MATH 110: LINEAR ALGEBRA HOMEWORK #9

CHU-WEE LIM

(1) We have already proven in the lecture that det A B 0 C = det(A) det(C)

if A, B and C are n1 n1 , n1 n2 and n2 n2 matrices respectively. Now we can use this to prove problem 1. Indeed, suppose X, Y and Z are n1 n1 , n2 n1 and n2 n2 matrices respectively. Then we have: X 0 det Y Z X 0 = det Y Z
t

= det

Xt Y t 0 Zt

= det(X t ) det(Z t ) = det(X) det(Z). (2) The proof is by induction on the size of A. When n = 1, there is nothing to prove. Now we assume: if B is an (n 1) (n 1) upper-triangular matrix, then det(B) is the product of the diagonal entries of B. Let A be an n n upper-triangular matrix. The last row of A consists of all zeros except the entry Ann . Hence, expanding along the last row gives: det(A) = (1)n+n Ann det(Ann ) = Ann det(Ann ), where Ann is the submatrix obtained by deleting the last row and last column of A. nn is an upper-triangular (n 1) (n 1) matrix, its determinant is the Since A product of the diagonal entries A11 A22 . . . An1,n1 . Thus, det(A) = A11 A22 . . . Ann , as desired. For the case of lower-triangular matrix A, we note that At is upper-triangular, so we can apply the above result: det(A) = det(At ) = A11 A22 . . . Ann . 4.2: Determinants of Order n Problem 1. (a) False, e.g. for the 2 2 identity matrix I, we have det(2I) = 4 = 2 = det(I) + det(I). (b) True, see theorem 4.4 (pg 215). (c) True, for we can subtract one row from the other to get a row of zeros. (d) True, see rule (a) on page 217. (e) False, if we multiply the row by 0, then det(B) = 0 regardless of what A is.
Date: November 1.
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CHU-WEE LIM

(f) False, suppose k = 0, A = I. Then adding 0 R2 to R1 has no eect, and so det(B) = det(A) = 1 = 0 det(A). (g) False. Quite the opposite: A has rank n if and only if A is invertible if and only if det(A) = 0. (h) True: we just proved it above. Problem 26. Using Q25, we get det(A) = (1)n det(A). Hence det(A) = det(A) i det(A) = (1)n det(A). Now, if n is even then (1)n = +1 so equality clearly holds. Also, if char(F ) = 2, then 1F + 1F = 0F and so (1)n = 1 = 1 regardless of the parity of n. Hence equality still holds. Finally, suppose n is odd and char(F ) = 2. Then we get det(A) = det(A), which gives 2 det(A) = 0. Since char(F ) = 2, we get det(A) = 0. Hence det(A) = det(A) if and only if at least one of the following is true: (i) char(F ) = 2; (ii) n is even; (iii) det(A) = 0. Problem 30. If we exchange two rows of A, then we ip the sign of det(A). Also, B is obtained from A by exchanging the i-th row and the (n + 1 i)-th row, for i = 1, 2, . . . , [ n ]. 2 n Here, [x] is the greatest integer x. Hence, we see that det(B) = (1)[ 2 ] det(A). n(n1) As an alternative, you can also write (1) 2 det(A). This can be seen by performing (n 1) row-exchanges to move the bottom row to the top; folowed by (n 2) row-exchanges to move the bottom row to the second, and so on. This gives us 1 + 2 + + (n 1) = n(n1) 2 row-exchanges.

4.3: Properties of Determinants Problem 1. (a) False, an elementary matrix of type (b) is not of determinant 1 in general. (b) True, by theorem 4.7, page 223. (c) False. In fact M is invertible if and only if det(M) = 0. See corollary on page 223. (d) True, since M has rank n if and only if it is invertible. (e) False. In fact, det(At ) = det(A) by theorem 4.8, page 224. (f) True, using the fact that we can perform cofactor expansion, and that det(At ) = det(A). (g) False. E.g. 0x = 0 cannot be solved by Cramers rule. (h) False. E.g. try solving x1 = 2, x1 + 2x2 = 0 by this new Cramers rule. We get A = ( 1 0 ) and Mk = ( 1 0 ), and so det(Mk )/ det(A) = 0 = x2 . 1 2 2 0 Problem 10. If M is nilpotent, then M k = 0 for some k. So det(M)k = det(M k ) = 0, and hence det(M) = 0.

