0 0 0


(c) Using the matrix of Part (a), we obtain

(d) Since 6 6x + 24x2 = p1 p2 + 3p3, we have

or

D(6 6x + 24x2) = 13(2) 16(2 3x) = 6 + 48x

19. (c) Since D(f1) = 2f1, D(f2) = f1 + 2f2, and D(f3) = 2f2 + 2f3, we have the matrix

20. The upper left-hand corner is all vectors in the space V. The upper right-hand corner is allvectors in the range of T. Thus it is a subspace of W. Since V is a real vector space, thelower left-hand corner is all of R4. The lower right-hand corner is the range of thetransformation obtained by multiplying every element of R4 by the matrix [T]B,B. It is thusa subspace of R7.

2 1 0

0 2 2

0 0 2

D x xB

( )6 6 24

03

2

23

6

0 016

30 0 0

2 + =

=

1

1

3

13

166

0

B

D x x( )6 6 24

0 1 0

0 0 2

0 0 0

6

6

24

2 + =

=

= +

6

48

0

6 48x


EXERCISE SET 8.5

1. First, we nd the matrix of T with respect to B. Since

and

then

In order to nd P, we note that v1 = 2u1 + u2 and v2 = 3u1 + 4u2. Hence the transitionmatrix from B to B is

Thus

P =

1

4

11

3

111

11

2

11

P =

2 3

1 4

A TB

= =

1 2

0 1

T( )uu2 =

2

1

T( )uu11

0=

407

and therefore

2. In order to compute A = [T]B, we note that

and

Hence

In order to nd P, we note that

v1 = (1.3)u1 + (0.4)u2

v2 = (0.5)u1

Hence

P =

. .

.

1 3 0 5

0 4 0

A TB

= =

0 8 6 1

3 6 3 8

. .

. .

T( ) ( . ) ( . )uu uu uu2 1 23

166 1 3 8=

= +

T( ) ( . ) ( . )uu uu uu1 1 216

20 8 3 6=

= +

= = =

A T P T PB B[ ] [ ]1 1

11

4 3

1 2

1 2

0 1

=

2 3

1 4

3

11

56

112

11

3

11


and

It then follows that

3. Since T(u1) = (1/ 2, 1/ 2) and T(u2) = (1/ 2, 1/ 2), then the matrix of T withrespect to B is

From Exercise 1, we know that

Thus

5. Since T(e1) = (1, 0, 0), T(e2) = (0, 1, 0), and T(e3) = (0, 0, 0), we have

In order to compute P, we note that v1 = e1, v2 = e1 + e2, and v3 = e1 + e2 + e3. Hence,

A TB

= =

1 0 0

0 1 0

0 0 0

A T P APB

= = =

'1 1

11 2

13 25

5 9

P P=

=

2 3

1 4

1

11

4 3

1 21and

A TB

= =

1 2 1 2

1 2 1 2

A T P APB

= = =

1 15 5 4 5

37 5 12 5

. .

. .

P =

1 0 2 5

2 6 5

.

.


and

Thus

7. Since

and

we have

We note that qq pp pp qq pp pp1 1 2 2 1 22

9

1

3

7

9

1

6= + = .and Hence

TB

=

2

3

2

91

2

4

3

T x( )pp pp pp2 1 212 22

9

4

3= + = +

T x( )pp pp pp1 1 29 32

3

1

2= + = +

TB

=

1 1 0

0 1 1

0 0 1

1 0 0

0 11 0

0 0 0

1 1 1

0 1 1

0 0 1

1 0 0

0

= 11 1

0 0 0

P =

11 1 0

0 1 1

0 0 1

P =

1 1 1

0 1 1

0 0 1


and

Therefore

8. (a) Since T(1, 0) = (3, 1) and T(0, 1) = (4, 7), we have

(c) Since T(1) = 1, T(x) = x 1, and T(x2) = (x 1)2, we have

9. (a) If A and C are similar n n matrices, then there exists an invertible n n matrix Psuch that A = P1CP. We can interpret P as being the transition matrix from a basis Bfor Rn to a basis B. Moreover, C induces a linear transformation T Rn Rn where C= [T]B. Hence A = [T]B. Thus A and C are matrices for the same transformation withrespect to different bases. But from Theorem 8.2.2, we know that the rank of T isequal to the rank of C and hence to the rank of A.

det( )T =

=

1 1 1

0 1 2

0 0 1

1

det( )T =

=

3 4

1 717

TB

=

3

4

7

23

21

2

3

2

91

2

4

3

=

2

9

7

91

3

1

6

1 1

0 1

P =

1

3

4

7

23

21

P =

2

9

7

91

3

1

6


Alternate Solution: We observe that if P is an invertible n n matrix, then Prepresents a linear transformation of Rn onto Rn. Thus the rank of the transformationrepresented by the matrix CP is the same as that of C. Since P1 is also invertible, itsnull space contains only the zero vector, and hence the rank of the transformationrepresented by the matrix P1 CP is also the same as that of C. Thus the ranks of Aand C are equal. Again we use the result of Theorem 8.2.2 to equate the rank of alinear transformation with the rank of a matrix which represents it.

