Determinants II Linear Algebra, Fall 2008

Transcription

1 Determinants II Linear Algebra, Fall Basic Properties of Determinants Here are the basic properties of determinants which you proved in the exercises to the previous handout Theorem 1 Let A be an n n matrix Then the following are all true (a) det(a) = det(a T ) (b) If some row or some column of A consists entirely of zeros, then det A = 0 (c) If A is an upper-triangular or lower-triangular matrix, then det A is just the product of the entries on the diagonal In particular, det I = 1 (d) If you multiply one row, or one column, of matrix A by a scalar k, then the determinant of the resulting matrix is k det A (e) If you multiply A by the scalar k, then det(ka) = k n det A Note that if A and B are n n matrices, then det(a + B) is not necessarily equal to det(a) + det(b) In fact, there is no rule at all for finding det(a + B) from det A and det B 2 The Main Theorem We are interested in determinants for one main reason: we want a function that will tell us whether a matrix is invertible More precisely, we want a function f that takes in a matrix and gives a real number, such that from the value f(a) we can tell whether matrix A is invertible It turns out that the determinant is precisely the function we need, but to prove that takes a few steps You may have noticed that Theorem 1(d) tells us that when we do the row operation kr i to A, the determinant is multiplied by k Now let us figure out what happens with the other two types of row operations We ll look first at a special case: swapping two adjacent rows

2 Determinants II Linear Algebra, Fall 2008 Page 2 of 7 Lemma 2 Let A be an n n matrix, and let B the matrix obtained by swapping rows r and r + 1 of A Then det B = det A Proof To help visualize the situation, consider the following example, in which n = 6 and r = 3 a 11 a 12 a 13 a 14 a 15 a 16 a 11 a 12 a 13 a 14 a 15 a 16 a 21 a 22 a 23 a 24 a 25 a 26 a 21 a 22 a 23 a 24 a 25 a 26 B = a 41 a 42 a 43 a 44 a 45 a 46 a 31 a 32 a 33 a 34 a 35 a 36 where A = a 31 a 32 a 33 a 34 a 35 a 36 a 41 a 42 a 43 a 44 a 45 a 46 a 51 a 52 a 53 a 54 a 55 a 56 a 51 a 52 a 53 a 54 a 55 a 56 a 61 a 62 a 63 a 64 a 65 a 66 a 61 a 62 a 63 a 64 a 65 a 66 Recall that to get the determinant of B we sum up the (signed) products corresponding to all rook-arrangements one entry from each row and column of B Consider a particular rook-arrangement σ, and let σ be the rook-arrangement that swaps the positions of the rooks in rows r and r + 1 Example: σ = [1, 4, 3, 5, 6, 2] and σ = [1, 4, 5, 3, 6, 2] rz0z0z 2 Z0ZrZ0 3 0ZRZ0Z 4 Z0Z0S0 5 0Z0Z0s 6 ZrZ0Z rz0z0z 2 Z0ZrZ0 3 0Z0ZRZ 4 Z0S0Z0 5 0Z0Z0s 6 ZrZ0Z0 Then the product of the entries of B under the rooks of σ is exactly the same as the product of the entries of A under the rooks of σ, since both the entries and rooks are moving in the same way: a a Product of B under σ: Product of A under σ : a 24 a 24 a 43 a 35 a 35 a 43 a 56 a 56 Therefore a 62 a 11 a 62 a 43 a 24 a 35 a 56 = a 11 a 62 a 43 a 24 a 35 a 56 a 62 det B = σ S n sign(σ)(product of B under σ) = σ S n sign(σ)(product of A under σ ) We next must determine how sign(σ ) and sign(σ) are related We ll do this by looking at how the number of northeast pairs changes between σ and σ Note that there are only two rooks that move between σ and σ, and the rest are stationary

