MA 1B ANALYTIC - HOMEWORK SET 7 SOLUTIONS

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1 MA 1B ANALYTIC - HOMEWORK SET 7 SOLUTIONS 1. (7 pts)[apostol IV.8., 13, 14] (.) Let A be an n n matrix with characteristic polynomial f(λ). Prove (by induction) that the coefficient of λ n 1 in f(λ) is tr A. Proof. Let A be (a ij ) n n, we will prove the statement by induction on n. When n = 1, f(λ) = λ a 11 and tr(a) = a 11. The statement is true. Now we assume that when k = n 1 (n 2) the statement is true, then we consider the case when k = n. Notice that f(λ) = det(λi A). We denote M = λi A, then the bottom row of M is ( a n,1,, a n,n 1, λ a n,n ). Use the Theorem3.9 in the textbook, we have that det M = ( 1) n a n,1 det(m n,1 ) + a n,n 1 det(m n,n 1 ) (a n,n λ) det(m n,n ). Since the only entries of M that contain λ are in the diagonal, so det(m n,i ) (1 i n 1) is at most a degree (n 2) polynomial on λ as M n,i won t contain λ a n,n. Then det(m n,i ) (1 i n 1) will contribute nothing in the coefficient of λ n 1. Observe that M n,n = λi (a ij ) (n 1) (n 1), so we can use our assumption on the characteristic polynomial of (a ij ) (n 1) (n 1). Then we get that the coefficient of λ n 2 in det M n,n is n 1 a ii. We also know that characteristic polynomial is monic, so det M n,n = λ n 1 n 1 a iiλ n 2 + some lower terms. We plug in the equation of det M above, we get that the coefficient of λ n 1 in det M is which completes the induction. a ii a nn = tra, n 1 (13.) Let A be B be n n matrices with det A = det B and tr A = tr B. Prove that A and B have the same characteristic polynomial if n = 2 but that this need not be the case if n > 2. Proof. Let X be any matrix, we denote f X (λ) the characteristic polynomial of X. Then we know that f X is monic, the coefficient of λ n 1 is trx by the previous problem and the constant term in f X is ( 1) n det(x). When n = 2, then f A = λ 2 tr(a)λ + det(a) and f B = λ 2 tr(b)λ + det(b). Since det A = det B and tr(a) = tr(b), so f A (λ) = f B (λ). In other words, A and B have the same characteristic polynomials. When n > 2, Let A = diag(2,,, ) be an n n matrix and B = diag(1, 1,,, ) be an n n matrix. Since n > 2, both matrices have determinants and traces 2. But f A (λ) = (λ 2)λ n 1 is different from f B (λ) = (λ 1) 2 λ n 2. Hence the statement is false when n > 2. (14.) Prove each of the following statements about the trace. Date: February 29, 21. 1

2 2 MA 1B ANALYTIC - HOMEWORK SET 7 SOLUTIONS (a) tr (A + B) = tr A + tr B. (b) tr (ca) = c tr A. (c) tr (AB) = tr (BA). (d) tr A t = tr A. Proof. Denote A = (a ij ) n n and B = (b ij ) n n. (a) By definition, tr(a) = a ii, tr(b) = b ii and tr(a + B) = (a ii + b ii ), so we have tr(a + B) = tr(a) + tr(b). (b) Matrix ca = (ca ij ) n n, then we have tr(ca) = (c a ii ) = c a ii = c tr(a). (c) Denote AB = (c ij ) n n and BA = (d ij ) n n. Then we have c ij = a ik b kj, So we can compute the traces: d ij = k=1 b ik a kj. k=1 tr(ab) = c ii = tr(ba) = d ii = k=1 k=1 a ik b ki, b ik a ki. We switch i and k in the equation of tr(ab), then get tr(ab) = a ki b ik = b ik a ki. k=1 k=1 Next, we change the order of the summation in the above equation and get tr(ab) = b ik a ki = tr(ba). k=1 Remark. Actually, AB and BA have the same characteristic polynomials. When A is invertible, AB and BA are similar matrices so they have the same characteristic polynomials. When A is not invertible, the characteristic polynomials are still the same by the continuity. (d) Notice that the (i, j)-th entry of A t is a ji, so tr(a t ) = a ii = tr(a). 2. ( pts)[apostol I.13.11] In the linear space of all real polynomials, define (f, g) = e t f(t)g(t) dt. (a) Prove that this improper integral converges absolutely for all polynomials f and g.