Second Alternative: Since the assertion that similar matrices have the same rank dealsonly with matrices and not with transformations, we outline a proof which involvesonly matrices. If A = P1 CP, then P1 and P can be expressed as products ofelementary matrices. But multiplication of the matrix C by an elementary matrix isequivalent to performing an elementary row or column operation on C. From Section5.5, we know that such operations do not change the rank of C. Thus A and C musthave the same rank.

10. (a) We use the standard basis for P4. Since T(1) = 1, T(x) = 2x + 1, T(x2) = (2x + 1)2,

T(x3) = (2x + 1)3, and T(x4) = (2x + 1)4, we have

(b) Thus rank(T) = 5, or [T] is invertible, so, by Theorem 8.4.3, T is one-to-one.

11. (a) The matrix for T relative to the standard basis B is

The eigenvalues of [T]B are = 2 and = 3, while corresponding eigenvectors are(1, 1) and (1, 2), respectively. If we let

P P=

=

1 1

1 2

2 1

1 1then 1

TB

=

1 1

2 4

det( )T =

1 1 1 1 1

0 2 4 6 8

0 0 4 12 24

0 0 0 8 32

0 0 0 0 16

1024=


and

is diagonal. Since P represents the transition matrix from the basis B to the standardbasis B, we have

as a basis which produces a diagonal matrix for [T]B.

12. (b) The matrix for T relative to the standard basis B is

The eigenvalues of [T]B are = 1 and = 2. The eigenspace corresponding to = 1

is spanned by the vectors and and the eigenspace corresponding to = 2

is spanned by the vector . If we let

then P1[T]BP will be diagonal. Hence the basis

will produce a diagonal matrix for [T]B.

B =

1

1

0

1

0

1

1

, , 11

1

P =

1 1 1

1 0 1

0 1 1

1

1

1

1

0

1

1

1

0

TB

=

0 1 1

1 0 1

1 1 0

B = ,1

1

1

2

P T PB

=

1 2 0

0 3[ ]


13. (a) The matrix of T with respect to the standard basis for P2 is

The characteristic equation of A is

3 22 15 + 36 = ( 3)2( + 4) = 0

and the eigenvalues are therefore = 4 and = 3.

(b) If we set = 4, then (I A)x = 0 becomes

The augmented matrix reduces to

and hence x1 = 2s, x2 = 83s, and x3 = s. Therefore the vector

is a basis for the eigenspace associated with = 4. In P2, this vector represents thepolynomial 2 + 8

3x + x2.

If we set = 3 and carry out the above procedure, we nd that x1 = 5 s, x2 = 2s,and x3 = s. Thus the polynomial 5 2x + x

2 is a basis for the eigenspace associatedwith = 3.

2

8 3

1

1 0 2 0

0 1 8 3 0

0 0 0 0

9 6 2

0 3 8

1 0 2

1

2

3

x

x

x

=

0

0

0

A =

5 6 2

0 1 8

1 0 2


14. (a) We look for values of such that

or

If we equate corresponding entries, we nd that

a 2c = 0

a + b c = 0(*)

b + ( + 2)c = 0

( 1)d= 0

This system of equations has a nontrivial solution for a, b, c, and d only if

or

( 1)( + 2)(2 1) = 0

Therefore the eigenvalues are = 1, = 2, and = 1.

(b) We find a basis only for the eigenspace of T associated with = 1. The basesassociated with the other eigenvalues are found in a similar way. If = 1, the equationT(x) = x becomes T(x) = x and the augmented matrix for the system of equations(*) above is

1 0 2 0 0

1 1 1 0 0

0 1 3 0 0

0 0 0 0 0

det

0 2 0

1 1 0

0 1 2 0

0 0 0 1

+

= 0

2

2

c a c

b c d

a b

c d

+

=

Ta b

c d

a b

c d

=


This reduces to

and hence a = 2t, b = 3t, c = t, and d = s. Therefore the matrices

form a basis for the eigenspace associated with = 1.

15. If v is an eigenvector of T corresponding to , then v is a nonzero vector which satisesthe equation T(v) = v or (I T)v = 0. Thus I T maps v to 0, or v is in the kernel ofI T.

17. Since C[x]B = D[x]B for all x in V, we can, in particular, let x = vi for each of the basisvectors v1, , vn of V. Since [vi]B = ei for each i where {e1, , en} is the standard basis forRn, this yields Cei = Dei for i = 1, , n. But Cei and Dei are just the i

th columns of C andD, respectively. Since corresponding columns of C and D are all equal, we have C = D.

