ISM: Linear Algebra
Section 4.1
Chapter 4 4.1
1. Not a subspace since it does not contain the neutral element, that is, the function f (t) = 0, for all t. 2. This subset V is a subspace of P2 : • The neutral element f (t) = 0 (for all t) is in V . • If f and g are in V (so that f (2) = g(2) = 0), then (f + g)(2) = f (2) + g(2) = 0 + 0 = 0, so that f + g is in V . • If f is in V (so that f (2) = 0), and k is any constant, then (kf )(2) = kf (2) = 0, so that kf is in V . A polynomial f (t) = a + bt + ct2 is in V if f (2) = a + 2b + 4c = 0, or a = −2b − 4c. The general element of V is of the form f (t) = (−2b − 4c) + bt + ct2 = b(t − 2) + c(t2 − 4), so that t − 2, t2 − 4 is a basis of V . 3. This subset V is a subspace of P2 : • The neutral element f (t) = 0 (for all t) is in V since f (1) = f (2) = 0. • If f and g are in V (so that f (1) = f (2) and g (1) = g(2)), then (f + g) (1) = (f + g )(1) = f (1) + g (1) = f (2) + g(2) = (f + g)(2), so that f + g is in V . • If f is in V (so that f (1) = f (2)) and k is any constant, then (kf ) (1) = (kf )(1) = kf (1) = kf (2) = (kf )(2), so that kf is in V . If f (t) = a + bt + ct2 then f (t) = b + 2ct, and f is in V if f (1) = b + 2c = a + 2b + 4c = f (2), or a + b + 2c = 0. The general element of V is of the form f (t) = (−b − 2c) + bt + ct2 = b(t − 1) + c(t2 − 2), so that t − 1, t2 − 2 is a basis of V . 4. This subset V is a subspace of P2 :
1
• The neutral element f (t) = 0 (for all t) is in V since
1 1
0 dt = 0.
0 1 1 1
• If f and g are in V that f + g is in V .
so that
0
f=
0
g=0
then
0
(f + g) =
0
f+
0
g = 0, so
179
Chapter 4
1 1
ISM: Linear Algebra
1
• If f is in V kf is in V .
so that
0
f =0
and k is any constant, then
0
kf = k
0
f = 0, so that
1
If f (t) = a + bt + ct2 then
0
b c f (t)dt = at + t2 + t3 2 3 b c − − 2 3
1
= a+
0
b c b c + = 0 if a = − − . 2 3 2 3 1 2 + c t2 − 1 , so 3
The general element of V is f (t) = 1 1 that t − , t2 − is a basis of V . 2 3
+ bt + ct2 = b t −
5. If p(t) = a + bt + ct2 then p(−t) = a − bt + ct2 and −p(−t) = −a + bt − ct2 . Comparing coefficients we see that p(t) = −p(−t) for all t if (and only if) a = c = 0. The general element of the subset is of the form p(t) = bt. These polynomials form a subspace of P2 , with basis t. 6. Not a subspace, since I3 and −I3 are invertible, but their sum is not. 7. The set V of diagonal 3 × 3 matrices is a subspace of R3×3 : 0 0 0 a. The zero matrix 0 0 0 is in V , 0 0 0
p 0 0 a 0 0 b. If A = 0 b 0 and B = 0 q 0 are in V , then so is their sum 0 0 r 0 0 c a+p 0 0 b+q 0 . A+B = 0 0 0 c+r ka a 0 0 c. If A = 0 b 0 is in V , then so is kA = 0 0 0 0 c
0 0 kb 0 , for all constants k. 0 kc
8. This is a subspace; the justification is analogous to Exercise 7. 180
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Section 4.1
9. Not a subspace; consider multiplication with a negative scalar. I3 belongs to the set, but −I3 doesn’t. 1 10. Let v = 2 . Let V be the set of all 3 × 3 matrices A such that Av = 0. Then V is a 3 subspace of R3×3 :
a. The zero matrix 0 is in V , since 0v = 0. b. If A and B are in V , then so is A + B, since (A + B)v = Av + Bv = 0 + 0 = 0. c. If A is in V , then so is kA for all scalars k, since (kA)v = k(Av) = k 0 = 0.
11. Not a subspace: I3 is in rref, but the scalar multiple 2I3 isn’t. 12. Yes, the set W of all arithmetic sequences is a subspace of V . Use the fact that a sequence (x0 , x1 , x2 , . . .) is arithmetic if xn = x0 + kn for some constant k.
• The sequence (0, 0, 0, . . .) is an arithmetic sequence, with k = 0. • If (xn ) and (yn ) are arithmetic sequences (with xn = x0 + pn and yn = y0 + qn), then xn + yn = x0 + y0 + (p + q)n, so that (xn + yn ) is an arithmetic sequence as well. • If (xn ) is an arithmetic sequence (with xn = x0 + pn) and k is an arbitrary constant, then kxn = kx0 + (kp)n, so that (kxn ) is an arithmetic sequence as well. 13. Not a subspace: (1, 2, 4, 8, . . .) and (1, 1, 1, 1, . . .) are both geometric sequences, but their sum (2, 3, 5, 9, . . .) is not, since the ratios of consecutive terms fail to be equal, for example, 5 3 2 = 3. 14. Yes
• (0, 0, 0, . . . , 0, . . .) converges to 0. • If limn→∞ xn = 0 and limn→∞ yn = 0, then limn→∞ (xn + yn ) = limn→∞ xn + limn→∞ yn = 0. • If limn→∞ xn = 0 and k is any constant, then limn→∞ (kxn ) = k limn→∞ xn = 0. 15. The set W of all square-summable sequences is a subspace of V : 181
Chapter 4 • The sequence (0, 0, 0, . . .) is in W .
ISM: Linear Algebra
2 • Suppose (xn ) and (yn ) are in W . Note that the inequality (xn + yn )2 ≤ 2x2 + 2yn holds n 2 2 2 2 2 2 for all n, since 2xn + 2yn − (xn + yn ) = xn + yn − 2xn yn = (xn − yn ) ≥ 0. Thus ∞ ∞ ∞ 2 2 2 n=1 (xn + yn ) ≤ 2 n=1 xn + 2 n=1 yn converges, so that the sequence (xn + yn ) is in W as well.
• If (xn ) is in W so that k, since
∞ n=1
x2 converges n
∞ n=1
, then (kxn ) is in W as well, for any constant
∞ n=1
(kxn )2 = k 2
x2 n
will converge.
a b 1 0 0 1 0 0 0 0 0 16. c d = a 0 0 + b 0 0 + c 1 0 + d 0 1 + e 0 e f 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 The matrices 0 0 , 0 0 , 1 0 , 0 1 , 0 0 , 0 0 1 0 0 0 0 0 0 0 0 0 R3×2 , so that dim(R3×2 ) = 6.
0 0 0 + f 0 0 0 0 0 form a 1
0 0 1 basis of
17. Let Eij be the n × m matrix with a 1 as its ijth entry, and zeros everywhere else. Any A in Rn×m can be written as the sum of all aij Eij , and the Eij are linearly independent, so that they form a basis of Rn×m . Thus dim(Rn×m ) = nm. 18. Any f in Pn can be written as a linear combination of 1, t, t2 , . . . , tn , by definition of Pn . Also, 1, t, . . . , tn are linearly independent; to see this consider a relation c0 + c1 t + · · · + cn tn = 0; since the polynomial c0 + c1 t + · · · + cn tn has more than n zeros, we must have c0 = c1 = · · · = cn = 0, as claimed. Thus, dim(Pn ) = n + 1. 19. 0 0 i 1 a + bi +d +c +b =a i 1 0 0 c + di The vectors dim(C2 ) = 4. 20. We use Summary 4.1.6. We have a = −d, so that the general element of the subspace is −1 0 0 0 0 1 −d b . +d +c =b 0 1 1 0 0 0 c d 182 0 0 i 1 , , , i 1 0 0 form a basis of C2 as a real linear space, so that
ISM: Linear Algebra 0 1 0 0 −1 0 , , 0 0 1 0 0 1
Section 4.1
Thus
is a basis of the subspace; the dimensions is 3.
21. Use Summary 4.1.6. The general element of the subspace is a 0 1 0 0 0 =a +d . Thus 0 d 0 0 0 1 dimension is 2. 1 0 0 0 , 0 0 0 1 is a basis of the subspace; the
22. Using Exercise 21 as a guide, we find the basis E11 , E22 , . . . , Enn , where Eii is the n × n matrix with all 0 entries, except for a 1 at the ith place on the diagonal. The dimension of this space is n. 23. Proceeding as in Exercise 21, we find the basis is 3. 24. Proceeding as in Exercise 21, we find the basis E11 , E12 , E13 , E22 , E23 , E33 . Here Eij is the 3 × 3 matrix with all 0 entries, except for a 1 in the ith component of the jth column; the dimension is 6. 25. A polynomial f (t) = a + bt + ct2 is in this subspace if f (1) = a + b + c = 0, or a = −b − c. The polynomials in the subspace are of the form f (t) = (−b − c) + bt + ct2 = b(t − 1) + c(t2 − 1), so that t − 1, t2 − 1 is a basis of the subspace, whose dimension is 2. 26. Denote the subspace by W . A polynomial f (t) = a + bt + ct2 + dt3 is in W if f (1) = a + b + c + d = 0 and c d b f (t)dt = at + t2 + t3 + t4 2 3 4 −1 The system 1 −3c 2 − c −d . 3 c d a+b+c+d 2a +
2 3c 1 1 −1
1 0 0 0 0 0 , , ; the dimension 0 0 1 0 0 1
2 = 2a + c = 0. 3
1 a + 3c
=0 =0
reduces to
=0 =0
b+
2 3c
+d
, with general solution
1 The polynomials in W are of the form f (t) = − 3 c− 3 d(t − t), so that
2 3c
2 + d t+ct2 +dt3 = c t2 − 3 t −
1 3
+
t2 − 2 t − 1 , t3 − t is a basis of W , and dim(W ) = 2. 3 3 183
Chapter 4 a c a 2c a 0 is 2. b d a c b d 1 0 0 2 a c 2b 2d
ISM: Linear Algebra 1 0 0 2 a c b d
27.
is in the subspace if
=
equals
=
b , which is the case if b = c = 0. The matrices in the subspace are of the form 2d 0 1 0 0 0 1 0 0 0 =a +d , so that , is a basis, and the dimension d 0 0 0 1 0 0 0 1 equals 1 1 0 1 a c b = d
28.
a a+b 1 1 a b a b = is in the subspace if c c+d 0 1 c d c d a+c b+d which is the case if c = 0 and a = d. c d The matrices in the subspace are of the form 1 0 0 1 , 0 1 0 0 a b 0 a =a
0 1 1 0 , so that +b 0 0 0 1
is a basis, and the dimension is 2. a c a b such that c d b d
29. We are looking for the matrices 0 0 0 0 −1 b 0
a+b a+b 1 1 = = c+d c+d 1 1 −b b . It is required that a = −b and c = −d. The general element is = −d d 1 0 0 −1 1 0 0 +d . Thus , is a basis, and the dimension is 2. 0 −1 1 0 0 −1 1 a c b 1 2 such that d 3 6 a c b a + 2c b + 2d = = d 3a + 6c 3b + 6d
30. We are looking for the matrices
0 0 . It is required that a = −2c and b = −2d. Thus the general element is 0 0 −2c −2d −2 0 0 −2 =c +d . c d 1 0 0 1 Thus −2 0 0 −2 , 1 0 0 1 is a basis, and the dimension is 2. a c b d such that 0 1 1 0 a c a b = c d b d 1 0 , 0 −1
31. We are looking for the matrices or, c a d a = b c
−b . It is required that a = c and b = −d. −d
0 −1 1 0 c −d . Thus +d =c 0 1 1 0 c d basis, and the dimension is 2. The general element is 184
0 −1 1 0 is a , 0 1 1 0
ISM: Linear Algebra a c b d 1 1 1 1 a c b a = d c b d
Section 4.1 2 0 , 0 0
32. We are looking for the matrices or,
such that
a+c b+d 2a 0 = . It is required that a = c and b = −d. a+c b+d 2c 0
a b 1 0 0 1 1 0 0 1 =a +b . Thus , is a a −b 1 0 0 −1 1 0 0 −1 basis, and the dimension is 2. The general element is 33. Let S = a c b . Then d 1 1 1 1 a c a b = c d b , meaning d
a b a+c b+d . So a + c = a, b + d = b, a + c = c and b + d = d. These imply, = c d a+c b+d respectively, that c = 0, d = 0, a = 0 and b = 0. Thus, S can only equal 34. Let S = a c 0 0 , and the basis is ∅. 0 0 3 2 4 5 a c b a = d c b , meaning d
b . We want d
a b 3a + 2c 3b + 2d . So 3a + 2c = a, 3b + 2d = b, 4a + 5c = c and 4b + 5d = d. = c d 4a + 5c 4b + 5d These imply that a = −c and b = −d. 0 1 1 0 0 1 1 0 a b , . Thus +b =a 0 −1 −1 0 0 −1 −1 0 −a −b is a basis, and the dimension is 2. a b c 35. Let A = d e f . We want AB = BA, or g h i a b c 2 0 0 2 0 0 a b c d e f 0 3 0 = 0 3 0 d e f , or g h i 0 0 4 0 0 4 g h i 2a 3b 4c 2a 2b 2c 2d 3e 4f = 3d 3e 3f . 2g 3h 4i 4g 4h 4i So the general element is We note that b, c, d, f, g, h must be zero, but a, e, and i are chosen freely. So, our space, a 0 0 V , consists of all matrices of the form 0 e 0 0 0 i 185
Chapter 4
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1 0 0 0 0 0 0 0 0 = a 0 0 0 + e 0 1 0 + i 0 0 0 . 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 Thus, 0 0 0 , 0 1 0 , 0 0 0 is a basis of V , and dim(V ) = 3. 0 0 0 0 0 0 0 0 1 a b c 36. Let A = d e f . We want AB = BA, or g h i a b c 2 0 0 2 0 0 a b c d e f 0 3 0 = 0 3 0 d e f , or g h i 0 0 3 0 0 3 g h i 2a 2b 2c 2a 3b 3c 2d 3e 3f = 3d 3e 3f . 3g 3h 3i 2g 3h 3i
We note that b, c, d, g must be zero, but a, e, f, h and i are chosen freely. So, our space, a 0 0 V , consists of all matrices of the form 0 e f 0 h i 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 = a 0 0 0 + e 0 1 0 + f 0 0 1 + h 0 0 0 + i 0 0 0 . 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Thus, 0 0 0 , 0 1 0 , 0 0 1 , 0 0 0 , 0 0 0 is a basis of V , 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 and dim(V ) = 5.
