MATH 110 LINEAR ALGEBRA HOMEWORK 9 CHU WEE LIM 1 We have already proven in the lecture that A B det det A det C 0 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 t t X Yt X 0 X 0 det det det Y Z Y Z 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 A nn Ann det A nn where A nn is the submatrix obtained by deleting the last row and last column of A Since A nn is an upper triangular n 1 n 1 matrix its determinant is the product of the diagonal entries A11 A22 An 1 n 1 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 1 2 CHU WEE LIM f False suppose k 0 A I Then adding 0 R2 to R1 has no e ect 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 n2 n Here x is the greatest integer x Hence we see that det B 1 2 det A n n 1 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 n 1 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 Cramer s rule h False E g try solving x1 2 x1 2x2 0 by this new Cramer s rule We get A 11 02 and Mk 12 00 and so det Mk det A 0 x2 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 3 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 1 On the other hand 1 0 is an example of a skew symmetric invertible 2 2 matrix 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 t 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 Q 1 AQ for some invertible Q Hence 1 det A det Q det A det B det Q 1 AQ det Q 1 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 de ne 1 x0 x20 xn0 1 x1 x21 xn1 M x0 x1 xn 1 xn x2n xnn and let P x0 x1 xn be the …