18. Let B be the standard basis for R2, and let

B = {(cos , sin ), (sin , cos )} = {v1, v2}

be the basis consisting of the unit vector v1 lying along and the unit vector v2perpendicular to . Note that to change basis from B to B, we rotate through an angle .Now relative to B, T is just the orthogonal projection on the v1-axis, which is accomplishedby setting the v2-coordinate equal to zero. Then to return to the basis B, we rotate throughan angle . Thus

2 3

1 0

0 0

0 1

and

1 0 2 0 0

0 1 3 0 0

0 0 0 0 0

0 0 0 0 0


19. (a) False. Every matrix is similar to itself, since A = I1 AI.

(b) True. Suppose that A = P1 BP and B = Q1 CQ. Then

A = P1(Q1 CQ)P = (P1Q1)C(QP) = (QP)1C(QP)

Therefore A and C are similar.

(c) True. By Table 1, A is invertible if and only if B is invertible, which guarantees that Ais singular if and only if B is singular.

Alternatively, if A = P1 BP, then B = PAP1. Thus, if B is singular, then so is A.Otherwise, B would be the product of 3 invertible matrices.

(d) True. If A = P1 BP, then A1=(P1 BP)1 = P1 B1(P1)1 = P1B1P, so A1 and B1 aresimilar.

20. If A is invertible and B is singular, then it follows from Exercise 19(c)that A and B cannotbe similar. Hence

are not similar.

Moreover, if A and B are both invertible but have different determinants, then it followsfrom Table 1 that they cannot be similar. Hence.

are not similar.

A B=

=

1 0

0 1

2 0

0 1and

A B=

=

1 0

0 1

1 0

0 0and

Tx

yT T I

x

yB B B B B

=

[ ] [ ] [ ], ,

=

cos sin

sin cos

1 0

0 0

cos sin

sin cos

x

y

=

cos sin cos

sin cos sin

2

2

x

y


22. Let B = P1 AP so that A = PBP1. Let be an eigenvalue common to A and B and let x bean eigenvector of A corresponding to . Then Ax = x, or

Ax = (PBP1)x = x

so that

BP1x = P1 x

Thus

B(P1x) = (P1x)

That is, P1x is an eigenvector of B corresponding to .

24. For every f in V, T(f) = f(x 2). Thus, if f is an eigenvector with eigenvalue = 1, wehave T(f) = 1 f, so f(x 2) = f(x) for all x. Thus, the eigenspace of = 1 is precisely theset of all functions f that are periodic with periods that evenly divide 2.

25. First, we need to prove that for any square matrices A and B, the trace satises tr(A) =tr(B). Let

Then,

[ ]AB a b a b a b a b an n jj

11 11 11 12 21 13 31 1 1 11

= + + + + ==

nn

j

n n

b

AB a b a b a b a b

= + + + + =

1

22 21 12 22 22 23 32 2 2[ ] aa b

AB a b a b a b

jj

n

j

nn n n n n n n

21

2

1 1 2 2 3 3

=

= + + + +

[ ] aa b a bnn nn njj

n

jn=

=

1

.

A

a a a

a a a

a a a

n

n

n n nn

=

11 12 121 22 2

1 2

=

and

b11

B

b b

b b bn

n

12 1

21 22 2

b b bn n nn1 2


Thus,

tr(AB) = [AB]11 + [AB]22 + + [AB]nn

Reversing the order of summation and the order of multiplication, we have

Now, we show that the trace is a similarity invariant. Let B = P1 AP. Then

tr(B) = tr(P1 AP)

= tr((P1 A)P)

= tr(P(P1 A))

= tr(PP1)A)

= tr(I A)

= tr(A).

tr( )AB b a

b a b

jkk

n

jkj

n

kk

n

k kk

n

=

= +

==

= =

11

11

1 21

+ += + + +

=

=

a b a

BA BA BA

k nkk

n

kn

nn

21

11 22

[ ] [ ] [ ]

ttr( ).BA

= + + +

=

= = =

a b a b a b

a

jj

n

j j jj

n

nj jnj

n

k

11

1 2 21 1

jjj

n

kjk

n

b==

11

.


EXERCISE SET 8.6

1. (a) This transformation is onto because for any ordered pair (a, b) in R2, T(b, a) = (a, b).

(b) We use a counterexample to show that this transformation is not onto. Since there isno pair (x, y) that satises T(x, y) = (1, 0), T is not onto.

(c) This transformation is onto. For any ordered pair (a, b) in R2, T = (a, b).

(d) This is not an onto transformation. For example, there is nopair (x, y) that satises T(x, y) = (1, 1, 0).

(e) The image of this transformation is all vectors in R3 of the form (a, a, 2a). Thus, theimage of T is a one-dimensional subspace of R3 and cannot be all of R3. In particular,there is no vector (x, y) that satises T(x, y) = (1, 1, 0), and this transformation is notonto.

(f) This is an onto transformation. For any point (a, b) in R2, there are an innite number

of points that map to it. One such example is T = (a, b).

