VIEWS: 4 PAGES: 6 POSTED ON: 3/25/2010 Public Domain
Answers to Exercise Set II.1. Drills 1. (Outline only) (a) Suppose a(1, 1, 0) + b(1, 2, 0) = (0, 0, 0). Then a + b = 0, a + 2b = 0, which gives a = 0 and b = 0. (b) Suppose a(1, 1, i)+b(1, i, 1)+c(i, 1, 1) = (0, 0, 0). Then a+b+ic = 0, a+ib+c = 0, ia + b + c = 0. Adding all three equations, we have (2 + i)(a + b + c) = 0 and hence a + b + c = 0. Now there is no diﬃculty to get a = 0, b = 0 and c = 0. (c) Suppose ap(x) + bp′ (x) + cp′′ (x) = 0, or a(1 + x + x2 ) + b(1 + 2x) + 2c = 0, or a + b + 2c + (a + 2b)x + ax2 = 0. Thus a + b + 2c = 0, a + 2b = 0 and a = 0. So a = 0, b = 0 and c = 0. (d) Suppose ap(x) + bp(x + 1) + cp(x + 2) = 0. Then ax2 + b(x + 1)2 + c(x + 2)2 = 0, or (a + b + c)x2 + (2b + 4c)x + b + 4c = 0. Thus a + b + c = 0, 2b + 4c = 0 and 4c = 0, which give a = 0, b = 0 and c = 0. . (e) Suppose aA + bA2 = O. Then 1 −1 1 1 0 0 a +b = . 0 −2 0 4 0 0 Thus we have a + b = 0, −a + b = 0 and −2a + 4b = 0, which give a = 0, b = 0 and c = 0. 2. (Outline only) (a) 87(1, 1, 0) + (1, 2, 0) − (88, 89, 0) = (0, 0, 0). (b) (1, i, 1) − i(i, 1, i) − 2(1, 0, 1) = (0, 0, 0). (c) p(x) − 3p(x + 1) + 3p(x + 2) − p(x + 3) = 0. 1 −1 1 1 1 −3 0 0 (d) −2A + A2 + A3 = O, or −2 + + = . 0 −2 0 4 0 −8 0 0 3. (Outline only) (a) Suppose a(u + v) + b(u − v) = 0, or (a + b)u + (a − b)v = 0. Hence a + b = 0 and a − b = 0, which gives a = b = 0. (b) Suppse a(u + v) + b(u + w) + c(v + w) = 0, or (a + b)u + (a + c)v + (b + c)w = 0. Thus a + b = 0, a + c = 0, b + c = 0, which give a = 0, b = 0 and c = 0. 1 (c) Suppose a(u + iv) + b(iu + v) = 0, or (a + bi)(u + (ai + b)v = 0. So a + bi = 0 and i(a − ib) = ai + b = 0. Hence a = b = 0. 4. True or False: (a) False. Need the condition v = 0 to turn it into a true statement. (b) True. (c) False. The following correction turns it into a true statement: if two nonzero vectors are linearly dependent, then each of them is a scalar multiple of the other. (d) False. (e) True. (f) False. 5. (Outline only) (a) Suppose a1 A1 + a2 A2 + a3 A3 = O. Then a1 BA1 + a2 BA2 + a3 BA3 = B(a1 A1 + a2 A2 + a3 A3 ) = O and hence a1 = a2 = a3 = 0. (b) Suppose a1 p1 (x) + a2 p2 (x) + a3 p3 (x) = 0. Then d a1 p′ (x) + a2 p′ (x) + a3 p′ (x) = 1 2 3 (a1 p1 (x) + a2 p2 (x) + a3 p3 (x)) = 0 dx and hence a1 = a2 = a3 = 0. 6. Find the leading vectors and express the other vectors as their linear combinations (a) Using the standard basis {1, x, x2 }, the coordinate vectors of the given polyno- mials are arranged as the columns of 1 2 −1 0 0 1 1 2 0 1 0 4 1 2 1 2 −2 5 ∼ 0 0 1 1 0 3 . 0 0 0 0 1 1 0 0 0 0 1 1 Hence the leading vectors are x + 1, x − 1 and x2 − 2x. Furthermore, 2x + 2 = 2(x + 1), 2x = (x + 1) + (x − 1), x2 + 5x + 1 = 4(x + 1) + 3(x − 1) + (x2 − 2x). (b) Arranged the given vectors as columns of 1+i i 1 1 1 (1 + i)/2 0 (3 + 4i)/4 ∼ . 1−i 1 i 2 0 0 1 (1 − 2i)/2 Leading vectors are (1 + i, 1 − i) and (1, i). Also, 1+i 3+i 1 − 2i (i, 1) = (1 + i, 1 − i), (1, 2) = (1 + i, 1 − i) + (1, i). 2 4 2 2 (c) Leading vectors are C1 , C3 , C4 . Also, C2 = 2C1 , C5 = 5C1 + 2C3 + 6C4 , C6 = 3C1 + 4C3 + 7C4 . 