37. If all diagonal entries of B are different, then dim(V ) = 3, as in Exercise 35. If two of the diagonal entries are equal, then dim(V ) = 5, as in Example 36. If all three entries are equal, then B is a scalar multiple of I3 , and will commute with all 3x3 matrices, so that dim(V ) = dim(R3×3 ) = 9. So, in summary, dim(V ) could be 3, 5 or 9. 38. We look at similar cases mentioned in Exercise 37, and see that the different possibilities occur when all four entries are different (dim(V ) = 4), when exactly two are the same, but the other two are different (dim(V ) = 6), when exactly three are the same (dim(V ) = 10), when all four are the same (dim(V ) = 16) and when two of the terms of B are equal, and the other two diagonal terms of B are also equal, but different from the first pair (dim(V ) = 8). 186
ISM: Linear Algebra a11 0 39. An upper-triangular matrix has the form: . . . 1 0 ··· 0 0 ··· = a11 . . . . . . . . . 0 0 ··· 0 0 ··· 0 1 ··· + a22 . . . . . . . . . a12 a22 . . . ··· ··· .. . a1n a2n . . .
Section 4.1
0 0 ··· 0
So the dimension of this space is n + (n − 1) + · · · + 1 =
0 0 · · · ann 0 0 ··· 0 1 ··· 0 0 0 0 ··· 0 0 ··· 0 0 . + · · · + a1n . . . . . + a12 . . . . . . . . . . . . . . . . . . 0 0 ··· 0 0 ··· 0 0 0 0 0 0 ··· 0 0 0 0 · · · 1 0 . + · · · + a2n . . . . . + · · · + ann . . . . . . . . . . . . 0 0 ··· 0
n(n+1) . 2
0
0 ··· 0 0 · · · 0 . .. . . . . . . . 0 0 ··· 1
1 0 . . .
40. Let V be the space of all n × n matrices A such that Av = 0. We look at some possibilities for v. If v = 0, then any matrix A will work, and dim(V ) = n2 . Now assume that v = 0, and suppose that the ith component vi is nonzero. If we denote the columns of A by w1 , . . . , wn , then the condition Av = 0 means that v1 w1 + · · · + vi wi + · · · + vn wn = 0. We can solve this relation for wi and express wi in terms of the other n − 1 column vectors of A. This means that we can choose n − 1 columns (or n(n − 1) entries) of A freely; the column wi is then determined by these choices. Thus dim(V ) = n(n − 1) in this case. In summary, we have dim(V ) = n2 if v = 0, and dim(V ) = n(n − 1) otherwise. 41. Let V be the space of all matrices A such that BA = 0. Note that A is in V if (and only if) all the columns of A are in the kernel of B. Since the columns of A can be chosen independently, it is plausible that dim(V ) = 3dim(ker(B)). We show this more clearly by investigating the different possibilities for ker(B). In the case where ker(B) = {0}, let v1 , . . . , vn be a basis of the kernel of B, and note that n = nullity(B) can be either 1, 2 or 3. Then, the general element of V is of the form: [ a1 v1 + . . . + an vn b1 v1 + . . . + bn vn c1 v1 . . . + cn vn ], which has the 3n arbitrary constants a1 , . . . , an , b1 , . . . , bn , c1 , . . . , cn . Using Summary 4.1.6, we can construct a basis of V with 3n elements, proving our claim that dim(V ) = 3dim(ker(B)). Thus, dim(V ) can be 0, 3, 6 or 9. 42. Let B be a matrix such that dim(ker(B))= k. Then, it is required that the columns of A contain only vectors in the kernel of B. Thus, each column of A can be written as: c1 v1 + c2 v2 + · · · + ck vk , where the vectors vi form as basis of the kernel of B. Thus, each of the n columns in A involves k arbitrary constants, and matrix A involves nk arbitrary constants overall. The space of matrices A has dimension nk, where k is an integer in the range [0, n]. 187
Chapter 4
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43. Let V be the space of all matrices S such that AS = S
1 0 . Using the terminology 0 −1 1 0 introduced in the hint, we want A [ v w ] = [ v w ] or [ Av Aw ] = [ v −w ]. 0 −1 Thus, v must be parallel to L, and w must be perpendicular to L. If v1 is a nonzero vector parallel to L, and v2 a nonzero vector perpendicular to L, then the general element of V is of the form S = [ v w ] = [ av1 bv2 ] = a [ v1 0 ] + b [ 0 v2 ] , where a and b are arbitrary constants. We see that [ v1 0 ] , [ 0 v2 ] is a basis of V and dim(V ) = 2. 1 0 0 44. Let V be the space of all matrices S such that AS = S 0 1 0 . Let’s denote the 0 0 0 1 0 0 column vectors of S by u, v and w. The condition AS = S 0 1 0 means that 0 0 0 Au = u, Av = v and Aw = 0. This in turn means that the vectors u and v have to be on the plane V, while w is perpendicular to V . If we choose a basis v1 , v2 of V and a nonzero vector v3 perpendicular to V, then we can write u = av1 + bv2 , v = cv1 + dv2 , w = ev3 , and S = [u v w] = [av1 +bv2 cv1 +dv2 ev3 ] = a[v1 0 0]+b[v2 0 0]+c[0 v1 0]+d[0 v2 0]+e[0 0v3 ]. Thus dim(V ) = 5; the five matrices in the linear combination above form a basis of V . a b c 45. Let A = d e f . We want AB = BA, or g h i a b c 0 1 0 0 1 0 a b c d e f 0 0 1 = 0 0 1 d e f , or g h i 0 0 0 0 0 0 g h i d e f 0 a b 0 d e = g h i . 0 0 0 0 g h So d, g, h = 0, a e = = 1 0 a b c 0 a b = a0 1 0 0 0 0 a 1 0 0 0 Thus, 0 1 0 , 0 0 0 1 0
i and f b. Our = space V consists of all matrices of the form 0 0 1 0 1 0 0 0 + b 0 0 1 + c 0 0 0 . 0 0 0 0 0 0 1 1 0 0 0 1 0 1 , 0 0 0 is a basis of V and dim(V ) = 3. 0 0 0 0 0 188
ISM: Linear Algebra
Section 4.1
46. The arithmetic sequences are of the form (a, a + k, a + 2k, a + 3k, . . .) = a(1, 1, 1, 1, . . .) + k(0, 1, 2, 3, . . .), so that the sequences (1, 1, 1, . . .) (all 1’s) and (0, 1, 2, 3, . . .) (the nth entry is n) form a basis of this space, whose dimension is 2. 47. We show that the set of all even functions is a subspace of F (R, R): • If f (t) = 0 for all t, then f (−t) = f (t) = 0 for all t. • If f and g are even (that is, f (−t) = f (t) and g(−t) = g(t) for all t), then (f + g)(−t) = f (−t) + g(−t) = f (t) + g(t) = (f + g)(t), so that f + g is even as well. • If f is even and k is any constant, then (kf )(−t) = kf (−t) = kf (t) = (kf )(t), so that kf is even as well. An analogous proof shows that the odd functions form a subspace of F (R, R). 48. If f (t) = a + bt + ct2 + dt3 + et4 , then f (−t) = a − bt + ct2 − dt3 + et4 and −f (−t) = −a + bt − ct2 + dt3 − et4 . a. f is even if f (−t) = f (t) for all t. Comparing coefficients we find that b = d = 0, so that f (t) is of the form f (t) = a + ct2 + et4 , with basis 1, t2 , t4 . The dimension is 3. b. f is odd if f (−t) = −f (t), which is the case if a = c = e = 0. The odd polynomials are of the form f (t) = bt + dt3 , with basis t, t3 and dimension 2.
49. We show that L(Rm , Rn ) is a subspace of F (Rm , Rn ): • The zero transformation T (x) = 0 (for all x) is linear, represented by the zero matrix. • If T and S are linear transformations from Rm to Rn (with T (x) = Ax and S(x) = Bx for some n × m matrices A and B), then (T + S)(x) = T (x) + S(x) = Ax + Bx = (A + B)x, so that T + S is linear as well, given by the matrix A + B. • If T is a linear transformation from Rm to Rn (with T (x) = Ax) and k is any constant, then (kT )(x) = kT (x) = kAx = (kA)x, so that kT is linear as well, with matrix kA. 50. Using Example 18 as a guide, we first look for solutions of the form f (x) = e kx . It is required that f (x) + 8f (x) − 20f (x) = k 2 ekx + 8kekx − 20ekx = 0 for all x, or k 2 + 8k − 20 = (k − 2)(k + 10) = 0. Thus k = 2 or k = −10. By Fact 4.1.5, the solutions of the differential equation are of the form f (x) = c1 e2x + c2 e−10x , where c1 and c2 are arbitrary constants. 189
Chapter 4
ISM: Linear Algebra
51. Using Example 18 as a guide, we first look for solutions of the form f (x) = e kx . It is required that f (x) − 7f (x) + 12f (x) = k 2 ekx − 7kekx + 12ekx = 0 for all x, or k 2 − 7k + 12 = (k − 3)(k − 4) = 0. Thus k = 3 or k = 4. By Fact 4.1.7, the solutions of the differential equation are of the form f (x) = c1 e3x + c2 e4x , where c1 and c2 are arbitrary constants. 52. We have to find constants a and b such that the functions e−x and e−5x are solutions of the differential equation f (x) + af (x) + bf (x) = 0. Thus it is required that e−x − ae−x + be−x = 0, or 1 − a + b = 0, and also that 25 − 5a + b = 0. The solution of this system of two equations in two unknowns is a = 6, b = 5, so that the desired differential equation is f (x) + 6f (x) + 5f (x) = 0. 53. Let B = (f1 , . . . , fn ) be a basis of V and suppose that the elements g1 , . . . , gm in V are linearly independent. In the proof of Fact 4.1.5, we show that the coordinate vectors [g1 ]B , . . . , [gm ]B in Rn are linearly independent, so that m ≤ n by Fact 3.2.8. 54. We can adapt the answer to Exercise 3.2.38a. Let m be the largest number of linearly independent elements we can find in W ; note that m ≤ n, by Exercise 53. Choose linearly independent elements f1 , . . . , fm in W . We claim that the elements f1 , . . . , fm span W . Indeed, if f is any element of W, then the m + 1 elements f1 , . . . , fm , f are linearly dependent, so that f is redundant: f is a linear combination of f1 , . . . , fm . It follows that f1 , . . . , fm is a basis of W, so that dim(W ) = m ≤ n =dim(V ), as claimed. 55. We will argue indirectly, assuming that F (R, R) is n-dimensional for some n. Now, the n + 1 polynomials 1, x, x2 , . . . , xn in F (R, R) are linearly independent, contradicting the fact that we can find at most n linearly independent elements in an n-dimensional space (see Exercise 53). We conclude that F (R, R) is infinite dimensional, as claimed. 56. Argue indirectly and assume that the space V of infinite sequences is finite-dimensional, with dim(V ) = n. According to the solution to Exercise 57, there can be at most n linearly independent elements in V . But here is our contradiction: It is easy to give n + 1 linearly independent infinite sequences, namely, (1, 0, 0, 0, . . .), (0, 1, 0, 0, . . .), (0, 0, 1, 0, . . .), . . . , (0, 0, 0, . . . , 0, 1, 0, . . .); in the last sequence the 1 is in the (n + 1)th place. 57. We can construct a basis of V by omitting the redundant elements from the list g1 , . . . , gm . It follows that V is finite-dimensional, and, in fact, dim(V ) ≤ m, since our basis is a “sublist” of the original list g1 , . . . , gm .
58. a. Let g(x) be a function in V . Thus, g (x) = −g(x). Now, if f (x) = g(x)2 + g (x)2 , then f (x) = 2(g(x))g (x) + 2(g (x))g (x) = 2(g(x))g (x) − 2(g(x))g (x) = 0. So, f (x) = g(x)2 + g (x)2 is a constant function. 190
ISM: Linear Algebra
Section 4.2
b. Let g(x) be a function in V such that g(0) = g (0) = 0. From part a we know that g(x)2 + g (x)2 = k, a constant. Now g(0)2 + g (0)2 = 02 + 02 = 0, so that k = 0. The equation g(x)2 + g (x)2 = 0 means that g(x) = g (x) = 0 for all x, as claimed. c. First we note that g(x) = f (x) − f (0) cos(x) − f (0) sin(x) is in V, since the functions f (x), cos(x) and sin(x) are all in V, and V is a subspace of F (R, R). Note that g(0) = f (0) − f (0) cos(0) − f (0) sin(0) = f (0) − f (0) = 0. Also, g (x) = f (x) + f (0) sin(x) − f (0) cos(x), so that g (0) = f (0) − f (0) = 0. By part b, we can conclude that g(x) = 0 for all x, so that f (x) = f (0) cos(x) + f (0) sin(x), as claimed.