2. The transformation TA is not onto if the image of TA is not all of Rn. Thus, rank(TA) < n, or

equivalently, null(T) > 0. This is equivalent to the matrix A being singular.

3. (a) We nd that rank(A) = 2, so the image of T is a two-dimensional subspace of R3. Thus,T is not onto.

(b) We nd that rank(A) = 3, so the image of T is all of R3. Thus, T is not onto.

(c) We nd that rank(A) = 3, so the image of T is all of R3. Thus, T is onto.

(d) We nd that rank(A) = 3, so the image of T is all of R3. Thus, T is onto.

4. (a) We nd that rank(A) = 1 and the codomain is R3, so this is not onto.

(b) We nd that rank(A) = 3 and the codomain is R3, so this is an onto transformation.

(c) We nd that rank(A) = 2 and the codomain is R3, so this is not onto.

a b a b+

2 2 0, ,

a b a b+

2 2,

421

5. (a) The transformation T is not a bijection because it is not onto. There is no p(x) inP2(x) so that xp(x) = 1.

(b) The transformation T(A) = AT is one-to-one, onto, and linear, so it is a bijection.

(c) By Theorem 8.6.1, there is no bijection between R4 and R3, so T cannot be a bijection. Inparticular, it fails being one-to-one. As an example, T(1, 1, 2, 2) = T(1, 1, 0, 0) = (1, 1, 0).

(d) Because dim P3 = 4 and dim R3 = 3, Theorem 8.6.1 states that there is no bijection

between P3 and R3, so T cannot be a bijection. In particular, it fails being one-to-one.

As an example, T(x + x2 + x3) = T(1 + x + x2 + x3) = (1, 1, 1).

6. Let T V W be a bijection. Then T is one-to-one, onto, and invertible.First, we show that T1 W V is one-to-one. If T1(w1) = T

1(w2), then T(T1(w1)) =

T(T1(w2)) so w1 = w2, and T1 is one-to-one. Second, we show that T1 is onto. Let v be any

vector in V, and let w = T(v). Then T1(w) = T1(T(v)) = v, so v is in the image of T1 forany v in V, and T1 is onto. Thus, the inverse of a bijection is again a bijection.

Now, we want to show that if T V W is a bijective linear transformation, thenT1 W V is also a bijective linear transformation.

Because T is onto, for any w in W, there is a vector v in V so that w = T(v). Then,

T1(cw) = T1(cT(v)).

Since T is a linear transformation, cT(v) = T(cv), so

T1(cw) = T1(T(cv))

= cv

= cT1(w).

Also, for any two vectors w1 and w2 in W, there exist vectors v1 and v2 in V so that w1 =T(v1) and w2 = T(v2), since T is onto. Thus,

T1(w1 + w2) = T1(v1) + T(v2)).

Since T is a linear transformation, T(v1) + T(v2) = T(v1 + v2), and we have

T1(w1 + w2) = T1(v1 + v2))

= v1 + v2

= T1(w1) + T1(w2).

Thus, T1 is a linear transformation.

7. Assume there exists a surjective (onto) linear transformation T V W, where dim W >dim V. Let m = dim V and n = dim W, with m < n. Then, the matrix AT of thetransformation is an n m matrix, with m < n. The maximal rank of AT is m, so thedimension of the image of T is at most m. Since the dimension of the image of T is smallerthan the dimension of the codomain Rn, T is not onto. Thus, there cannot be a surjectivetransformation from V onto W if dim V < dim W.


If n = dim W dim V = m, then the matrix AT of the transformation is an n m matrixwith maximim possible rank n. If rank(AT) = n, then T is a surjective transformation.

Thus, it is only possible for T V W to be a surjective linear transformation if dim W dim V.

8. Let V be the vector of all symmetric 3 3 matrices, and let T V R6 be dened by

Clearly, T is one-to-one and onto, so T is a bijection. Because T is the coordinate mapfrom V to R6, it is a linear transformation (refer to Example 7 in Section 8.1). Thus, T is abijective linear transformation.

9. Let T V Rn be dened by T(v) = (v)S, where S = {u1, u2, , un} is a basis of V. Weknow from Example 7 in Section 8.1 that the coordinate map is a linear transformation.Let (a1, a2, , an) be any point in R

n. Then, for the vector v = a1u1 + a2u2 + + anun, wehave

T(v) = T(a1u1 + a2u2 + + anun) = (a1, a2, , an)

so T is onto.

Also, let v1 = a1u1 + a2u2 + + anun and v2 = b1u1 + b2u2 + + bnun. If T(v1) =T(v2), then (v1)S = (v2)S, and thus (a1, a2, , an) = (b1, b2, , bn). It follows that a1 = b1,a2 = b2, , an = bn, and thus

v1 = a1u1 + a2u2 + + anun = v2.

So, T is one-to-one and is thus an isomorphism.