7. Invariant subspaces in Example 1.3.1 for operators D and Ta . (a) Invariance of V1 for D: D(1) = 0, D(x) = 1, D(x2 ) = 2x are in V1 . (b) V1 Invariant for Ta : Ta (1) = 1, Ta (x) = a + x, Ta (x2 ) = a2 + 2ax + x2 are in V1 . (c) Invariance of V2 for D: D(ekx ) = kekx , D(xekx ) = ekx + kxekx , D(x2 ekx ) = 2xekx + kx2 ekx are vectors in V2 . (d) Invariance of V2 for Ta : Ta (ekx ) = eka ekx , Ta (xekx ) = (aeka )ekx + (eka )xekx , Ta (x2 ekx ) = (a2 eka )ekx + (2aeka )(xekx ) + eka (x2 ekx ), which are vectors in V2 . (e) Invariance of V4 for D: D(cos bx) = (−b) sin bx, D(sin bx) = b cos bx, D(x cos bx) = cos bx − bx sin bx D(x sin bx) = sin bx + bx cos bx D(x2 cos bx) = 2x cos bx − bx2 sin bx D(x2 sin bx) = 2x sin bx + bx2 cos bx which are vectors in V4 . (f) Invariance of V4 for Ta : Ta (cos bx) = (cos ba) cos bx + (− sin ba) sin bx Ta (sin bx) = (sin ba) cos bx + (cos ba) sin bx Ta (x cos bx) = (a cos ba) cos bx + (−a sin ba) sin bx + (cos ba)x cos bx + (− sin ba)x sin bx Ta (x sin bx) = (a sin ba) cos bx + (a cos ba) sin bx + (sin ba)x cos bx + (cos ba)x sin bx Ta (x2 cos bx) = (a2 cos ba) cos bx + (−a2 sin ba) sin bx + (2a cos ba)x cos bx + (−2a sin ba)x sin bx + (cos ba)x2 cos bx + (− sin ba)x2 sin bx Ta (x2 sin bx) = (a2 sin ba) cos bx + (a2 cos ba) sin bx + (2a sin ba)x cos bx + (2a cos ba)x sin bx + (sin ba)x2 cos bx + (cos ba)x2 sin bx 3 which are vectors in V4 . Exercises 1. Suppose a1 v1 + a2 v2 + · · · + ar vr = 0. Then a1 T v1 + a2 T v2 + · · · + ar T vr = T (a1 v1 + a2 v2 + · · · + ar vr ) = 0. Since T v1 , T v2 , . . . , T vr are linearly independent, we have a1 = a2 = · · · = ar = 0. This shows the independence of v1 , v2 , . . . , vr . Take any set of linearly independent set of vectors v1 , v2 , . . . vr , and let T = O, then . Then T v1 , T v2 , . . . T vr are zero vectors and hence are linearly dependent. This shows that the converse of this statement is false. 2. Suppose that v1 , v2 , . . . , vr are linearly independent. Then v1 = 0. For each k with 2 ≤ k ≤ r, vk is not in the linear span of v1 , v2 , . . . , vk−1 . Indeed, if this is not the case, we can write vk = b1 v1 +b2 v2 +· · ·+bk−1 vk−1 for some scalars b1 , b2 , . . . , bk−1 . Thus we have the nontrivial linear relation a1 v1 + a2 v2 + · · · + ar vr = 0 where aj = bj for j < k, ak = −1 and aj = 0 for j > k. This contradicts the linear independence of v1 , v2 , . . . , vr . Now we prove the “if” part. Assume the contrary that v1 , v2 , . . . , vr are linearly dependent, say, it satisﬁes a nontrivial linear relation a1 v1 + a2 v2 + · · · + ar vr = 0. At least one of aj is nonzero. Let k be the largest index for which ak = 0. It follows that aj = 0 for j > k and hence a1 v1 + a2 v2 + · · · + ak vk = 0 with ak = 0. If k = 1, we have a1 v1 = 0 and since a1 = ak = 0, we have v1 = 0. So we assume k > 0. In this case we have vk = b1 v1 + b2 v2 + · · · + bk−1 vk−1 where bj = −aj /ak . Thus vk can be expressed as a linear combination of v1 , v2 , . . . , vk−1 , contradicting our given condition. 3. (a) Suppose a0 v + a1 T v + a2 T 2 v + · · · + an T n v = 0. (∗) Applying T n to this identity, we get a0 T n v +a1 T n+ 1 v +a2 T n+ 2 v +· · ·+an T 2n v = 0, which becomes a0 T n v = 0 in view of T n+ 1 v = 0. Since T n v = 0, we have a0 = 0. Suppose that we already have a0 = a1 = · · · = ak−1 = 0 for some k ≤ n. Then (∗) becomes ak T k v + ak+ 1 T k+ 1 v + · · · + an T n v = 0. Applying T n−k to the last identity, 4 we obtain ak T n v + ak+ 1 T n+ 1 v + · · · + an T 2n−k v = 0, which can be rewritten as ak T n v = 0 in view of T n+ 1 v = 0. Since T n v = 0, we have ak = 0. Now it is clear that a0 = a1 = · · · = an = 0. (b) When p(x) is a polynomial of degree n, then the degree of p′ (x) is n − 1, the degree of p′′ (x) is n − 2 and so forth. Finaly the degree of p(n − 1)(x) is 1, p(n) (x) is a nonzero constant and p(n+ 1) (x) = 0. Thus part (a) is applicable. 