4.2
1. Fails to be linear, since T (A + B) = A + B + I2 doesn’t equal T (A) + T (B) = A + I2 + B + I2 = A + B + 2I2 . 2. Linear, since T (A + B) = 7(A + B) = 7A + 7B equals T (A) + T (B) = 7A + 7B, and T (kA) = 7kA equals kT (A) = k(7A) = 7kA.
1 Yes, T is an isomorphism, with T −1 (A) = 7 A.
3. Linear, since T a c a c
a c
b p q + d r s
=T
a+p c+r
b+q =a+p+d+s d+s k a c b d =T ka kb = ka+kd kc kd
equals T equals kT
p q b = a+d+p+s, and T +T r s d b = k(a + d) = ka + kd. d
No, T fails to be an isomorphism, since 4 = dim(R2×2 ) = dim(R) = 1; see Fact 4.2.4b. 4. Fails to be linear, since T (2I2 ) = det(2I2 ) = 4 does not equal 2T (I2 ) = 2 det(I2 ) = 2. 5. Fails to be linear, since T (2I2 ) = 4I2 does not equal 2T (I2 ) = 2I2 . 6. Let P = 1 2 . Transformation T is linear, since 3 6
T (A + B) = (A + B)P = AP + BP equals T (A) + T (B) = AP + BP , and 191
Chapter 4 T (kA) = (kA)P = kAP equals kT (A) = kAP . No, T isn’t an isomorphism, since ker(T ) = {0}; the matrix A = kernel of T (see Fact 4.2.4a). 7. Linear, since T (M + N ) =
ISM: Linear Algebra
3 −1 0 0
is in the
1 2 1 2 1 2 N = T (M ) + T (N ). M+ (M + N ) = 3 4 3 4 3 4 1 2 1 2 Also, T (kM ) = (kM ) = k M = kT (M ). 3 4 3 4 This is also an isomorphism. Solve the equation N = M= 1 2 3 4
−1
1 2 M for M to find the inverse 3 4
N=
1 2
−4 2 N. 3 −1 1 2 1 2 (kM ) = k 2 M M = k 2 T (M ) = kT (M ), 3 4 3 4
8. Not linear, since T (kM ) = (kM ) in general.
9. Linear, since T (A + B) = S −1 (A + B)S = S −1 AS + S −1 BS equals T (A) + T (B) = S −1 AS + S −1 BS, and T (kA) = S −1 (kA)S = kS −1 AS equals kT (A) = kS −1 AS. Yes, T is an isomorphism. Solve the equation B = S −1 AS for A to find the inverse A = SBS −1 . 10. Linear, since T (A + B) = P (A + B)P −1 = P AP −1 + P BP −1 equals T (A) + T (B) = P AP −1 + P BP −1 , and T (kA) = P (kA)P −1 = kP AP −1 equals kT (A) = kP AP −1 . Yes, T is an isomorphism. Solve the equation B = P AP −1 for A to find the inverse A = P −1 BP . 11. Linear, since T (M +N ) = P (M +N )Q = (P M +P N )Q = P M Q+P N Q = T (M )+T (N ). Also, T (kM ) = P (kM )Q = kP M Q = kT (M ). This is also an isomorphism. Solve the equation N = P M Q for M to find the inverse M = P −1 N Q−1 . 192
ISM: Linear Algebra
Section 4.2 2 3 . Then T (c + d) = (c + d)A = cA + dA equals T (c) + T (d) = 4 5
12. Linear. Let A = cA + dA, and
T (kc) = kcA equals kT (c) = kcA. No, T isn’t an isomorphism, since domain and codomain have different dimensions. 13. Let Q = 1 2 . Transformation T is linear, since 0 1
T (M + N ) = (M + N )Q − Q(M + N ) = M Q + N Q − QM − QN equals T (M )+T (N ) = M Q−QM +N Q−QN , and T (kM ) = (kM )Q−Q(kM ) = kM Q−kQM equals kT (M ) = k(M Q − QM ) = kM Q − kQM . No, T isn’t an isomorphism, since A = I2 is in ker(T ); see Fact 4.2.4a. 14. Let Q = 2 3 . Transformation T is linear, since 5 7
T (M + N ) = Q(M + N ) − (M + N )Q = QM + QN − M Q − N Q equals T (M )+T (N ) = QM −M Q+QN −N Q, and T (kM ) = Q(kM )−(kM )Q = kQM −kM Q equals kT (M ) = k(QM − M Q) = kQM − kM Q. No, T isn’t an isomorphism, since A = I2 is in ker(T ); see Fact 4.2.4a. 15. Linear, since T (M +N ) = M 4 0 4 0 −N 0 5 0 5 T (N ). 2 0 2 = 0 0 (M +N )−(M +N ) 3 4 0 0 + M −M 0 5 3 4 0 2 0 2 0 2 0 0 N− M+ = 0 3 0 3 5 4 0 0 = T (M ) + N −N 0 5 3
Also, T (kM ) =
2 0 4 0 2 0 4 0 (kM ) − (kM ) =k M − kM = kT (M ). 0 3 0 5 0 3 0 5
This is also an isomorphism. We first see that the dimensions of V and W are equal, then a b 2 0 a b − = show that the kernel contains only zero. Let T (M ) = T c d 0 3 c d −2a −3b 4a 5b 2a 2b 4 0 a b . Clearly the only time this = − = −c −2d 4c 5d 3c 3d 0 5 c d matrix will equal zero is when a, b, c, d = 0. Thus, ker(T ) = { 0}. 193
Chapter 4 2 0 3 0 − (M +N ) = M 0 3 0 4 3 0 3 0 2 0 3 0 2 0 M− N =M − M +N − 0 4 0 4 0 3 0 4 0 3 T (N ).
ISM: Linear Algebra 2 0 3 0 0 2 0 +N − 3 0 3 0 N = T (M ) + 4
16. Linear, since T (M +N ) = (M +N )
Also, T (kM ) = (kM )
2 0 3 0 2 0 3 0 − (kM ) = k M − M 0 3 0 4 0 3 0 4
= kT (M ).
To determine whether T is an isomorphism, we find the kernel of T . Now, T (M ) = a b a b 2 0 3 0 a b 2a 3b 3a 3b −a 0 T = − = − = . c d c d 0 3 0 4 c d 2c 3d 4c 4d −2c −d a b is in the kernel of T if a = c = d = 0. Thus, the kernel consists We see that M = c d 0 b of all matrices , where b is arbitrary. Since the kernel is nonzero, T fails to be an 0 0 isomorphism. 17. T is linear: T ((a + ib) + (c + id)) = T (a + c + i(b + d)) = a + c = T (a + ib) + T (c + id), and T (k(x + iy)) = T (kx + iky) = kx = kT (x + iy). However, T (5i) = 0, so T fails to be an isomorphism. 18. This transformation fails to be linear, since 2T (3) = 2(9) = 18 = T (6) = 36. 19. T is linear: if w and z are complex numbers, then T (w + z) = i(w + z) = iw + iz = T (w) + T (z) and T (kz) = i(kz) = k(iz) = kT (z). T is also an isomorphism. Solve the equation w = iz for z to find that z = We have used the fact that i2 = −1, so that 1 = −i. i T (x + iy) + T (z + it) = x − iy + z − it = x + z − i(y + t) and T (k(x + iy)) = T (kx + iky) = kx − iky equals kT (x + iy) = k(x − iy) = kx − iky. Yes, T is an isomorphism; it’s its own inverse, since T (T (x + iy)) = T (x − iy) = x + iy. 21. Linear, since T ((x + iy) + (z + it)) = T (x + z + i(y + t)) = y + t + i(x + z) equals T (x + iy) + T (z + it) = y + ix + t + iz = y + t + i(x + z), and T (k(x + iy)) = T (kx + iky) = ky + ikx equals kT (x + iy) = k(y + ix) = ky + ikx. Yes, T is an isomorphism; it’s its own inverse, since T (T (x + iy)) = T (y + ix) = x + iy. 194
1 iw
= −iw.
20. Linear, since T ((x + iy) + (z + it)) = T (x + z + i(y + t)) = x + z − i(y + t) equals
ISM: Linear Algebra
3 3 3
Section 4.2
22. Linear, since T (f + g) =
−2 3 3
(f + g) =
−2
f+
−2 3
g equals
3 3
T (f ) + T (g) =
−2
f+
−2
g, and T (kf ) =
−2
kf = k
−2
f equals kT (f ) = k
−2
f.
No, T isn’t an isomorphism, since domain and codomain have different dimensions. 23. T is linear, because T (f (t) + g(t)) = f (7) + g(7) = T (f (t)) + T (g(t)), and T (kf (t)) = kf (7) = kT (f (t)). However, T cannot be an isomorphism, because the dimensions of the domain and codomain fail to be equal. 24. T is not linear. 2T (t2 ) = 2(2(t2 )) = 4t2 = T (2t2 ) = 4(2t2 ) = 8t2 . 25. Linear, since T (f + g) = (f + g) + 4(f + g) = f + g + 4f + 4g equals T (f ) + T (g) = f + 4f + g + 4g , and T (kf ) = (kf ) + 4(kf ) = kf + 4kf equals kT (f ) = k(f + 4f ) = kf + 4kf . No, it isn’t an isomorphism, since the constant function f (x) = 1 is in ker(T ). 26. Linear, since T (f (t) + g(t)) = f (−t) + g(−t) equals T (f (t)) + T (g(t)) = f (−t) + g(−t), and T (kf (t)) = kf (−t) equals kT (f (t)) = kf (−t). Yes, T is an isomorphism; it’s its own inverse, since T (T (f (t))) = T (f (−t)) = f (t). 27. Linear, since T (f (t) + g(t)) = f (2t) + g(2t) equals T (f (t)) + T (g(t)) = f (2t) + g(2t), and T (kf (t)) = kf (2t) equals kT (f (t)) = kf (2t). Yes, T is an isomorphism; the inverse is T −1 (g(t)) = g
t 2
.