10. T1 and T2 are both bijective linear transformations, we know from Theorem 8.1.2 thatT1 T2 is also a linear transformation. Let T1 V W and T2 U V. Because T1 is onto,we know that for any w in W, there exists a v in V so that T1(v) = w. Similarly, because T2is onto, we know that for any v in V, there exists a u so that T2(u) = v. Then

(T1 T2)(u) = T1(T2(u)) = T1(v) = w,

and T1 T2 is onto.

If there are two vectors u1 and u2 so that (T1 T2)(u1) = (T1 T2)(u2), then T1(T2(u1))= T1(T2(u2)). Since T1 is one-to-one, this means that T2(u1) = T2(u2), and since T2 is one-to-one, this means that u1 = u2. Thus, the composition T1 T2 is one-to-one.

Thus, the composition of two bijective linear transformations is again a bijective lineartransformation.

T

a b c

b d e

c e f

a

b

c

d

e

f

=


11. Let V = Span{1, sin x, cos x, sin 2x, cos 2x}. Differentiation is a linear transformation (seeExample 11, Section 8.1). In this case, D maps functions in V into other functions in V. Toconstruct the matrix of the linear transformation with respect to the basis B = {1, sin x,cos x, sin 2x, cos 2x}, we look at coordinate vectors of the derivatives of the basis vectors:

D(1) = 0 D(sin x) = cos x D(cos x) = sin x D(sin 2x) = 2 cos 2xD(cos 2x) = 2 sin 2x

The coordinate matrices are:

Thus, the matrix of the transformation is

Then, differentiation of a function in V can be accomplished by matrix multiplication by theformula

[D(f)]B = AD[f]B.

The nal vector, once transformed back to V from coordinates in R5, will be the desiredderivative.

For example,

Thus, D(3 4 sin x + sin 2 x + 5 cos 2x) = 4 cos x 10 sin 2x + 2 cos 2x.

12. (a) We have

uu uu uu uu uu uuV V W WT T T= = =, ( ), ( ) ( )/ /1 2 1 2

[ ( sin sin cos )]D x x x AB D3 4 2 5 2

3

4

+ + =

0

1

5

0 0 0 0 0

0 0 1 0

=

0

0 1 0 0 0

0 0 0 0 2

0 0 0 2 0

=

3

4

0

1

5

0

.

0

4

10

2

AD =

0 0 0 0 0

0 0 1 0 0

0 1 0 0 0

0 0 0 0 2

0 0 00 2 0

[ ( )] [ (sin )]D D xB B1

0

0

0

0

0

0

0

1=

=

00

0

0

1

0

0

=

[ (cos )]D x B

00

2

0

0

0

0

2

=

[ (sin )]D x B

=

[ (cos )]D x B2

0

0

0

2

0

.


so an inner product space isomorphism preserves lengths. The angle V between uand v in V is given by

and the angle W between T(u) and T(v) in W is given by

Thus, because the inner product space isomorphism preserves lengths,

and the inner product space isomorphism also preserves angles.

(b) If B = {v1, v2, , v2} is an orthonormal set in V, then

Thus, the set {T(v1, T(v2), , T(vn)} = is an orthonormal set in W.

However, in general this is not true. Consider the linear transformation T R2 R2 dened by

This transformation is a change of basis transformation, and is thus one-to-one andonto, but the vectors T(e1) and T(e2) are not orthonormal. In fact, T(e2) = 2, andT(e1), T(e2) = 1 0. Thus, length and angles are not preserved with thistransformation.

(c) The set {e1, e2, , en} is an orthonormal basis of W. By the Gram-Schmidt process, wecan construct an orthonormal basis {v1, v2, , vn} of V. Dene a linear transformationT V W by

T(vj) = ej, j = 1, 2, , n.

T e T e( ) ( )1 21

0

1

1=

=

and

( ( ), ( )) ( , ) T T i ji ji j i jv v v v

if1 if

= =

=

0

cos( ), ( )

( ) ( )

,cW

W V

T T

T T= = =

uu vv

uu vv

uu vv

uu vvW V

VW

oos .V

cos( ), ( )

( ) ( ).W W

W W

T T

T T=

uu vv

uu vv

cos,

V VV V

=

( )uu vvuu vv


Let u = a1v1 + a2v2 + + anvn and v = b1v1 + b2v2 + + bnvn. Then

T(u) = a1T(v1) + a2T(v2) + + anT(vn)

= a1e1 + a2e2 + + anen

Similarly,

T(v) = b1T(v1) + b2T(v2) + + bnT(vn)

= b1e1 + b2e2 + + bnen

So, if T(u) = T(v), then and a1 = b1, a2 = b2, , an = bn. Thus, u = v

and T is one-to-one.

For any vector in W, we have

T(c1v1 + c2v2 + + cnvn) = = w.

so T is onto.

c

c

cn

1

2

w =

c

c

cn

1

2

a

a

a

b

b

bn n

1

2

1

2

=

=

b

b

bn

1

2

=

a

a

an

1

2


We now need to show that T preserves the inner product. We have

u, vV = a1v1 + a2v2 + + anvn, b1v1 + b2v2 + + bnvn

= ajbk vj, vk.