4. We have 0 1 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 0 0 0 P = P2 = P3 = . 0 0 0 1 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 1 0 Suppose aI + bP + cP 2 + dP 3 = O. Then we can rewrite this identity as a b c d 0 0 0 0 d a b c 0 0 0 0 = c d a b 0 0 0 0 b c d a 1 0 0 0 and hence a = b = c = d = 0. 5. (a) An example of three linearly dependent vectors v1 , v2 , v3 in R3 such that every pair of them form a linearly independent set: v1 = (1, 0, 0), v2 = (0, 1, 0), v3 = (1, 1, 0). (b) An example of two linear operators S and T on R2 such that, as vectors in L (V ), S and T are linearly independent, but, for every v in V , the vectors Sv and T v are linearly dependent: S = MA and T = MB with 1 0 1 1 A= and B = . 0 0 0 0 Notice that aA + bB = O gives a+b b 0 0 = 0 0 0 0 and hence a + b = 0, b = 0, which give a = b = 0. For each vetor v = (x, y), we have Sv = (x, 0) and T v = (x + y, 0), which are linearly dependent: there exists a, b, not both zeros, such that a(x, 0) + b(x + y, 0) = (0, 0). Indeed, when x = 0, we may let a = 1 and b = 0. When x = 0, we may let a = (x + y)/x and b = −1. 6. Assume that H ∩ K = {0}. Suppose a1 h1 + a2 h2 + · · · + ar hr + b1 k1 + b2 k2 + · · · + bs ks = 0. 5 Let v = a1 h1 +a2 h2 +· · ·+ar hr ≡ −b1 k1 −b2 k2 +· · ·−bs ks . Then v is in both H and K. Since H ∩ K = {0}, we must have v = 0. Thus a1 h1 + a2 h2 + · · · + ar hr = 0 and b1 k1 +b2 k2 +· · ·+bs ks = 0. Since SH is linearly independent, we have a1 = a2 = · · · = ar = 0. Since SK is linearly independent, we have b1 = b2 = · · · = bs = 0. This shows that S is linearly independent. Conversely, suppose that S is linearly independent. We have to show H ∩ K = {0}. Take any vector in H ∩ K. That v is in H means that we can write v as a linear combination of vectors in SH , say v = a1 h1 + a2 h2 + · · · + ar hr . Similarly, since v is in K, we can write v = b1 k1 + b2 k2 + · · · + bs ks . Thus we have a1 h1 + a2 h2 + · · · + ar hr − b1 k1 − b2 k2 − · · · − bs ks = 0. Since h1 , h2 , . . . , hr , k1 , k2 , . . . , ks are assumed to be linearly independent, we have a1 = a2 = · · · = ar = b1 = b2 = · · · = bs = 0. Hence v = a1 h1 + a2 h2 + · · · + ar hr = 0. This shows H ∩ K = {0}. 7. Take a basis SH ≡ {h1 , h2 , . . . , hr } in H. Extend this basis to a basis of V , say S ≡ {h1 , h2 , . . . , hr , k1 , k2 , . . . , ks }; (such a basis of V exists, according to Corollary 1.4.4). Let SK ≡ {k1 , k2 , . . . , ks } and let K be the subspace spanned by SK . By the last exercise we see that H ∩ K = {0}. Next, let v be any vector in V . Since S is a basis of V , we can express v as a linear combination of vectors in S, say v = a1 h1 + a2 h2 + · · · + ar hr + b1 k1 + b2 k2 + · · · + bs ks . Thus we have v = h + k, where h = a1 h1 + a2 h2 + · · · + ar hr is in H and k = b1 k1 + b2 k2 + · · · + bs ks is in K. This shows V = H + K. 6