28. T is linear, since T (f (t) + g(t)) = f (2t) + g(2t) − f (t) − g(t) = f (2t) − f (t) + g(2t) − g(t) = T (f (t)) + T (g(t)), and T (kf (t)) = kf (2t) − kf (t) = k(f (2t) − f (t)) = kT (f (t)). T is not an isomorphism, however, since T (3) = 3 − 3 = 0. 29. Linear, because T (f (t)+g(t)) = f (t)+g (t) = T (f (t))+T (g(t)) and T (kf (t)) = kf (t) = kT (f (t)). However, since T (5) = 0, the kernel is nonzero, and T fails to be an isomorphism. 30. Linear, since T (f (t) + g(t)) = t(f (t) + g (t)) = t(f (t)) + t(g (t)) equals T (f (t)) + T (g(t)) = t(f (t)) + t(g (t)), and T (kf (t)) = t(kf (t)) = kt(f (t)) equals kT (f (t)) = kt(f (t)). 195
Chapter 4
ISM: Linear Algebra
No, T isn’t an isomorphism, since the constant function f (t) = 1 is in ker(T ). 31. T is linear: T (f (t)+g(t)) = g(0) g(1) f (0) f (1) f (0) + g(0) f (1) + g(1) + = g(2) g(3) f (2) f (3) f (2) + g(2) f (3) + g(3) f (0) f (1) kf (0) kf (1) = kT (f (t)). =k f (2) f (3) kf (2) kf (3)
= T (f (t)) + T (g(t)), and T (kf (t)) =
However, the dimensions here are different, so that T fails to be an isomorphism. 32. T is not linear: T (2) = 0 + t2 = 2T (1) = 2(0 + t2 ). 33. Linear, since T ((x0 , x1 , x2 , . . .) + (y0 , y1 , y2 , . . .)) = T (x0 + y0 , x1 + y1 , x2 + y2 , . . .) = (x0 +y0 , x2 +y2 , . . .) equals T (x0 , x1 , x2 , . . .)+T (y0 , y1 , y2 , . . .) = (x0 , x2 , . . .)+(y0 , y2 , . . .) = (x0 + y0 , x2 + y2 , . . .), and T (k(x0 , x1 , x2 , . . .)) = T (kx0 , kx1 , kx2 , . . .) = (kx0 , kx2 , . . .) equals kT (x0 , x1 , x2 , . . .) = k(x0 , x2 , . . .) = (kx0 , kx2 , . . .). No, T isn’t an isomorphism, since (0, 1, 0, 0, 0, . . .) is in ker(T ). 34. Linear, since T ((x0 , x1 , x2 , . . .) + (y0 , y1 , y2 , . . .)) = T (x0 + y0 , x1 + y1 , x2 + y2 , . . .) = (0, x0 + y0 , x1 + y1 , x2 + y2 , . . .) equals T (x0 , x1 , x2 , . . .) + T (y0 , y1 , y2 , . . .) = (0, x0 , x1 , x2 , . . .) + (0, y0 , y1 , y2 , . . .) = (0, x0 + y0 , x1 + y1 , x2 + y2 , . . .), and T (k(x0 , x1 , x2 , . . .)) = T (kx0 , kx1 , kx2 , . . .) = (0, kx0 , kx1 , kx2 , . . .) equals kT (x0 , x1 , x2 , . . .) = k(0, x0 , x1 , x2 , . . .) = (0, kx0 , kx1 , kx2 , . . .). No, T isn’t an isomorphism, since (1, 0, 0, 0, . . .) isn’t in im(T ). 35. Linear, since T (f (t) + g(t)) = (f (0) + g(0), f (0) + g (0), · · ·) = (f (0), f (0), · · ·) + (g(0), g (0), · · ·) = T (f (t)) + T (g(t)) and T (kf (t)) = (kf (0), kf (0), · · ·) = k(f (0), f (0), · · ·) = kT (f (t)). T fails to be an isomorphism. Note that the sequences in the image of T have only finitely many nonzero entries, so that a sequence like (1, 1, 1, 1, . . .), with all 1’s, fails to be in the image of T . Now use Fact 4.2.4a. 36. T is linear: T (f (t)+g(t)) = (f (0)+g(0), f (1)+g(1), ...) = (f (0), f (1), ...)+(g(0), g(1), ...) = T (f (t)) + T (g(t)), and T (kf (t)) = (kf (0), kf (1), ...) = k(f (0), f (1), ...) = kT (f (t)). 196
ISM: Linear Algebra
Section 4.2
We will show that the image of T isn’t all of V , so that T fails to be an isomorphism. More specifically, we claim that the sequence (1, 0, 0, 0, . . .), a 1 followed by all 0’s, fails to be in the image. We make an attempt to find a polynomial f (t) such that T (f (t)) = (f (0), f (1), f (2), . . .) = (1, 0, 0, . . .). This polynomial f (t) is required to have infinitely many zeros, at t = 1, 2, 3, . . . , so that f (t) must be the zero polynomial, and the equation f (0) = 1 isn’t satisfied. Thus there is no polynomial f (t) such that T (f (t)) = (1, 0, 0, 0, . . .). 37. Linear, since T (f + g) = f + g + f + g = f + f + g + g = T (f ) + T (g) and T (kf ) = kf + kf = k(f + f ) = kT (f ). However, T (e−x ) = e−x − e−x = 0, so T fails to be an isomorphism. 38. Linear, just as in Exercise 37. T is not an isomorphism, since T (sin(x)) = sin(x) − sin(x) = 0. 39. Linear; the proof is analogous to Exercise 25. No, T isn’t an isomorphism, since the kernel of T is two-dimensional, by Fact 4.1.7. 40. Same answer as in Exercise 39. 41. Not linear, because T (f (t) + g(t)) = f (t) + g(t) + f (t) + g (t) + sin(t) does not equal T (f (t)) + T (g(t)) = f (t) + f (t) + sin(t) + g(t) + g (t) + sin(t). 42. Linear, since T (f (t) + g(t)) = f (7) + g(7) f (11) + g(11) equals T (f (t)) + T (g(t)) = equals
f (7) g(7) f (7) + g(7) kf (7) + = , and T (kf (t)) = f (11) g(11) f (11) + g(11) kf (11) kT (f (t)) = k kf (7) f (7) . = kf (11) f (11)
Not an isomorphism, since domain and codomain have different dimensions. g(5) f (5) f (5) + g(5) 43. Linear, since T (f (t) + g(t)) = f (7) + g(7) = f (7) + g(7) = T (f (t)) + g(11) f (11) f + (11) g(11) kf (5) f (5) T (g(t)), and T (kf (t)) = kf (7) = k f (7) = kT (f (t)). kf (11) f (11) T is an isomorphism; the proof is analogous to Example 6b. 197
Chapter 4
ISM: Linear Algebra
f (1) + g(1) f (1) g(1) 44. Linear, since T (f (t)+g(t)) = f (2) + g (2) = f (2) + g (2) = T (f (t))+T (g(t)), f (3) + g(3) f (3) g(3) and kf (1) f (1) T (kf (t)) = kf (2) = k f (2) = kT (f (t)). kf (3) f (3) Not an isomorphism, since T ((t − 1)(t − 3)) = T (t2 − 4t + 3) = 0. 45. Linear, since T (f (t) + g(t)) = t(f (t) + g(t)) = tf (t) + tg(t) = T (f (t)) + T (g(t)) and T (kf (t)) = t(kf (t)) = ktf (t) = kT (f (t)). This is not an isomorphism, since the constant function f (t) = 1 isn’t in the image. 46. Linear, since T (f (t) + g(t)) = (t − 1)(f (t) + g(t)) = (t − 1)f (t) + (t − 1)g(t) = T (f (t)) + T (g(t)) and T (kf (t)) = (t − 1)(kf (t)) = k(t − 1)f (t) = kT (f (t)). This is not an isomorphism, since the constant function f (t) = 1 isn’t in the image. 47. T is linear, because T (f (t) + g(t)) = 0 (f (x) + g(x))dx = 0 f (x)dx + t t T (f (t)) + T (g(t)). Also, T (kf (t)) = 0 kf (x)dx = k 0 f (x)dx = kT (f (t)). However, there is no polynomial f (t) such that T (f (t)) = isomorphism.
t 0 t t t 0
g(x)dx =
f (x) = 6. Thus, T is not an
48. Linear, since T (f (t)+g(t)) = (f +g) (t) = f (t)+g (t) = T (f (t))+T (g(t)) and T (kf (t)) = kf (t) = kT (f (t)). This is not an isomorphism, however, since T (5) = 0. 49. Linear, since T (f (t) + g(t)) = f (t2 ) + g(t2 ) = T (f (t)) + T (g(t)) and T (kf (t)) = kf (t2 ) = kT (f (t)). However, there is no f (t) in P such that T (f (t)) = f (t2 ) = t. Thus, the image of T fails to be all of P , and T fails to be an isomorphism. 50. Linear. T (f (t) + g(t)) = f (t+2)+g(t+2)−f (t)−g(t) = f (t+2)−f (t) + 2 2 T (g(t)), and T (kf (t) = kf (t+2)−kf (t) = k f (t+2)−f (t) = kT (f (t)). 2 2 This is not an isomorphism, however, since T (5) = 51. We need to find the matrices M = x y z t 198
5−5 2 g(t+2)−g(t) 2
= T (f (t)) +
= 0. 1 2 , that is, 0 1
that commute with
ISM: Linear Algebra x z y t 1 2 1 2 = 0 1 0 1 x y , or, z t x 2x + y z 2z + t x + 2z z y + 2t . t
Section 4.2
=
It follows that z = 0 and x = t, so that the kernel of T consists of all matrices of the form t y . The nullity (i.e., the dimension of the kernel of T ) is 2. 0 t 52. We need to find the matrices A = x z y t 1 2 3 6 = x + 3y z + 3t x z y t = such that
0 0 . It is required that x = −3y and 0 0 −3y y . z = −3t, so that the kernel of T consists of all matrices of the form −3t t The nullity (i.e., the dimension of the kernel of T ) is 2. 53. Note that T (a + bt + ct2 ) = 2c + 4b + 8ct. Thus the kernel consists of all constant polynomials f (t) = a(when b = c = 0), and the nullity is 1. The image consists of all linear polynomials f (t) = p + qt, and the rank is 2.
5 54. Use calculus to see that T (a + bt + ct2 ) = 5a + 2 b + 35 3 c.
2x + 6y 2z + 6t
The image is all of
R, so that the rank is 1. The kernel consists of all polynomials of the form
1 7 f (t) = − 2 b − 3 c + bt + ct2 , and thus the nullity is 2.
55. The kernel consists of all infinite sequences (x0 , x1 , x2 , x3 , . . .) such that T (x0 , x1 , x2 , x3 , x4 , . . .) = (x0 , x2 , x4 , . . .) = (0, 0, 0, . . .), that is, all terms xk with even k must be 0. Thus the kernel consists of all sequences of the form (0, x1 , 0, x3 , 0, . . .). The image consists of all infinite sequences (y0 , y1 , y2 , . . .), since (y0 , y1 , y2 , . . .) = T (y0 , 0, y1 , 0, y2 , 0, . . .) for example. 56. Note that T (a + bt + ct2) = bt + 2ct2. Thus the kernel consists of all constant polynomials f (t) = a (when b = c = 0), and the nullity is 1. The image consists of all polynomials of the form f (t) = pt + qt2 , and the rank is 2. 57. The kernel consists of the solutions of the differential equation f (t) − 5f (t) + 6f (t) = 0. Using the approach outlined in Example 18 of Section 4.1 (involving a trial solution f (t) = ekt ), we find the general solution f (t) = c1 e2t + c2 e3t ; thus the nullity is 2. 58. The kernel consists of all infinite sequences such that 199
Chapter 4
ISM: Linear Algebra
T (x0 , x1 , x2 , . . .) = (0, x0 , x1 , x2 , . . .) = (0, 0, 0, 0, . . .), that is, all terms xk must be 0. Thus the kernel consists of the zero sequence (0, 0, 0, . . .) alone. The image consists of all infinite sequences of the form (0, x0 , x1 , x2 , . . .). 59. To find the kernel, we solve the equation T (f (t)) = T (a + bt + ct2 ) = a + 7b + 49c = 0. It follows that a = −7b−49c, and the general element of the kernel is (−7b−49c)+bt+ct 2 = b(−7 + t) + c(−49 + t2 ). Then a basis of the kernel is −7 + t, −49 + t2 , and the nullity of T is 2. Now the rank of T must be 1, and the image is all of R. 60. Note that T (a + bt + ct2 ) = system a + 7b + 49c . To find the kernel, solve the linear a + 11b + 121c
a + 7b + 49c 0 = . The solution is a = 77c, b = −18c, so that the a + 11b + 121c 0
kernel consists of all polynomials of the form f (t) = c(77 − 18t + t2 ) = c(t − 11)(t − 7). You can also see directly that the quadratic polynomials f (t) with f (7) = f (11) = 0 are of this form. The nullity is 1. The image consists of all of R2 , so that the rank is 2. 61. The kernel consists of all polynomials f (t) such that t(f (t)) = 0 for all t, that is, the zero polynomial f (t) = 0 alone. The image consists of all polynomials g(t) that can be written as g(t) = t(f (t)), meaning that we can factor out a t. These are the polynomials with constant term 0, of the form g(t) = a1 t + a2 t2 + · · · + an tn . 62. The image of this transformation consists of all polynomials, since any polynomial is the derivative of another. The kernel of this transformation consists of all constant functions, or the span of the function f (t) = 1. 63. This is impossible, since dim(P3 ) = 4 and dim(R3 ) = 3. See Fact 4.2.4b. 64. Consider T (a + bt + ct2 + dt3 ) = a c b , for example. d
65. a. First, we need to show that T (A + B) = T (A) + T (B) for all n × m matrices A and B, that is (T (A + B))(v) = (T (A) + T (B))(v) for all v in Rm . Indeed (T (A + B))(v) = (A + B)v = Av + Bv equals (T (A) + T (B))(v) = T (A)(v) + T (B)(v) = Av + Bv. Also 200
ISM: Linear Algebra (T (kA))(v) = (kA)v = kAv equals (kT (A))(v) = k(T (A))(v) = kAv.
Section 4.2
b. The kernel of T consists of all n×m matrices A such that T (A) = 0, that is (T (A))(v) = Av = 0 for all v in Rm . This holds for the zero matrix only. Thus ker(T ) = {0}. c. This is true by definition of a linear transformation (Definition 2.1.1). d. Note that T gives an isomorphism from Rn×m to L(Rm , Rn ), by parts a, b, and c. Since dim(Rn×m ) = nm, by Exercise 17 of Seciton 4.1, we have dim(L(Rm , Rn )) = nm, by Fact 4.2.4b.
66. The kernel of T consists of all smooth functions f (t) such that T (f (t)) = f (t) − f (t) = 0, or f (t) = f (t). As you may recall from a discussion of exponential functions in calculus, those are the functions of the form f (t) = Cet , where C is a constant. Thus the nullity of T is 1. 67. To show that T is linear, proceed as in Exercise 15. Now let M = Then T (M ) = = 2 3 0 4 a c b a − d c b d 3 0 0 k a c b . d
2a + 3c 2b + 3d 3a − 4c 4d 3c a c b d
kb −a + 3c (2 − k)b + 3d = . kd c (4 − k)d
The matrix
is in the kernel of T if a = c = 0, (2 − k)b + 3d = 0 and (4 − k)d = 0.
If k is neither 2 nor 4, then the equation (4−k)d = 0 implies that d = 0, and (2−k)b+3d = 0 then implies that b = 0. Thus the kernel of T is 0, and T is an isomorphism. If k is 2, then b is arbitrary (while a = c = d = 0), and T fails to be an isomorphism. A 0 1 in this case. nonzero matrix in the kernel is 0 0 If k is 4, then b is arbitrary, and d = 2 b, so that, again, the kernel contains nonzero 3 0 3 , and T fails to be an isomorphism. matrices for example, 0 2 In summary, T is an isomorphism if (and only if) k is neither 2 nor 4. 201
Chapter 4
ISM: Linear Algebra a b c d a b 5a b 2a 2b 3a = − = c d 5c d kc kd (5 − k)c 0 0 0 = . Similarly, if k = 5, T 0 0 1
68. We find the constants k which make the kernel of T nonzero. Let M = T (M ) = We 0 0 For a c b d
. Then
5 0 2 0 −b − . 0 1 0 k (1 − k)d 0 0 0 see that if k = 1, T = 0 1 0 0 . Thus, for k = 1, 5, the kernel of T is non-zero, and T is not an isomorphism. 0 all other values of k, however, T is an isomorphism.