Recall that if j k, then vj, vk = 0. We then haveu, vV = a1b1 v1, v1 + a2b2 v2, v2 + + anbn vn, vn.

Now recall that vj, vj = 1. Thenu, v = a1b1 + a2b2 + + anbn

We also have

T(u), T(v)W = a1e1 + a2e2 + + anen, b1e1 + b2e2 + + bnen

[b1 b2 bn]

= a1b1 + a2b2 + + anbn.

Thus, T(u), T(v)W = u, vV and T is an inner product space isomorphism.

(d) Let dim V = dim W = n. Then by Part (c), there is an inner product isomorphism T1 V Rn and another inner product isomorphism T2 W R

n. Then T21 Rn W is

an isomorphism by Problem 6. And, since

u, v = T2(u), T2(v)Rn,we rename x = T2(u) and y = T2(v) and we have

T21(x), T21(y)W = x, yRn.Thus, T2

1 Rn W is another inner product space isomorphism. Then, thecomposition (T2

1 T1) V W is an isomorphism by Problem 10. We have

u, vV = T1(u), T1(v)Rn= T21(T1(u)), T21(T1(v))W.= (T21 T1)(u), (T21 T1)(v)W.

so T21 T1 is an inner product isomorphism from V to W.

=

a

a

an

1

2

k

n

k

n

==

11


(e) An orthonormal basis for P5 is {1, x, x2, x3, x4, x5}, using the inner product

a0 + a1x + a1x2 + + a5x5, b0 + b1x + b2x2 + + b5x5P5

= a0b0 + a1b1 + + a5b5.

An orthonormal basis for M22 is

using the inner product

A, BM23

= tr(AT B).

We can dene a linear transformation T P5 M23 by

The transformation T is one-to-one and onto. To see that is an inner productisomorphism, we need to compute u, vP5 and T(u), T(v)M23. Let u = a0 + a1x +a2x

2 + + a5x5, and let v = b0 + b1x + b2x

2 + + b5x5. Then we have

u, vP5

= a0b0 + a1b1 + + a5b5

and

Thus, u, vP5

= T(u),T(v)M23

, and T is an inner product space isomorphism.

T Ta a a

a a a

b b bM

( ), ( ) ,uu vv23

0 1 2

3 4 5

0 1=

22

3 4 5

0 0 1 1 5 5

b b b

a b a b a b

= + + + + .

T T x T11 0 0

0 0 0

0 1 0

0 0 0( ) =

( ) =

, , xx

T x T

2

3

0 0 1

0 0 0

0 0 0

1 0 0

( ) = ( ) =

,

, xx T x4 50 0 0

0 1 0

0 0 0

0 0 1( ) = ( ) = , .

1 0 0

0 0 0

0 1 0

0 0 0

0 0 1

0 0 0

0

, , ,00 0

1 0 0

0 0 0

0 1 0

0 0 0

0 0 1

, ,

,


SUPPLEMENTARY EXERCISES 8

3. By the properties of an inner product, we have

T(v + w) = v + w, v0v0= (v, v0 + w, v0)v0= v, v0v0 + w, v0v0= T(v) + T(w)

and

T(kv) = kv, v0v0 = kv, v0v0 = kT(v)

Thus T is a linear operator on V.

4. (a) By direct computation, we have

T(x + y) = ((x + y) v1, , (x + y) vm)

= (x v1 + y v1, , x vm + y vm)

= (x v1, , x vm) + (y v1, , y vm)

= T(x) + T(y)

and

T(kx) = ((kx) v1, , (kx) vm)

= (k(x v1), , k(x vm))

= kT(x)

429

4. (b) If {e1, , en} is the standard basis for Rn, and if vi = (a1i, a2i, , ani) for i = 1, , m,

then

T(ej) = (aj1, aj2, , ajm)

But T(ej), interpreted as a column vector, is just the jth column of the standard matrix

for T. Thus the ith row of this matrix is (a1i, a2i, , ani), which is just vi.

5. (a) The matrix for T with respect to the standard basis is

We rst look for a basis for the range of T; that is, for the space of vectors B such thatAx = b. If we solve the system of equations

x + z + w = b1

2x + y + 3x + w = b2

x + w = b3

we nd that = z = b1 b3 and that any one of x, y, or w will determine the other two.Thus, T(e3) and any two of the remaining three columns of A is a basis for R(T).

Alternate Solution :We can use the method of Section 5.5 to find a basis for thecolumn space of A by reducing AT to row-echelon form. This yields

so that the three vectors

form a basis for the column space of T and hence for its range.