3 Determinants II Linear Algebra, Fall 2008 Page 3 of 7 X 1 Y 1 Z 1 0Z0S S0Z0 Z 2 Y 2 X 2 X 1 Y 1 Z 1 RZ0Z Z0ZR Z 2 Y 2 X 2 (a) For every pair of stationary rooks, changing from σ to σ doesn t change whether they form a NE pair or not If it was a NE pair in σ it still is in σ, and if it was not a NE pair in σ then it still will not be in σ (b) Each stationary rook in regions X 1 and X 2 does not form a NE pair with either moving rook, whether in σ or in σ (c) Each stationary rook in regions Y 1 and Y 2 forms a NE pair with just one moving rook in σ, and it forms a NE pair with just one moving rook in σ too (d) Each stationary rook in regions Z 1 and Z 2 forms a NE pair with both moving rooks in σ, and it forms a NE pair with both moving rooks in σ too (e) Finally, if the two moving rooks form a NE pair in σ, then they will not form a NE pair in σ ; if they were not a NE pair in σ, then they will be in σ So the only change between σ and σ in the number of NE pairs occurs in case (e), where it changes by one Thus if the total number of NE pairs was even in σ, it will be odd in σ, and vice versa Thus sign(σ ) = ( 1) sign(σ) So we can compute that det B = σ S n sign(σ)(product of B under σ) = σ S n sign(σ)(product of A under σ ) = σ S n sign(σ )(product of A under σ ) = σ S n sign(σ )(product of A under σ ) = det A Using this lemma, we can show what happens with the remaining two types of row (or column) operations

4 Determinants II Linear Algebra, Fall 2008 Page 4 of 7 Theorem 3 Let A be an n n matrix Then the following are true (a) Let B be the matrix obtained by swapping two rows (or two columns) of A det B = det A Then (b) If A has two identical rows (or two identical columns), then det A = 0 (c) Let B be the matrix obtained from A by the row operation R i + kr j for some k 0 Then det B = det A Proof For part (a), first consider the case of swapping two rows, say rows r and r + t where t > 0 Now swapping rows r and t can be accomplished by swapping adjacent rows 2t 1 times, as follows First, we swap adjacent rows t times to move row r down to position r +t Step 1 Step 2 Step (t 1) Step t r r + 1 r + 1 r + 1 r + 1 r + 1 r r + 2 r + 2 r + 2 r + 2 r + 2 r r + 3 r + 3 r + t 1 r + t 1 r + t 1 r r + t r + t r + t r + t r + t r Now we swap adjacent rows t 1 times to move row r + t from its current location back up to position r Step 1 Step (t 2) Step (t 1) r + 1 r + 1 r + 1 r+t r + 2 r + 2 r+t r + 1 r + t 1 r+t r + t 2 r + t 2 r+t r + t 1 r + t 1 r + t 1 r r r r So we can transform matrix A into matrix B by swapping adjacent rows 2t 1 times With each such swap we multiply the determinant by 1, so the total effect is to multiply the determinant by ( 1) 2t 1 = 1 Thus det B = det A, as desired What if B is obtained by swapping two columns of A? That s easy, thanks to Theorem 1(a) We can get B by taking the transpose of A, swapping rows, and taking the transpose back: transpose swap rows transpose Matrix: A A T B T B Determinant: det A det A det A det A So whether we swap two rows or two columns, det B = det A, proving part (a) Now that we have proved part (a), proving part (b) is easy Suppose rows (or columns) r and s of matrix A are identical Let B be the matrix you get by swapping those two rows (or columns); then on the one hand det B = det A by part (a), but on the other hand B equals A so det B = det A! Thus det A = det A, so det A = 0