69. No, since A is similar to B, there exists an invertible S such that AS = SB. Now T (S) = AS − SB = 0, so that the nonzero matrix S is in the kernel of T . 70. Since the dimensions of the domain and codomain are equal, it suffices to examine when the kernel is zero. Now f (t) is in the kernel of T if f (c0 ) = f (c1 ) = · · · = f (cn ) = 0. Recall that a nonzero polynomial of degree ≤ n has at most n zeros. If the n + 1 numbers c0 , c1 , . . . , cn are all different, then the only polynomial in Pn with f (c0 ) = f (c1 ) = . . . . = f (cn ) = 0 is the zero polynomial, so that the kernel of T is zero, and T is an isomorphism. However, if some of the numbers are equal, for example, cp = cq , then there will be nonzero polynomials in the kernel (for example, the product of all (t − ci ) where i = q), so that T fails to be an isomorphism. 71. Exercise 70 tells us that T is an isomorphism so long as c0 , c1 , · · · , cn are all different. f (2) f (3) This condition is met here, so T (f (t)) = f (5) is an isomorphism from P4 to R5 . f (7) f (11) Thus, there is exactly one polynomial f (t) with the required properties. 72. We satisfy all the requirements of Definition 4.1.2. Clearly 0 is an element of Z n . Let h = f + g, where f and g are elements of Zn . Then h(0) = f (0) + g(0) = 0 + 0 = 0. Also, if f is in Zn , then kf (0) = k(0) = 0. We notice that the space Zn has the basis, t, t2 , . . . , tn . Thus, the dimension of Zn is n. 73. Yes. T is an isomorphism; the inverse transformation is D(f (t)) = f (t) = df , the dt derivative. We will check that the composite of T with D is the identity, in either order. Indeed d t d f (x)dx = f (t) D(T (f (t))) = (T (f (t)) = dt dt 0 by what is sometimes called the first fundamental theorem. And
t
T (D(f (t))) = T (f (t)) =
0
f (x)dx = f (t) − f (0) = f (t) for all f (t) in Zn 202
ISM: Linear Algebra by the ”second” fundamental theorem.
Section 4.2
74. Let T (f (t)) = f (t). This is an isomorphism, since the spaces have the same dimension, and the kernel consists only of the zero polynomial. 75. To see that 0 + 0 = 0, apply the formula f + 0 = f to f = 0 (see Definition 4.1.1). Then 0 = k0 − k0 = k(0 + 0) − k0 = k0 + k0 − k0 = k0. 76. 0W = T (0V ) − T (0V ) = T (0V + 0V ) − T (0V ) = T (0V ) + T (0V ) − T (0V ) = T (0V ) We have used the equation 0V + 0V = 0V derived in Exercise 41.
Figure 4.1: for Problem 4.2.77. 77. If T and L are linear, then (L ◦ T )(f + g) = L(T (f + g)) = L(T (f ) + T (g)) = L(T (f )) + L(T (g)) = (L ◦ T )(f ) + (L ◦ T )(g), and (L ◦ T )(kf ) = L(T (kf )) = L(kT (f )) = kL(T (f )) = k(L ◦ T )(f ), so that L ◦ T is linear as well. If T and L are isomorphism, then L ◦ T is an isomorphism as well, since the composite of invertible functions is invertible. (See Figure 4.1.)
78. a. Check all the conditions in Definition 4.1.1. A basis is 2. b. T (x ⊕ y) = T (xy) = ln(xy) = ln(x) + ln(y) = T (x) + T (y) and T (k x) = T (xk ) = ln(xk ) = k ln(x) = kT (x). The inverse of T is L(y) = ey , so that T is indeed an isomorphism.
79. Yes; let T be an invertible function from R2 to R. On R2 we can define the “exotic” operations v ⊕ w = T −1 (T (v) + T (w)) and k v = T −1 (kT (v)) (check that the conditions of Definition 4.1.1 hold). Then T is a linear transformation from R2 (with these exotic operations) to R (with the usual operations): T (v ⊕ w) = T (T −1(T (v) + T (w))) = T (v) + T (w), and T (k v) = kT (v). Now T is an isomorphism, so that the dimension of R2 (with our exotic operations) equals the dimension of the “usual” R, which is 1. 203
Chapter 4
ISM: Linear Algebra
80. If a real linear space V has more than one element, then it is infinite; to see this, note that the scalar multiples kf of a nonzero f in V are all distinct (think about it!). Since there is more than one student in your class, but the number of students is finite, we cannot make the set X into a real linear space.
81. a. ker(T ) is a subspace of V , so by Exercise 4.1.54, ker(T ) must also be finite-dimensional. Also, im(T ) is finite-dimensional, because it is finitely-generated by some elements T (f1 ), T (f2 ), · · · , T (fk ), where f1 , f2 , · · · , fk is a basis of V . b. Following the hint: T (c1 u1 + · · · + cr ur + d1 v1 + · · · + dn vn ) = T (0), so c1 T (u1 ) + · · · + cr T (ur ) + d1 T (v1 ) + · · · + dn T (vn ) = c1 w1 + · · · + cr wr + 0 + · · · + 0 = 0. So, since w1 , · · · , wr is a basis and must be linearly independent, c1 , · · · cr must all be zero. Now, d1 v1 + · · · + dn vn = 0, but v1 , · · · , vn is also a basis, so d1 , · · · dn also must all equal zero. Thus, the vectors u1 , · · · , ur , v1 , · · · , vn are linearly independent. c. The hint guides us right along here. Let v be in V . Then T (v) is in im(T ), so T (v) = d1 w1 + · · · + dr wr . Now, T (v − d1 u1 − · · · − dr ur ) = T (v) − d1 T (u1 ) − · · · − dr T (ur ) = d1 w1 + · · · + dr wr − d1 w1 − · · · − dr wr = 0. Thus, v − d1 u1 − · · · − dr ur is in the kernel of T , and there are some c1 , · · · , cn such that v − d1 u1 − · · · − dr ur = c1 v1 + · · · + cn vn . Thus, v = d1 u1 + · · · + dr ur + c1 v1 + · · · + cn vn . Thus the vectors u1 , · · · , ur , v1 , · · · , vn span V .
4.3
1. Let B be the standard basis of P2 : 1, t, t2 . Then the coordinates of the given poly 7 9 3 nomials with respect to B are [f ]B = 3 , [g]B = 9 , [h]B = 2 . Finding 1 4 1 7 9 3 rref 3 9 2 = I3 , we conclude that [f ]B , [g]B , [h]B are linearly independent, hence 1 4 1 so are f, g, h, since the coordinate transformation is an isomorphism. 2. Let B be the basis 1 0 0 1 0 0 0 0 , , , of R2×2 . Then the coordinates of 0 0 0 0 1 0 0 1 1 1 1 1 2 2 3 1 1 2 = = , = , the given matrices with respect to B are 5 7 B 3 4 B 3 1 1 1 B 4 1 204
ISM: Linear Algebra 2 3 1 4 , 6 8 5 7
Section 4.3
1 0 0 −1 1 1 2 1 1 4 0 1 0 1 2 3 4 4 = . Finding rref = I4 , we = 0 0 1 −1 1 3 5 6 6 B 0 0 0 0 1 4 7 8 8 1 1 2 1 1 2 3 4 conclude that the four vectors , , , are linearly dependent, and so are 1 3 5 6 1 4 7 8 1 4 1 1 1 2 2 3 the four given matrices. In fact =− +4 − . 6 8 1 1 3 4 5 7 1 2 3. We proceed as in Exercise 1. Since rref 9 1 do form a basis of P3 . 1 7 0 7 1 8 1 5 1 8 = I4 , the four given polynomials 4 8
4. Consider the coordinate vectors of the 3 given polynomials with respect to the standard basis of P2 : 1, t, t2 . 1 0 2k 1,1,2 + k 0 1 1 1 Since matrix 1 0 of R3 unless k = unless k = 1. 5. Use a diagram: a b + 2c a b = 0 3c 0 c a −→ − b + 2c A 3c a a 1 0 0 The matrix A that transforms b into b + 2c is A = 0 1 2 . c 3c 0 0 3 a b 0 c a b c −→ − T 1 2 0 3 205
1 0 2k 0 2k 1 2 + k reduces to 0 1 2 − k , these three vectors form a basis 0 0 k−1 1 1 1. Therefore, the three polynomials f (t), tf (t), g(t) form a basis of P2
Chapter 4 A is invertible, so T is an isomorphism. 6. Use Fact 4.3.2. to construct matrix B column by column: B= T 1 0 0 0 1 0 0 0 T
B
ISM: Linear Algebra
0 1 0 0 0 3 0 3
T
B
0 1 0 1
=
B
0 1 0 0
B
B
This matrix is invertible, so T is an isomorphism. 7. Use Fact 4.3.2 to construct B column by column: B= T 1 0 0 1 T
B
1 0 0 = 0 1 0 . 0 0 3
B
0 1 0 0
T
B
1 0 0 −1
=
B
0 0 0 0
B
0 0 0 0
B
0 4 0 0
B
0 0 0 1 0 0 = 0 0 4 . 0 , 1 is a basis of the kernel of B, and 1 is a basis of the image 0 0 0 0 0 0 1 0 0 1 0 1 of B. This implies that , is a basis of the kernel of T , and is a 0 1 0 0 0 0 basis of the image of T . Thus, T fails to be an isomorphism. 8. Use a diagram: 0 2a − 2c 0 0 0 −→ − 2a − 2c A 0 a 0 0 0 0 The matrix that transforms b into 2a − 2c is A = 2 0 −2 . c 0 0 0 0 0 1 0 We see that a basis of the kernel of A is 1 , 0 and a basis of the image of A is 2 0 1 0 0 2 0 1 . Thus, , I2 while a basis of the image is Thus, a basis of the kernel of T is 0 0 0 0 the rank of T is 1 and T is not an isomorphism. a b 0 c a b c −→ − T 206
ISM: Linear Algebra 9. Use a diagram: a b 0 c a b c −→ − T a 2b 0 c a 2b c
Section 4.3
−→ − A
A is invertible, so T is an isomorphism. 10. Use a diagram: a b 0 c a b c −→ − T
1 3
a a 1 0 0 The matrix A that transforms b into 2b is A = 0 2 0 . c c 0 0 1
3 −2 0 1
a b 0 c
−→ − A
Since A is invertible, T is an isomorphism. 11. Use Fact 4.3.2. to construct matrix B column by column: B= T 1 −1 0 0 T
B
1 0 0 The matrix is A = 2 3 −2 . 0 0 1
1 2 a 2a + 3b − 2c = 0 3 0 c a 2a + 3b − 2c c
0 1 0 1
T
B
0 1 0 0
=
B
1 −1 0 0
B
0 1 0 1
B
0 3 0 0
B
A is invertible, so T is an isomorphism. 12. Use a diagram as in Definition 4.3.2: 207
1 0 0 = 0 1 0 . 0 0 3
Chapter 4 a b cd −→ − T 2a 3b 2c 3d 2a 3b 2c 3d 0 3 0 0
ISM: Linear Algebra
a b c d
−→ − A
2 0 The matrix is A = 0 0
0 0 2 0
0 0 . 0 3
Since A is invertible, T is an isomorphism. 13. Use a diagram: a b cd −→ − T a+c b+d 2a + 2c 2b + 2d a+c b+d 2a + 2c 2b + 2d 0 1 0 2 1 0 2 0
a b c d
−→ − A
1 0 The matrix is A = 2 0
0 1 . 0 2
1 0 1 0 0 1 0 1 We find , to be a basis of the kernel of A, and , to be a basis of −1 0 2 0 0 −1 0 2 0 1 1 0 , a basis of the , the image of A. Thus, a basis of the kernel of T is 0 −1 −1 0 1 0 0 1 image of T is , , and T fails to be an isomorphism. 2 0 0 2 14. Use Fact 4.3.2 to construct matrix B column by column: 208
ISM: Linear Algebra 0 0 = 0 0 0 0 0 0 0 0 3 0 0 0 0 3
Section 4.3
B=
0 0 0 0
B
0 0 0 0
B
3 0 6 0
B
0 3 0 6
B
We find bases of the kernel and image of B, and use them to find that a basis of the 1 0 0 1 1 0 0 1 kernel of T is , , and a basis of the image is , . Thus −1 0 0 −1 2 0 0 2 the rank of T is 2, and T is not an isomorphism. 15. We use a diagram again: −→ − x + iy T x − iy x y −→ − A x −y 1 0 . 0 −1 0 1 . Since B is invertible, T must be an isomorphism. 1 0
Thus A =
A is invertible, so T is an isomorphism. 16. B = [[1 − i]B [1 + i]B ] = 17. Another diagram: −→ − x + iy T −y ix + x y −→ − A −y x 0 −1 . 1 0
A=
A is invertible, so T is an isomorphism. 18. x + iy x y −→ − T 2x − 3y + i(3x + 2y) 2x − 3y 3x + 2y
.
−→ − A
209
Chapter 4 2 −3 . Since A is invertible, T must be an isomorphism. 3 2
ISM: Linear Algebra
A=
19. We use a diagram to show our work: −→ − x + iy T px − qy + i(qx + py) x y −→ − A px − qy qx + py
A=
p −q . If p = q = 0, then T (z) = 0 for all z, so that the kernel is all of C, while q p the image is {0}. Otherwise, T is an isomorphism.