1

2

1

0

1

0

0

0

1

1 1

0 1 0

0 0 1

0 0 0

2

A =

1 0 1 1

2 1 3 1

1 0 0 1

430 Supplementary Exercises 8

Second Alternative: Note that since rank(A) = 3, then R(T) is a 3-dimensionalsubspace of R3 and hence is all of R3. Thus the standard basis for R3 is also a basis forR(T).

To nd a basis for the kernel of T, we consider the solution space of Ax = 0. If we setb1 = b2 = b3 = 0 in the above system of equations, we nd that z = 0, x = w, and y =w. Thus the vector (1, 1, 0, 1) forms a basis for the kernel.

7. (a) We know that T can be thought of as multiplication by the matrix

where reduction to row-echelon form easily shows that rank([T]B) = 2. Therefore therank of T is 2 and the nullity of T is 4 2 = 2.

(b) Since [T]B is not invertible, T is not one-to-one.

9. (a) If A = P1 BP, then

AT = (P1 BP)T

= PT BT (P1)T

= ((PT)1)1 BT (P1)T

= ((P1)T)T B (P1)T

Therefore AT and BT are similar. You should verify that if P is invertible, then so is PT

and that (PT)1 = (P1)T.

10. If statement (i) holds, then the range of T is V and hence rank(T) = n. Therefore, by theDimension Theorem, the nullity of T is zero, so that statement (ii) cannot hold. Thus (i)and (ii) cannot hold simultaneously.

Now if statement (i) does not hold, then the range of T is a proper subspace of V andrank(T) < n. The Dimension Theorem then implies that the nullity of T is greater thanzero. Thus statement (ii) must hold. Hence exactly one of the two statements must alwayshold.

[ ]T B =

1 1 2 21 1 4 6

1 2 5 6

3 2 3 2

Supplementary Exercises 8 431

11. If we let then we have

The matrix X is in the kernel of T if and only if T(X) = 0, i.e., if and only if

a + b + c = 0

2b + d = 0

d = 0

Hence

The space of all such matrices X is spanned by the matrix and therefore has

dimension 1. Thus the nullity is 1. Since the dimension of M22 is 4, the rank of T musttherefore be 3.

Alternate Solution. Using the computations done above, we have that the matrix for thistransformation with respect to the standard basis in M22 is

Since this matrix has rank 3, the rank of T is 3, and therefore the nullity must be 1.

1 1 1 0

0 2 0 1

0 0 0 1

0 0 0 1

,1 0

1 0

Xa

a=

0

0

Ta b

c d

a c b d b b

d d

=

+ +

+

0 0

=

+ + +

a b c b d

d d

2

Xa b

c d=

,


12. We are given that there exist invertible matrices P and Q such that A = P1 BP and B = Q1

CQ. Therefore

A = P1(Q1 CQ)P = (QP)1 C(QP)

That is, A and C are similar.

13. The standard basis for M22 is the set of matrices

If we think of the above matrices as the vectors

[1 0 0 0]T, [0 1 0 0]T, [0 0 1 0]T, [0 0 0 1]T

then L takes these vectors to

[1 0 0 0]T, [0 0 1 0]T, [0 1 0 0]T, [0 0 0 1]T

Therefore the desired matrix for L is

14. (a) Reading directly from P, we have

v1 = 2u1 + u2

v2 = u1 + u2 + u3

v3 = 3u1 + 4u2 + 2u3

1 0 0 0

0 0 1 0

0 1 0 0

0 0 0 1

1 0

0 0

0 1

0 0

0 0

1 0

0 0

0 1

, , ,


14. (b) Since

by direct calculation we have

u1 = 2v1 2v2 + v3

u2 = 5v1 + 4v2 2v3

u3 = 7v1 5v2 + 3v3

15. The transition matrix P from B to B is

Therefore, by Theorem 8.5.2, we have

Alternate Solution: We compute the above result more directly. It is easy to show that u1= v1, u2 = v1 + v2, and u3 = v2 + v3. So

T(v1) = T(u1) = 3u1 + u2 = 4v1 + v2

T(v2) = T(u1 + u2) = T(u1) + T(u2)

= u1 + u2 + u3 = v3

T(v3) = T(u1 + u2 + u3) = T(u1) + T(u2) + T(u3)

= 8u1 u2 + u3

= 9v1 2v2 + v3

[ ] [ ]T P T PB B

= =

1

4 0 9

1 0 2

0 1 1

P =

1 1 1

0 1 1

0 0 1

P =

12 5 7

2 4 5

1 2 3


16. Let and solve for a, b, c, and d so that

This yields the system of equations

a b = 2a + c

a + 4b = 2b + d

c d = a + 3c

c + 4d = b + 3c

which has solution b = a, c = 0, and d = a. Thus, for instance,

will work.

However, if we try the same procedure on the other pair of matrices, we nd that a =3d, b = d, and c = 3d, so that det(P) = 3d2 3d2 = 0. Thus P is not invertible.

Alternatively, if there exists an invertible matrix P such that

then we have a matrix with zero determinant on the left, and one with nonzero determinanton the right. Why? Thus the two matrices cannot be similar.