MATH 110: HOMEWORK #9

Problem 11. If M t = M, then taking the determinant gives det(M) = det(M t ) = det(M) = (1)n det(M). If n is odd, then det(M) = det(M) and so det(M) = 0 (recall that we are working over the complex eld C, so char = 2). Hence M is not invertible. 0 On the other hand ( 1 1 ) is an example of a skew-symmetric invertible 2 2 matrix. 0 Problem 12. We have 1 = det(I) = det(QQt ) = det(Q) det(Qt ) = det(Q) det(Q). Hence det(Q)2 = 1 and so det(Q) = 1. Problem 13. (a) Suppose M has the LU-decomposition: M = P1 LUP2 , where P1 and P2 are permutation matrices. Also, L is a unit lower-triangular matrix, while U is an uppertriangular matrix. Then det(M) = det(P1 ) det(L) det(U) det(P2 ) = det(P1 )U11 U22 . . . Unn det(P2 ). Then taking the conjugate gives M = P 1 LU P 2 = P1 LU P2 . Hence det(M) = det(P1 ) det(U) det(P2 ) = det(P1 )U 11 U 22 . . . U nn det(P2 ) = det(M), since det(P1 ) and det(P2 ) are 1. Alternative solution: use induction on the size of M. (b) We have 1 = det(I) = det(QQ ) = det(Q) det(Q ) = det(Q)det(Q) = | det(Q)|2 . Hence, | det(Q)| = 1. Problem 15. If A and B are similar, then B = Q1 AQ for some invertible Q. Hence 1 det(A) det(Q) = det(A). det(B) = det(Q1 AQ) = det(Q1 ) det(A) det(Q) = det(Q) Problem 17. Since AB = BA, taking the determinant gives det(A) det(B) = det(AB) = det(BA) = (1)n det(BA) = det(B) det(A), since n is odd. Thus, 2 det(A) det(B) = 0. Since char(F ) = 2, we have det(A) det(B) = 0. So det(A) = 0 or det(B) = 0, i.e. either A or B is not invertible. Problem 22(c). The proof I have in mind uses the polynomial factorizations. For variables x0 , x1 , . . . , xn , dene 1 x0 x2 . . . xn 0 0 1 x1 x2 . . . xn 1 1 M(x0 , x1 , . . . , xn ) = . . . , . .. . . . . . . . . . 1 xn x2 . . . xn n n and let P (x0 , x1 , . . . , xn ) be the determinant of M(x0 , . . . , xn ). Note that M(x0 , . . . , xn ) is a matrix whose entries are polynomials! Similarly, P (x0 , . . . , xn ) is a polynomial in the xi s. By expansion along the rightmost column, we see that P (x0 , . . . , xn ) =(1)n+1 xn P (x1 , . . . , xn ) + (1)n+2 xn P (x0 , x2 , x3 , . . . , xn ) 0 1 + + (1)2n xn P (x0 , x1 , x2 , . . . , xn1 ). n
t

CHU-WEE LIM

Hence, by induction, we can prove that P (x0 , . . . , xn ) is a homogeneous polynomial of degree 1 + 2 + + n = n(n+1) . 2 Now, suppose 0 i < j n and xi = xj . Then the matrix M(x0 , . . . , xn ) would have two identical rows, and so its determinant P (x0 , . . . , xn ) = 0. In short, whenever xj xi = 0, we have P (x0 , . . . , xn ) = 0. By the factor theorem for polynomials, xj xi is a factor of P (x0 , . . . , xn ). Thus we can write P (x0 , . . . , xn ) = Q(x0 , . . . , xn )
n(n+1) 2

(xj xi ).
0i<jn

as we noted, and the right hand side has degree deg(Q) The left-hand side has degree + number of pairs (i, j), 0 i < j n. This latter number is 1 + 2 + + n = n(n+1) , and 2 so we get deg(Q) = 0, i.e. Q is constant. To compute this constant, we note that the coecient of x1 x2 . . . xn in P (x0 , x1 , . . . , xn ) 1 2 n is 1. On the other hand, the corresponding coecient in i<j (xj xi ) is also 1 (since the only way to get x1 x2 . . . xn in the product is to take xn from xn x0 , xn x1 , . . . , xn xn1 ; 1 2 n and xn1 from xn1 x0 , xn1 x1 , . . . , xn1 xn2 etc). Problem 24. We wish to compute the determinant of: t 0 0 ... 0 a0 1 t 0 . . . 0 a1 a2 . B = A + tI = 0 1 t . . . 0 . . . .. . . . . . . . . . . . . . 0 0 0 . . . 1 an1 + t Let D(a0 , a1 , . . . , an1 ) = det(B). Expanding 1 t 0 . . . 0 1 t . . . det(B) = (1)n+1 a0 det . . . .. . . . . . . . 0 0 the rst row, we get t 0 1 0 . + t det . . . . . 0 0 t . . . 0 ... a1 0 ... a2 . . .. . . . . . . 0 0 . . . an1 + t

0 . . . 1

The rst matrix has determinant (1)n1 , while the second has determinant D(a1 , . . . , an1 ). Hence, D(a0 , a1 , . . . , an1 ) = a0 + t D(a1 , . . . , an1 ). Together with D(an1 ) = an1 + t, we get D(a0 , . . . , an1 ) = a0 + a1 t + a2 t2 + + an1 tn1 + tn .