20. −→ − b + 2ct a + bt + ct2 T a b −→ − b A 2c c 0 0 1 0 1 Thus A = 0 0 2 . We see that a basis of the kernel of A is 0 and a basis of the 0 0 0 0 0 1 image of A is 0 , 2 . Thus, a basis of the kernel of T is 1, while a basis of the image 0 0 of T is 1, 2t. Thus, the rank of T is 2, and T is not an isomorphism. 21. We use a diagram to show our work: −→ − a + bt + ct2 T b − 3a + (2c 3b)t − 3ct2 − a b − 3a −→ − b 2c − 3b A c −3c −3 1 0 Thus A = 0 −3 2 . 0 0 −3
210
ISM: Linear Algebra A is invertible, so T is an isomorphism. 22.
Section 4.3
−→ − a + bt + ct2 T 4b + 2c + 8ct a 4b + 2c −→ − b A 8c c 0 1 0 4 2 Thus A = 0 0 8 . We find a basis of the kernel of A to be 0 , a basis of the 0 0 0 0 2 4 image of A to be 0 , 8 . From this we see that a basis of kernel of T is 1, while a 0 0 basis of the image of T is 4, 2 + 8t. Thus, T is not an isomorphism, since the rank of T is only 2. 23. A diagram shows our work: −→ − a + 3b + 9c a + bt + ct2 T a a + 3b + 9c −→ − b A 0 c 0
1 3 9 Thus A = 0 0 0 . 0 0 0 1 9 3 We find −1 , 0 to be a basis of the kernel of A, and 0 to be a basis of the 0 −1 0 image of A. Thus, a basis of the kernel of T is 3 − t, 9 − t2 , a basis of the image of T is 1 (the image is R), and T fails to be an isomorphism. 1 0 0 24. B = [[1]B [0]B [0]B ] = 0 0 0 . A basis of the kernel and image of B are obvious, 0 0 0 0 0 1 1 , 0 and 0 respectively. 0 1 0 211
Chapter 4
ISM: Linear Algebra
So we find a basis of the kernel of T to be t − 3, (t − 3)2 , and a basis of the image of T to be 1. Thus T has a rank of 1, and is not an isomorphism. 25. We use the following diagram: −→ − a + bt + ct2 T a − bt + ct2 a a −→ − −b b A c c 1 0 0 Thus A = 0 −1 0 . 0 0 1 26. −→ − a + 2bt + 4ct2 a + bt + ct2 T a a −→ − b 2b A c 4c 1 0 0 Thus A = 0 2 0 . Since A is invertible, T must be an isomorphism. 0 0 4 a + b(2t − 1) + c(2t − 1)2 −→ − a + bt + ct2 T = a − b + c + − 4c)t + 4ct2 (2b a−b+c a −→ − 2b − 4c b A 4c c 1 −1 1 2 −4 . Thus A = 0 0 0 4 212
A is invertible, so T is an isomorphism.
27. We use a diagram:
A is invertible, so T is an isomorphism.
ISM: Linear Algebra
Section 4.3 1 0 0 = 0 2 0 . Since B is invertible, T will be an 0 0 4
28. B = [1]B [2t − 2]B 4(t − 1)2 isomorphism.
B
29. This diagram shows our work: −→ − 2a + 2b + 8c/3 a + bt + ct2 T 2a + 2b + 8c/3 a −→ − b 0 A 0 c 2 2 8/3 Thus A = 0 0 0 . 0 0 0 4 1 1 3 We find −1 , 0 to be a basis of the kernel of A, and 0 to be a basis of the −1 0 0 image of A. Thus, a basis of the kernel of T is 1 − t, 4 − t2 , a basis of the image of T is 1 3 (the image is R), and T fails to be an isomorphism. 30. a + bt + ct = b + ch + 2ct b + ch a −→ − 2c b A 0 c 0 1 h 1 Thus A = 0 0 2 . A basis of the kernel of A is 0 , while a basis of the image of 0 0 0 0 1 h A is 0 , 2 . Thus, a basis of the kernel of T is 1 while a basis of the image of T is 0 0 1, h + 2t. So the rank of T is 2, and it is not an isomorphism. Recall that in calculus
f (t+h)−f (t) h 2 a+b(t+h)+c(t+h)2 −a−bt−ct2 h
−→ − T
is called the difference quotient of function f (t).
Geometrically, it represents the slope of the secant through points (t, f (t)) and 213
Chapter 4
ISM: Linear Algebra
(t + h, f (t + h)). The limit of the difference quotient, as h approaches 0, is the derivative f (t). Compare with Exercise 20. 31. a + bt + ct a −→ − b A c 0 1 0 Thus A = 0 0 2 . 0 0 0
2
−→ − T
a+b(t+h)+c(t+h)2 −a−b(t−h)−c(t−h)2 2h
= b + 2ct b 2c 0
1 1 0 We find 0 to be a basis of the kernel of A, and 0 , 2 to be a basis of the image 0 0 0 of A. Thus, a basis of the kernel of T is 1, a basis of the image of T is 1, 2t, and T fails to be an isomorphism. Geometrically, f (t+h)−f (t−h) is the slope of the secant through points (t − h, f (t − h)) and 2h (t + h, f (t + h)). This exercise shows that for a quadratic polynomial f (t) this secant is parallel to the tangent at the midpoint (t, f (t)), a fact that was known to Archimedes. Draw a sketch! Thus T (f (t)) = 32. a + b + c + (b + 2c)(t − 1) −→ − a + bt + ct2 T = a − c (b + 2c)t + a a−c −→ − b b + 2c A c 0 1 0 −1 −1 Thus A = 0 1 2 . We find a basis of the kernel of A to be 2 , while a basis 0 0 0 −1 214
f (t+h)−f (t−h) 2h
= f (t) in this case. Compare with Exercise 20.
ISM: Linear Algebra
Section 4.3
1 0 of the image of A is 0 , 1 . Thus, a basis of the kernel of T is −1 + 2t − t2 and a 0 0 basis of the image of T is 1, t. The rank of T is only 2, and T is not an isomorphism. Note that T (f (t)) = f (1) + f (1)(t − 1) is the equation of the tangent to the graph of f (t) at the point (1, f (1)). 33. B = [[1]B [t − 1]B 1 0 0 [0]B ] = 0 1 0 . 0 0 0
0 1 0 We find 0 to be a basis of the kernel of B, and 0 , 1 to be a basis of the image 1 0 0 of B. Thus, a basis of the kernel of T is (t − 1)2 , a basis of the image of T is 1, t − 1, and T fails to be an isomorphism. 34. A = T 1 0 0 0 T
A
0 1 0 0
T
A
0 0 1 0
T
A
0 0 0 1
A
0 0 0 0 0 0 0 0 0 −3 0 0 0 −3 0 0 = = . 0 0 A 0 0 A 3 0 A 0 0 A 0 0 3 0 0 0 0 0 0 0 0 1 −3 0 0 0 We find , to be a basis of the kernel of A, and , to be a basis of the 3 0 0 0 0 0 1 0 1 0 0 0 image of A. Thus, a basis of the kernel of T is , , a basis of the image of T 0 0 0 1 0 0 0 −3 , meaning that T has a rank of 2 and T fails to be an isomorphism. , is 3 0 0 0 35. Again, we use a diagram to show our work: M= a c b d −→ − T T (M ) = 0 0 0 1 1 0 c d−a −c + (d − a) =c 0 1 0 0 0 0 0 −c c d −a [T (M )]A = 0 −c
a b [M ]A = c d
−→ − A
215
Chapter 4 0 −1 Thus, A = 0 0 0 1 0 0 0 0 0 −1 0 1 . 0 0
ISM: Linear Algebra
0 1 1 0 By inspection, a basis of the kernel and image of this matrix are , and 0 0 0 1 0 1 −1 0 , , respectively. Thus, we see that a basis of ker(T ) is 0 0 0 −1 and a basis of im(T ) is 0 1 1 0 , , 0 0 0 1
0 −1 1 0 , . Thus, T fails to be an isomorphism. 0 0 0 −1
36. We will build A column-by-column: A= T 1 0 0 0 T
A
0 1 0 0
T
A
0 0 1 0
T
A
0 0 0 1
A
=
0 −1 1 0
A
−1 0 0 1
A
1 0 0 −1
A
0 1 −1 0
A
0 −1 1 0 0 1 −1 0 = . 1 0 0 −1 0 1 −1 0
−1 0 0 1 −1 0 0 1 We find to be a basis , to be a basis of the kernel of A, and , 0 1 −1 0 1 0 0 −1 1 0 0 1 of the image of A. Thus, a basis of the kernel of T is , , a basis of the 0 −1 −1 0 −1 0 0 −1 , the rank of T is 2 and T fails to be an isomorphism. , image of T is 0 1 1 0 37. We will construct our matrix B column-by-column: 2 2 −2 −2 2 −2 −0 2 −2 −2 0 = 0 0 0 2 0 0 0 0 0 0 0 0 . 0 0
B=
0−
B
B
[0 − 0]B [0 − 0]B
216
ISM: Linear Algebra
Section 4.3
0 −2 0 0 0 2 0 0 We find , to be a basis of the kernel of B, and , to be a basis of 0 0 0 1 0 0 1 0 0 1 the image of B. Thus, a basis of the kernel of T is I2 , , and a basis of the image 1 0 2 −2 −2 −2 , and T fails to be an isomorphism. , of T is 2 −2 2 2 38. We will construct our matrix B column-by-column: 0 0 0 0 −2 0 0 2 0 −2 0 0 B = [0]B [0]B = . 2 0 B 0 2 B 0 0 2 0 0 0 0 0 0 0 0 1 −2 0 0 0 We find , to be a basis of the kernel of B, and , to be a basis of the 2 0 0 0 0 0 1 0 1 0 0 1 image of B. From this we conclude that a basis of ker(T ) is , , a basis 1 0 0 −1 −2 0 0 2 of im(T ) is , and the rank of T is 2. So, T fails to be an isomorphism. 2 0 0 2 39. We use a diagram to show our work: M= a c b d −→ − T T (M ) = c−a d−b a c b − d −
c−a −→ − d − b A [T (M )]A = a−c b−d −1 0 1 0 1 0 −1 0 Then our A-matrix of T is: . 1 0 −1 0 0 1 0 −1 1 0 −1 0 0 1 0 −1 Thus, , is a basis of the kernel, and , is a basis of the image. 1 0 1 0 0 1 0 1 a b [M ]A = c d 217
Let
Chapter 4 1 0 0 1 , 1 0 0 1
ISM: Linear Algebra
From this we conclude that a basis of ker(T ) is
and a basis of im(T ) is
−1 0 0 −1 , . So, T fails to be an isomorphism. 1 0 0 1 40. We will construct our matrix A column-by-column: −4 0 4 0 1
2
A=
A
0 2 0 4
A
2 0 −2 0
A
0 2 0 4
A
0 −4 0 0 2 0 1 We find , , to be a basis of to be a basis of the kernel of A, and 0 4 0 1 4 0 −1 0 the image of A. From this we conclude that a basis of ker(T ) is is 1
1 2
−4 0 = 4 0
0 2 0 2 0 2 . 0 −2 0 4 0 4
−4 0 0 2 , . So, the rank of T is 2 and T fails to be an isomorphism. 4 0 0 4
0 1 0 , 0 −1 0
and a basis of im(T )
41. a. In Exercise 5 we consider the standard basis A of U 2×2 , and in Exercise 6 we work with the alternative basis B consisting of 1 0 0 1 0 1 , , . 0 0 0 0 0 1 Now SB→A = 1 0 0 0 0 1 0 0 0 1 0 1 1 0 0 = 0 1 1 . 0 0 1
A
A
A
42. a. If A is the standard basis considered in Exercise 8 and B is the basis in Exercise 7, then 218
1 0 0 b. Check that AS = SB = 0 1 3 . 0 0 3 1 0 0 −1 c. SA→B = SB→A = 0 1 −1 . 0 0 1
ISM: Linear Algebra 1 0 1 = 0 1 0 . 1 0 −1
Section 4.3
S=
1 0 0 1
A
0 1 0 0
A
1 0 0 −1
A
44. a. If A is the standard basis considered in Exercise 13 and B is the basis in Exercise 14, then 1 0 1 0 1 0 0 1 1 0 0 1 1 0 1 0 S= = . −1 0 A 0 −1 A 2 0 A 0 2 A −1 0 2 0 0 −1 0 2 0 0 b. Check that AS = SB = 0 0 0 0 0 0 3 0 6 0 0 3 . 0 6
43. a. If A is the standard basis considered in Exercise 10 then 1 0 1 0 1 1 −1 = −1 SB→A = 0 0 A 0 1 A 0 0 A 0 1 0 0 b. Check that AS = SB = −1 1 3 . 0 1 0 1 0 0 −1 c. SA→B = SB→A = 0 0 1 . 1 1 −1
0 0 0 b. Check that AS = SB = 0 0 4 . 0 0 0 1 1 0 2 2 −1 c. SA→B = SB→A = 0 1 0 . 1 0 −1 2 2
and B is the basis in Exercise 11, 0 0 1 1 . 1 0
45. a. If A is the standard basis considered in Exercise 15 and B is the basis in Exercise 16, then SB→A = [[1 + i]A [1 − i]A ] = 1 1 . 1 −1 219
Chapter 4 0 1 1 0
ISM: Linear Algebra 1 0 , respectively. Now 0 −1
b. In Exercises 16 and 15 we found B = check that AS = SB =
−1 c. SA→B = SB→A = 1 2
and A =
1 1 . −1 1
1 1 . 1 −1
46. a. If A is the standard basis considered in Exercise 23 and B is the basis in Exercise 24, then 1 −3 9 S = [1]A [−3 + t]A 9 − 6t + t2 A = 0 1 −6 . 0 0 1 1 0 0 b. Check that AS = SB = 0 0 0 . 0 0 0
−1 c. SA→B = SB→A
47. a. If A is the standard basis considered in Exercise 27 and B is the basis in Exercise 28, then 1 −1 1 SB→A = [1]A [−1 + t]A 1 − 2t + t2 A = 0 1 −2 . 0 0 1 1 0 b. In Exercises 28 and 27 we found B = 0 2 0 0 1 −2 tively. Now check that AS = SB = 0 2 0 0
−1 c. SA→B = SB→A
1 = 0 0
3 −9 1 6. 0 1
1 1 1 = 0 1 2 . 0 0 1
0 1 −1 1 0 and A = 0 2 −4 , respec4 0 0 4 4 −8 . 4
48. Use a diagram: 220
ISM: Linear Algebra a cos(t) b sin(t) + a b −→ − T b cos(t) − a sin(t) b −a
Section 4.3
−− −→ B
Thus B =
0 1 . −1 0 2 2 . Yes, T is an isomorphism, −2 2 b−1 −a a . b−1
49. B = [[2 cos(t) − 2 sin(t)]B [2 cos(t) + 2 sin(t)]B ] = since matrix B is invertible.
50. B = [[(b − 1) cos(t) − a sin(t)]B [a cos(t) + (b − 1) sin(t)]B ] =
51. Note that cos(t − π/2) = sin(t) and sin(t − π/2) = − cos(t). Thus B = [[cos(t − π/2)]B [sin(t − π/2)]B ] = [[sin(t)]B [− cos(t)]B ] = isomorphism. 52. Recall that cos(t − δ) = cos(δ) cos(t) + sin(δ) sin(t) and sin(t − δ) = cos(δ) sin(t) − sin(δ) cos(t). Also, cos(π/4) = sin(π/4) = Thus B = [[cos(t − π/4)]B [sin(t − π/4)]B ] =
√ 2 2
0 −1 . Yes, T is an 1 0
√ 2/2.