17. Since

T ,

1

0

0

1

0

1

=

T

0

1

0

1

1

0

=

, and T

0

0

1

=

1

0

1

3 1

6 2

1 2

1 01

= P P

P P=

=

1 1

0 1

1 1

0 11and

1 1

1 4

2 1

1 3

1 1

1 41

=

P P Por

=

2 1

1 3P

Pa b

c d=


we have

In fact, this result can be read directly from [T(X)]B.

18. We know that det(T) 0 if and only if the matrix of T relative to any basis B has nonzerodeterminant; that is, if and only if every such matrix is invertible. Choose a particular basisB for V and let V have dimension n. Then the matrix [T]B represents T as a matrixtransformation from Rn to Rn. By Theorem 4.3.1, such a transformation is one-to-one if andonly if [T]B is invertible. Thus T is one-to-one if and only if det(T) 0.

19. (a) Recall that D(f + g) = (f(x) + g(x)) = f (x) + g(x) and D(cf) = (cf(x)) = cf (x).

(b) Recall that D(f) = 0 if and only if f(x) = a for some constant a if and only if f(x) = ax+ b for constants a and b. Since the functions f(x) = x and g(x) = 1 are linearlyindependent, they form a basis for the kernel of D.

(c) Since D(f) = f(x) if and only if f (x) = f(x) if and only if f(x) = aex + bex for a andb arbitrary constants, the functions f(x) = ex and g(x) = ex span the set of all suchfunctions. This is clearly a subspace of C2 (, ) (Why?), and to show that it hasdimension 2, we need only check that ex and ex are linearly independent functions.To this end, suppose that there exist constants c1 and c2 such that c1e

x + c2ex = 0. If

we let x = 0 and x = 1, we obtain the equations c1 + c2 = 0 and c1e + c2e1 = 0. These

imply that c1 = c2 = 0, so ex and ex are linearly independent.

21. (a) We have

and

T kp x

kp x

kp x

kp x

k( ( ))

( )

( )

( )

=

=

1

2

3

( )

( )

( )

( ( ))

p x

p x

p x

kT p x1

2

3

=

T p x q x

p x q x

p x q x( ( ) ( ))

( ) ( )

( ) ( )+ =

+

+1 1

2 2

pp x q x

p x

p x

p x( ) ( )

( )

( )

( )3 3

1

2

3+

=

+

=

( )

( )

( )

(

q x

q x

q x

T1

2

3

pp x T q x( )) ( ( ))+

[ ]T B =

1 1 1

0 1 0

1 0 1


(b) Since T is dened for quadratic polynomials only, and the numbers x1, x2, and x3 aredistinct, we can have p(x1) = p(x2) = p(x3) = 0 if and only if p is the zero polynomial.(Why?) Thus ker(T) = {0}, so T is one-to-one.

(c) We have

T(a1P1(x) + a2P2(x) + a3P3(x)) = a1T(P1(x)) + a2T(P2(x)) + a3T(P3(x))

(d) From the above calculations, we see that the points must lie on the curve.

23. Since

then

where the above vectors all have n + 1 components. Thus the matrix of D with respect toB is

[ ( )]( , , )

( , , , , ) , ,D x

k

k kk

B ==

=

0 0 0

0 0 1 2

if

if , n

kth component

D xk

kx k nk

k( )

, , ,=

=

=

0 0

1 21if

if

=

+

+

a a a1 2 3

1

0

0

0

1

0

0

0

1

=

a

a

a

1

2

3


24. Call the basis B and the vectors v0, v1, , vn. Notice that D(vi) = vi1 for i = 1, , n,while D(v0) = 0. That is,

where the above vectors all have n + 1 components. Thus the matrix of D with respect toB is

In fact, the differentiation operator maps Pn to Pn1.

25. Let Bn and Bn+1 denote the bases for Pn and Pn+1, respectively. Since

J(xk) = xk+1

k + 1

for k = 0, , n

0 1 0 0

0

0 0 1 0 0

0 0 0 1 0

0 0 0 0 1

0 0 0 0 0

[ ( )]( , , )

( , , , , )D

k

kk Bv

if

if=

=

=

0 0 0

0 1 0 1 ,, , ,2 n

kth componeent

0 1 0 0

0

0 0 2 0 0

0 0 0 3 0

0 0 0 0

0 0 0 0 0

n


we have

where [xk]Bn= [0, , 1, , 0]T with the entry 1 as the (k + 1)st component out of a total

of n + 1 components. Thus the matrix of J with respect to Bn+1 is

with n + 2 rows and n + 1 columns.

0 0 0 0

1 0 0 0

0 1 2 0 0

0 0 1 3 0

0 0 0 1 1( )

n +

[ ( )] , , , , (J x Bk

nk n+ = +

+1 0

1

10 2 )components

(

+k 22)nd component


antonela ism ch08

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