3 MA 1B ANALYTIC - HOMEWORK SET 7 SOLUTIONS 3 (b) If x n (t) = t n for n =, 1, 2,..., prove that (x n, x m ) = (m + n)!. (c) Compute (f, g) when f(t) = (t + 1) 2 and g(t) = t (d) Find all linear polynomials g(t) = a + bt orthogonal to f(t) = 1 + t. Also, compute an orthonormal basis for the subspace consisting of polynomials of degree at most 3. Proof. (a) Since f(t)g(t) is a polynomial, we can find N such that if t N we have f(t)g(t) e t 2. Then e t f(t)g(t) dt N N e t f(t)g(t) dt + e t f(t)g(t) dt + N N e t e t 2 dt e t 2 dt <. (b) We argue by induction. We have x, x = e t dt = 1 =!. Assume x m, x n = (m + n)!. Then integrating by parts we have x m, x n+1 = e t t m+n+1 dt ( d ( = e t )) t m+n+1 dt dt = [ e t t n+m+1] ( ) d e t dt tn+m+1 dt = (n + m + 1) = (n + m + 1) x n, x m = (n + m + 1)(n + m)! = (n + m + 1)!. e t t n+m dt (c) By linearity and (b) we have (t + 1) 2, t = t 2 + 2t + 1, t = t 2, t 2 + t 2, t, t t, 1 + 1, t 2 + 1, 1 = 4! + 2! + 2 3! + 2 1! + 2! +! = 43. (d) Consider the set {1 + t, 1} which spans the space of linear polynomials. We apply Gram-Schmidt to get an orthogonal basis {v 1, v 2 } for the same space, by

4 4 MA 1B ANALYTIC - HOMEWORK SET 7 SOLUTIONS letting v 1 = 1 + t and 1, 1 + t v 2 = 1 (1 + t) 1 + t, 1 + t 1, 1 + 1, t = 1 (1 + t) 1, , t + t, t! + 1! = 1 (1 + t)! + 2 1! + 2! = 1 2 (1 + t) 5 = 1 (3 2t). 5 Since the space of linear polynomials is two dimensional, the orthogonal complement of 1 + t is one-dimensional. Therefore every linear polynomial orthogonal to 1 + t is of the form c(3 2t) for some c R. An orthonormal basis for the subspace consisting of polynomials of degree at most 3 is { 1, t 1, 1 2 t2 2t + 3, 1 t3 3 } 2 t2 + 2t (5 pts) Let V be a vector space with an inner product (, ). Let {v 1, v 2,..., v n } be a generating set of V. Prove (a) if (x, v) = for all v V, then x =. (b) if (x, v i ) = for all i = 1,..., n, then x =. (c) if (x, v i ) = (y, v i ) for all i = 1,..., n, then x = y. Proof. (i) If x, then by the axiom of the inner product, x, x >. (ii) Let v V. Then v = n a iv i for some a i R. Then x, v = x, a i v i = x, a i v i = a i x, v i = By part (i), x =. (iii) The assumption implies that x y, v i = for all i. By part (ii), x y =. x = y. 4. (7 pts) Let V = R 4 and U = L(v 1, v 2 ) be the subspace of V generated by the vectors v 1 = (1, 3, 1, 1) and v 2 = (3, 2, 2, 1). (a) Find an orthogonal basis of U. Solution. Since the generating set for U is linearly independent, we can use the Gram Schmidt process to get an orthogonal basis. We set w 1 = (1, 3, 1, 1). Then we produce an element in U orthogonal to w 1 by taking w 2 = (3, 2, 2, 1) proj w1 (3, 2, 2, 1) = (3, 2, 2, 1) (1, 3, 1, 1) = (2, 1, 1, ),

5 MA 1B ANALYTIC - HOMEWORK SET 7 SOLUTIONS 5 where we used the calculation that proj w1 (3, 2, 2, 1) = w 1, (3, 2, 2, 1) w 1 w 1, w 1 = w 1 = w 1. Thus, {w 1, w 2 } is an orthogonal basis for U. (b) Find the orthogonal projection of v = (1, 1, 1, 1) T onto U. Solution. Since {w 1, w 2 } is an orthogonal basis for U, we have We calculate proj U (1, 1, 1, 1) = proj U (v) = v, w 1 w 1, w 1 w 1 + v, w 2 w 2, w 2 w 2. (1, 1, 1, 1), (1, 3, 1, 1) w 1 + w w 2 = = 1 2 (1, 3, 1, 1) + 1 (2, 1, 1, ) 3 = 3 (1, 3, 1, 1) + 2 (2, 1, 1, ) = (7/, 7/, 5/, 3/). (c) Find the distance of v = (1, 1,, ) T from U. (1, 1, 1, 1), (2, 1, 1, ) w 2 Solution. This is just finding v proj U (v). We compute proj U (1, 1,, ) = Therefore, (1, 1,, ), (1, 3, 1, 1) w 1 + = 2 (1, 3, 1, 1) + 1 (2, 1, 1, ) = (4/, 5/, 3/, 2/). (1, 1,, ), (2, 1, 1, ) w 2 proj U (1, 1,, ) (1, 1,, ) = (4/, 5/, 3/, 2/) (1, 1,, ) = ( 2/, 1/, 3/, 2/) = (4/3 + 1/3 + 9/3 + 4/3) 1/2 = (18/3) 1/2 = 1/ 2.