cos(t) +
√ 2 2
sin(t)
B
−
√ 2 2
cos(t) +
√ 2 2
sin(t)
B
=
√ 2 2
1 −1 1 1
53. Recall that cos(t − θ) = cos(θ) cos(t) + sin(θ) sin(t) and sin(t − θ) = cos(θ) sin(t) − sin(θ) cos(t). Thus B = [[cos(t − θ)]B [sin(t − θ)]B ] = [[cos(θ) cos(t) + sin(θ) sin(t)]B [− sin(θ) cos(t) + cos(θ) sin(t)]B ] = Yes, T is an isomorphism. Note that B is a rotation matrix. 221 cos(θ) sin(θ) − sin(θ) . cos(θ)
Chapter 4
ISM: Linear Algebra
1 5 1 54. Note that the two basis vectors 1 and −4 are perpendicular. Thus T 1 = −1 1 −1 0 1 0 5 1 1 and T −4 = 0 . Now B = 1 0 = 1 0 . 0 0 −1 B 0 B 0 1 −1 1 1 1 55. Let u = √6 −2 be the unit vector in the direction of −2 . 1 1 1 Now T 1 = −1 5 T −4 = 1 1 1 1 1 1 1 −2 · 1 −2 = − 3 −2 and 6 1 1 −1 1
−5 14 0 14 B = 4 14 = . −1 0 −1 B −14 B 1 −6 1 5 5 3 1 57. T 1 = 3 = − 1 − −4 and T −4 = 3 = 3 1 . −1 0 −1 1 1 −3 −1 −6 3 −1 3 Thus B = 3 3 = . −1 0 0 B −3 B 222
1 1 2 −14 1 1 Now B = − 3 −2 7 −2 = 9 . 3 −1 7 1 B 1 B 1 1 1 −5 5 1 5 14 56. T 1 = 2 × 1 = 4 and T −4 = 2 × −4 = 14 . Thus −1 3 −1 −1 1 3 1 −14
1 1 Also note that −2 = − 2 1 + 3 1 −1
1 5 1 1 −2 · −4 −2 = 6 1 1 1
1 5 1 −4 , so that −2 = 3 1 B 1
1 7 −2 , by Fact 2.2.1. 3 1
1 3
−2 . 1
ISM: Linear Algebra
Section 4.3
1 3 1 5 0 3 0 58. T 1 = 3 = 3 1 and T −4 = 0 . Thus B = 3 0 = −1 −3 −1 1 0 −3 B 0 B 3 0 . 0 0 59. T (f ) = t · f is linear and ker(T ) = {0}, but T is not an isomorphism since the constant function 1 is not in the image of T .
60. a. We will use Fact 4.3.3. SB→A
1 = 2 2 A
−1 b. Here, we know SA→B = SB→A =
1 −1 . 0 1
3 0 = 1 1 . 0 1 3 A
c. Fact 4.3.4 reveals that [ v1
v2 ] = [ u1
u2 ] SB→A =
A
61. a. We will use Fact 4.3.3. SB→A =
5 10 −10 A 5 and is thus a reflection combined with a scaling.
1 −9−16
−3 4 4 3
= 5
3 −5 4 5
4 5 3 5
,
−1 b. Here, we know SA→B = SB→A =
3 −4 3 −4 1 = − 25 = −4 −3 −4 −3 u2 ] SB→A
1 25
−3 4 . 4 3
c. Fact 4.3.4 reveals that [ v1
v2 ] = [ u1
62. a. Finding this basis is equivalent to finding a basis of the kernelof 1 −2 2 ] [ that 2 4 does not contain any zeroes. We can quickly spot the vectors 2 and 1 , so 1 −1 2 4 B = 2 , 1 , for example. 1 −1 2 4 1 = 2 1 . b. From Fact 4.3.3, SB→A = 2 1 −1 1 A −1 A
−1 c. SA→B = SB→A = 1 −3
−1 −1 = −1 2
1 3
1 1 . 1 −2
d. Fact 4.3.4 reveals that [ v1
v2 ] = [ u1
u2 ] SB→A
223
Chapter 4
ISM: Linear Algebra
63. a. Finding this basis is equivalent to finding a basis the kernel [ 1 3 −2 ]. We can of of 0 2 0 2 quickly spot the vectors 0 and 2 , so B = 0 , 2 , for example. 3 1 3 1 b. From Fact 4.3.3, SB→A
−1 c. SA→B = SB→A = 1 −3+1
2 = 0 1 A
0 2 = −1 −1 . 1 3 3 A
1 2
3 1 3 1 1 = −2 = −1 −1 −1 −1 v2 ] = [ u1 u2 ] SB→A
−3 −1 . 1 1
d. Fact 4.3.4 reveals that [ v1
64. a. We find A column-by-column: A = [I2 ]B [P ]B P 2
B
−3 b. We find a basis of the kernel of A to be 4 , and a basis of the image of A to be −1 1 1 0 , 2 . Thus, −3 4 is a basis of the kernel of T , and I2 , P is a basis of the 0 −1 1 3 image of T. 65. a. P 2 = a2 + bc ab + bd b = = ac + cd bc + d2 d 1 0 bc − ad (bc − ad) . So [P 2 ]B = . 0 1 a+d a c
2
1 1 1 = 0 2 8. 1 3 9
a2 + bc (a + d)c
(a + d)b bc + d2
= (a + d)
a c
b + d
b. We will do this column-by-column: B = [[T (I2 )]B [T (P )]B ] = [P ]B P 2 =
B
0 bc − ad , by part a. T is an isomorphism if bc − ad = 0, that is, if P is 1 a+d invertible. c. Assume that bc − ad = 0. Then the B-matrix of T is B = span 0 0 . So, ker(B) = 1 a+d
0 a+d . Thus a basis of ker(T ) is (a + d)I2 − P = and im(B) =span 1 −1 d −b , a basis of im(T ) is P , and the rank of T is 1. −c a 224
ISM: Linear Algebra
Section 4.3 0 −1 0 = 2 0 −2 . B 0 1 0 0 −1 Now im(B) = span 2 , 0 and 0 1
66. a. B =
T (x2 ) 1
B
[T (x1 x2 )]B T (x2 ) 2
1 0 −1 b. Note that rref(B) = 0 1 0 . 0 0 0 1 ker(B) = span 0 . Thus 2x1 x2 , −x2 + x2 is a basis of im(T ) and x2 + x2 is a basis 1 2 1 2 1 of ker(T ). [T (cos t)]B [T (sin t)]B = [0]B = 0 = [0]B = 0
67. a.
1 p 0 q b. The equation T (f ) = cos(t) corresponds to M x = , with solutions x = , 0 0 1 0 2 1 where p and q are arbitrary. Thus f (t) = p cos(t) + q sin(t) + 2 t sin(t). In Figure 4.2 we graph f (t) for p = q = 0; note that p cos(t) + q sin(t) is just a sinusoidal function
0 −2 [T (t cos(t))]B = [−2 sin(t)]B = 0 0 2 0 [T (t sin(t))]B = [2 cos(t)]B = 0 0 0 0 0 2 0 0 −2 0 so M = . 0 0 0 0 0 0 0 0
68. a. • The sequence (0, 0, 0, . . .) is in W . • If the sequences (xn ) and (yn ) are in W (that is, xn+2 = xn+1 + 6xn and yn+2 = yn+1 + 6yn for all n), then xn+2 + yn+2 = xn+1 + yn+1 + 6(xn + yn ) so that the sequence (xn + yn ) is in W as well. 225
Chapter 4
2π π –π –π –2π π
ISM: Linear Algebra
t 2
1 t sin (t) 2
2π
t –2
3π
t
Figure 4.2: for Problem 4.3.67b. • If the sequence (xn ) is in W and k is any constant, then kxn+2 = kxn+1 + 6kxn , so that the sequence (kxn ) is in W as well.
b. A sequence in W is determined by its first two components (a and b, say), which we can choose freely. All the later components can then be expressed in terms of a and b, since xn+2 = xn+1 + 6xn : (a, b, b + 6a, 6a + 7b, 42a + 13b, . . .) = a(1, 0, 6, 6, 42, . . .) + b(0, 1, 1, 7, 13, . . .). The two sequences (1, 0, 6, 6, 42, . . .) and (0, 1, 1, 7, 13, . . .) form a basis of W , so that dim(W ) = 2. c. It is required that cn+2 = cn+1 + 6cn for all n ≥ 0 or c2 = c + 6 or c2 − c − 6 = 0
or (c − 3)(c + 2) = 0 The solution are c1 = 3 and c2 = −2. The geometric sequences (1, 3, 9, 27, 81, . . .) and (1, −2, 4, −8, 16, . . .) are in W .
d. Yes, the two sequences we found in part c do the job, since dim(W ) = 2. e. (x0 , x1 , x2 , x3 , x4 , . . .) = (0, 1, 1, 7, 13, . . .). We are looking for constants p and q such that (x0 , x1 , x2 , x3 , . . .) = (0, 1, 1, 7, . . .) = p(1, 3, 9, 27, . . .) + q(1, −2, 4, −8, . . .). To find p and q, it suffices to consider the first two components: p=
1 5 1 1 and q = − 5 . Thus xn = 5 3n − 1 (−2)n . 5
0=p+q , so that 1 = 3p − 2q
226
ISM: Linear Algebra f (a ) 1 1 f1 (a2 ) 69. As the hint suggests, we find the kernel of M = . . . f (a ) c1 1 1 c2 f1 (a2 ) If M . = . . . . . 0 0 . , . . 0 cn f2 (a1 ) f2 (a2 ) . . . ··· ··· f2 (a1 ) f2 (a2 ) . . . ··· ···
Section 4.3 fn (a1 ) fn (a2 ) . . . .
f1 (an ) f2 (an )
· · · fn (an )
f1 (an ) f2 (an ) · · · fn (an )
fn (a1 ) c1 c1 f1 (a1 ) + · · · + cn fn (a1 ) fn (a2 ) c2 c1 f1 (a2 ) + · · · + cn fn (a2 ) = . . = . . . . . . . cn c1 f1 (an ) + · · · + cn fn (an )
then the polynomial c1 f1 + · · · + cn fn in Pn−1 has at least n zeros, namely, a1 , a2 , · · · , an . It follows that c1 f1 + c2 f2 + · · · + cn fn = 0 and therefore c1 = c2 = · · · = cn = 0 since the fi are linearly independent. We have shown that ker(M ) = {0}, so that M is invertible. 70. Yes; suppose dim(V ) = dim(W ) = n, and let B and C be bases of V and W , respectively. Then the coordinate transformation TB and TC define isomorphisms from V to Rn and from W to Rn , respectively.
−1 TC ◦ TB is an isomorphism from V to W (see Exercise 4.2.77).
71. We need to show that there are constants w1 , . . . , wn such that
1
w1 f1 (a1 ) + w2 f1 (a2 ) + · · · + wn f1 (an ) = w1 f2 (a1 ) + w2 f2 (a2 ) + · · · + wn f2 (an ) = . . . . . .
f1
−1 1
f2
−1 1
w1 fn (a1 ) + w2 fn (a2 ) + · · · + wn fn (an ) =
fn .
−1
In order to be able to use the matrix M introduced in Exercise 69, as the hint suggests, we write this system in a somewhat unusual way, as f1 (a1 ) f2 (a1 ) · · · fn (a1 ) 1 1 1 f (a ) f2 (a2 ) · · · fn (a2 ) [ w1 w2 . . . wn ] 1 2 f2 · · · fn f1 = ··· ··· ··· ··· −1 −1 −1 f1 (an ) f2 (an ) · · · fn (an )
M
227
Chapter 4 Since M is invertible (Exercise 69), we have the unique solution
1 1 1
ISM: Linear Algebra
[ w1
w2
. . . wn ] =
−1
f1
−1
f2
···
n
fn M −1 .
−1
Now if f is any polynomial in Pn−1 and f =
j=1 1 n 1 n n n
cj fj , then
n n
f=
−1 j=1
cj
−1
fj =
j=1
cj
i=1
wi fj (ai ) =
i=1
wi
j=1
cj fj (ai ) =
i=1
wi f (ai ),
as claimed. 72. If we work with the basis f1 (t) = 1, f2 (t) = t, and f3 (t) = t2 of P2 , then we have to solve the system w1 + w 2 + w 3 −w1 + w2 w1 + w 2 that
1
=2 = 0 (see Exercise 71). The solution is w1 = w3 = =2 3
1 3
4 and w2 = 3 , so
1 4 1 f (−1) + f (0) + f (1) for f in P2 . This is what you get when you apply 3 3 3 −1 Simpson’s Rule (with two subintervals) to f ; note that Simpson’s Rule gives the exact value of the integral in this case. f =
True or False
1. T; check the three properties listed in Definition 4.1.2. 2. T; by Definition 4.2.1. 3. F; A basis of R2×3 is 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 , , , , , 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0
0 0 0 , so it has a dimension of 6. 0 0 1 4. T; check with Definition 4.1.3c. 5. T; The linear transformation T (ax + b) = a + ib is an isomorphism from P1 to C, with the inverse T −1 (a + ib) = ax + b. 6. T; by Fact 4.2.4c. 228
ISM: Linear Algebra 7. T; This fits all properties of Definition 4.1.2. 8. F; The transformation T could be: T (f ) =
True or False
0 0 , in which case the kernel would be 0 0 all of P6 and the dimension of the kernel would be 7.
9. T; We are looking at P6 , with a basis 1, t, t2 , t3 , t4 , t5 , t6 , which has seven elements. 10. True, we can check both requirements of Definition 4.2.1. 11. F; dim(P3 )= 4, so the three given polynomials cannot span P3 . 12. T; We can construct a basis of V by omitting the redundant elements from a list of ten elements that span V . Thus dim(V ) ≤ 10. 13. F; det 1 0 0 0 + det 0 0 1 0 = 0 = det = 1. 0 1 0 1
14. F; For any matrix A, the space of matrices commuting with A is at least two-dimensional. Indeed, if A is a scalar multiple of I2 , then A commutes with all 2 × 2 matrices, and if A fails to be a scalar multiple of I2 , then A commutes with the linearly independent matrices A and I2 . 15. F; t3 , t3 + t2 , t3 + t, t3 + 1 is a basis of P3 . 16. T; If T is linear and invertible, then T −1 will be linear and invertible as well. 17. F; T (sin(x)) = sin(x) − sin(x) = 0. 18. F; T (f ) = 0 0 0 0 is not an isomorphism. 1 −1 . Now im(A) = ker(A) = span 1 −1 1 1 .
19. F; Let V =R2 , A =
20. T; the dimensions of both spaces are the same: 10. 21. T; If T (f (t)) = f (t2 ) = 0, f (t) must also be zero. 22. T; The inverse is T −1 (N ) = S −1 N S −1 . 23. T; Let A = 1 1 . Then we want 0 0 1 1 0 0 a c b a = d c b d 1 1 , or 0 0
a a a+c b+d . Thus, c = 0 and a = b + d. So our space is the span of = c c 0 0 1 0 0 0 and . 0 1 1 −1 229
Chapter 4
ISM: Linear Algebra
24. T; Let our basis be
1 0 1 0 0 1 0 1 , , , . Each matrix here is invertible, 0 1 0 −1 1 0 −1 0 and also clearly none are redundant.
25. F; T (f (t)) = f (t) is not an isomorphism. 26. T; We need only show that either the new list contains no redundant elements, or spans the whole space. The latter is slightly easier to show. Since f1 , f2 , f3 form a basis of V , it suffices to show that these three elements are in the span of f1 , f1 + f2 , f1 + f2 + f3 . This is simple to demonstrate: f2 = (f1 + f2 ) − f1 , and f3 = (f1 + f2 + f3 ) − (f1 + f2 ). 27. T; We show that none of the polynomials is redundant; let’s call them f (x), g(x) and h(x). Now g(x) isn’t a multiple of f (x) since f (b) = 0, but g(b) = 0. Likewise, h(x) isn’t a linear combination of f (x) and g(x) since f (c) = g(c) = 0, but h(c) = 0. 28. T; Make the substitution 4t − 3 = s to see that the inverse is T −1 (g(s)) = g( s+3 ). 4 29. F; T a c b d = 1 2 3 6 a c b a + 2c b + 2d = d 3a + 6c 3b + 6d 1 0 3 0 and 0 1 , 0 3
= (a + 2c)
0 1 1 0 . So the image is the span of + (b + 2d) 0 3 3 0 and rank(T )= 2.
30. T; If the basis B we consider is f1 , f2 , then the given matrix tells us that T (f1 ) = 3f1 and T (f2 ) = 5f1 + 4f2 . Thus f = f1 does the job. 31. F; The space spanned by 1 0 0 0 and 0 1 0 0 contains no invertible matrices.
32. F; This is the change of basis matrix from B to A. The change of basis matrix we are 1 −1 . looking for is: 0 1 33. F; Let B = (f, g) and C = (g, f ). The fact that 1 2 is the B- matrix of T implies that 3 4 3 1 , meaning that the second , or T (f ) = f + 3g. But then [T (f )]C = [T (f )]B = 1 3 2 1 3 fails to be the . This shows that the matrix column of the C-matrix of T is 4 3 1 C-matrix of T .
34. T; The image of T is Pn−1 , so that rank(T ) = dim (imT ) = dim(Pn−1 ) = n. 35. T; because the matrix is invertible. 230
ISM: Linear Algebra
True or False
36. T; The dimension of P9 is 10, and the dimension of R3×4 is 12. Thus, any 10-dimensional subspace of R3×4 will be acceptable. For example, we can consider the space of all 3 × 4 matrices A with a11 = a12 = 0. 37. T; let W1 be {0}. Then any other subspace W2 unioned with W1 will simply be W2 again, which we know is a subspace. 38. T; Let T (a0 + a1 t + a2 t2 + · · · + a5 t5 + · · ·) = a0 + a1 t + a2 t2 + · · · + a5 t5 . The image of this transformation is clearly all of P5 , and T satisfies the requirements of Definition 4.2.1. 39. F; P2 is a subspace of P , and P is infinite dimensional. a b c b c 40. T; Let T d e f = . We can easily see that the kernel and image of this 0 f g h i transformation are exactly as required. 41. T; there will be no redundant elements in this list. 42. F; The kernel of T consists of all constant functions. 43. T; We apply the rank-nullity theorem: dim(W ) = dim(im(T )) = dim(P4 )−dim(ker(T )) = 5 − dim(ker(T )) ≤ 5. 44. F; We can construct as many linearly independent elements in ker(T ) as we want, for 1 example, the polynomials f (t) = tn − n+1 , for all positive integers n. 45. T; 0 is in our set, and if f and g are in our set, then T (f + g) = T (f ) + T (g) = f + g so that f + g is in V as well. Also, if f is in V and k is an arbitrary scalar, then T (kf ) = kT (f ) = kf , so kf is in V as well. 46. F; Consider the transformation that drops the constant term: T (c0 + c1 t + · · · + c6 t6 ) = c 1 t + · · · + c 6 t6 . 47. T; Let P = I2 , Q = −I2 . Then T (M ) = I2 M − M (−I2 ) = 2M , which is an isomorphism. 48. F; We use dimension arithmetic here to show that this cannot happen. Any transformation T from P6 to C must have a kernel of at least 5 dimensions, since P6 is 7-dimensional and C is only a 2-dimensional space. Thus, any such kernel cannot be isomorphic to R2×2 , which is a 4-dimensional space. 49. F; If f = −f1 , then 0 is a member of the list! 50. T; Consider the space of all matrices of the form 231 a −b , for example. b a
Chapter 4
ISM: Linear Algebra
51. T; note that dim(P11 ) = 12 = dim(R3×4 ). The linear spaces P11 and R3×4 are both isomorphic to R12 , via the coordinate transformation, and thus they are isomorphic to each other. 52. F; Consider the linear transformation T (f (t)) = f (t) from P2 to P , for example. 53. T; We use the rank-nullity theorem: dim(V ) =dim(im(T ))+dim(ker(T ))=dim(im(T ))≤ dim(R2×2 ) = 4. 54. T; Using the fundamental theorem of calculus, we can write g(t) = T (f (t)) = 3f (3t + 4). Make the substitution 3t + 4 = s to see that the inverse is T −1 (g(s)) = g((s − 4)/3)/3. 55. T; Using a coordinate transformation, it suffices to show this for R4 . For every real number k, we define the three dimensional subspace Vk of R4 consisting of all vectors x such that x4 = kx3 . If c is different from k, then Vc and Vk will be different subspaces of 0 0 R4 , since Vk contains the vector , but Vc does not. Thus we have generated infinitely 1 k many distinct three-dimensional subspaces Vk of R4 , one for every real number k. 56. T; If the basis B we consider is f1 , f2 , then the given matrix tells us that T (f1 ) = 3f1 and T (f2 ) = 5f1 + 4f2 . We are looking for a nonzero f = af1 + bf2 such that T (f ) = 4f . Now T (f ) = aT (f1 ) + bT (f2 ) = 3af1 + 5bf1 + 4bf2 = (3a + 5b)f1 + 4bf2 must be equal to 4f = 4af1 + 4bf2 . Thus it is required that 3a + 5b = 4a, or a = 5b. For example, f = 5f1 + f2 does the job. 57. T; This is logically equivalent to the following statement: If the domain of T is finite dimensional, then so is the image of T . Compare with Exercises 4.2.81a and 4.1.57. 58. F; If A is a scalar multiple of I2 , then all 2×2 matrices commute with A, so that the space a b of commuting matrices is 4 - dimensional. If A = fails to be a scalar multiple c d a b x y x y a b , which amounts to the = of I2 , consider the equation c d z t z t c d system cy − bz = 0, bx + (d − a)y − bt = 0, cx + (d − a)z − ct = 0. If b = 0, then the first two equations are independent; if c = 0, then the first and the third equation are independent; and if a = d, then the second and the third equation are independent. Thus the rank of the system is at least two and the solution space is at most two-dimensional. (The solution space is in fact two -dimensional, since A and I2 are independent solutions.) 59. T; If A = 0, then we are done. If rank(A) = 1, then the image of the linear transformation T (M ) = AM from R2×2 to R2×2 is two dimensional (if v is a basis of im(A), then v 0 , 0 v is a basis of im(T )). Since the three matrices AB = T (B), AC = T (C) , and AD = T (D) are all in im(T ), they must be linearly dependent. 232
ISM: Linear Algebra
True or False
60. F; Consider two distinct three-dimensional subspaces W1 and W2 of P4 . Since the spaces W1 and W2 are distinct, neither of them is a subspace of the other, so that we can find a polynomial f1 that is in W1 but not in W2 as well as an f2 that is in W2 but not in W1 . Then f1 and f2 are both in the union of W1 and W2 , but f1 + f2 isn’t. 61. T; Pick the first redundant element fk in the list. Since the elements f1 , . . . , fk−1 are linearly independent, the representation of fk as a linear combination of the preceding elements will be unique. 62. F; T (I3 ) = P − P = 0, and T can never be an isomorphism. 63. T; Let W = span(f1 , f2 , f3 , f4 , f5 ) = span(f2 , f4 , f5 , f1 , f3 ). If we omit the two redundant elements from the first list, f1 , f2 , f3 , f4 , f5 , we end up with a basis of W with three elements, so that dim(W ) = 3. If we omit the redundant elements from the second list, f2 , f4 , f5 , f1 , f3 , we end up with a (possibly different) basis of W , but that basis must consist of 3 elements as well. Thus there must be two redundant elements in the second list. 64. F; The dimensions of the kernel and image would have to be equal, and both add up to the dimension of P6 , which is the odd number 7. 65. T; Consider the proof of the rank nullity theorem outlined in Exercise 4.2.81. In the proof, we use bases of ker(T ) and im(T ) to construct a basis of the domain. 66. F; If the basis B we consider is f1 , f2 , then the given matrix tells us that T (f1 ) = 3f1 and T (f2 ) = 5f1 + 4f2 . We are looking for a nonzero f = af1 + bf2 such that T (f ) = 5f . Now T (f ) = aT (f1 ) + bT (f2 ) = 3af1 + 5bf1 + 4bf2 = (3a + 5b)f1 + 4bf2 must be equal to 5f = 5af1 + 5bf2 . Thus it is required that 3a + 5b = 5a and 4b = 5b, implying that a = b = 0. We are unable to find a nonzero f with the desired property. 67. T; Consider a 3-dimensional subspace W of R2×2 . The matrices x y in W can be z t described by a single linear equation ax + by + cz + dt = 0 , where at least one of the coefficients is nonzero.
Suppose x is the leading variable (meaning that a = 0), and y, z and t are the free ∗ 1 in W variables. We can choose y = z = 1 and t = 0, and the resulting matrix 1 0 will be invertible. We represent x by a star, since its value does not affect the invertibility. If y is the leading variable and the other three are the free variables, then we can construct 1 ∗ the invertible matrix in W . If z is the leading variable, we have the invertible 0 1 0 1 1 0 . . Finally, for the leading variable t we have matrix 1 ∗ ∗ 1 233