5 Determinants II Linear Algebra, Fall 2008 Page 5 of 7 We move on to part (c) Suppose B is obtained from A by the row operation R i + kr j, where k 0 Then b ir = a ir + ka jr for every column r, but the entries of b in other rows equal the corresponding entries of a Then det B = σ S n sign(σ)b 1σ1 b iσi b nσn = σ S n sign(σ)a 1σ1 (a iσi + ka jσi ) a nσn = ( σ S n sign(σ)a 1σ1 a iσi a nσn ) + ( = det A + k σ S n sign(σ)a 1σ1 a jσi a nσn = det A + k det C, σ S n sign(σ)a 1σ1 (ka jσi ) a nσn where C is the matrix you get by replacing row i of A with row j But then rows i and j of matrix C are identical, so by part (b) we know that det C = 0 Thus det B = det A The proof for columns can be completed using transposes as above, and we have finished the proof of the theorem Theorem 3(c) is especially surprising, since one would expect the determinant to change after so drastic a change to the matrix Nevertheless it doesn t, so we see that the following happens: Doing this to A multiplies det A by where proved kr i k Theorem 1(d) R i R j 1 Theorem 3(a) R i + kr j 1 Theorem 3(c) Now let s consider the elementary matrices corresponding to each of these operations Now det E kri = det(e kri I) = k det(i) = k, since E kri I is just what we get from performing the row operation kr i on I Likewise det E Ri R j = det(e Ri R j I) = det I = 1 and det E Ri +kr j = det(e Ri +kr j I) = det I = 1 ) In summary, E corresponds to det E kr i k R i R j 1 R i + kr j 1

6 Determinants II Linear Algebra, Fall 2008 Page 6 of 7 We can now make the first step towards understanding the determinant of a product Lemma 4 Let A be an n n matrix, and let E be an elementary n n matrix Then det(ea) = det(e) det(a) Proof Suppose first that E corresponds to the row operation kr i ; then det E = k Moreover, EA is the matrix we get from A by multiplying its ith row by k, so by Theorem 1(d) we have det(ea) = k det A = (det E)(det A), as desired The proofs for the matrix operations R i R j and R i + kr j work in exactly the same way In fact, multiplying by lots of elementary matrices works in the same way: Corollary 5 Let C be an n n matrix and let E 1,, E r be elementary n n matrices Then det(e 1 E 2 E r C) = (det E 1 )(det E 2 ) (det E r )(det C) Proof Applying Lemma 4 many times, we have det(e 1 E 2 E 3 E r C) = (det E 1 ) det(e 2 E 3 E r C) = (det E 1 )(det E 2 ) det(e 3 E r C) = = (det E 1 )(det E 2 )(det E 3 ) (det E r )(det C) Now at last we can prove our most important theorem: Theorem 6 Let A be an n n matrix Then A is invertible if and only if det A 0 Proof ( ) First suppose A is invertible Then by the FTLA, A is the product of some elementary matrices E 1,, E r Then by Corollary 5 we have det(a) = det(e 1 E r I) = (det E 1 )(det E 2 ) (det E r ) det(i) Since the determinant of each E i is nonzero, and det(i) = 1, it follows that det A 0 ( ) Suppose A is not invertible, so by the FTLA, rank A < n Let B be the row-reduced form of A; so B has at least one zero row Thus det B = 0 by Theorem 1(b) On the other hand, since B and A are row-equivalent, there are elementary matrices E 1,, E s such that A = E 1 E 2 E s B; thus by Corollary 5, det A = det(e 1 E 2 E s B) = (det E 1 )(det E 2 ) (det E s )(det B) = (det E 1 )(det E 2 ) (det E s )(0) = 0 Thus if A is not invertible, det A = 0 We conclude that A is invertible if and only if det A = 0

7 Determinants II Linear Algebra, Fall 2008 Page 7 of 7 Not only is the previous theorem important in its own right, but it also leads to a nice corollary Although there was no nice formula for det(a + B), it turns out that there is a very nice formula for det(ab) Corollary 7 Let A and B be n n matrices Then det(ab) = (det A)(det B) Proof There are two possibilities: either A is invertible or it is not If A is invertible, then by the FTLA write A as a product of invertible matrices A = E 1 E 2 E r ; then applying Corollary 5 twice, we have det(ab) = det(e 1 E 2 E r B) = (det E 1 )(det E 2 ) (det E r )(det B) = (det A)(det B) On the other hand, if A is not invertible, then AB must not have an inverse either (Otherwise A(B(AB) 1 ) = (AB)(AB) 1 = I, so B(AB) 1 would be an inverse for A) Thus applying Theorem 6 twice, (det A)(det B) = (0)(det B) = 0 = det(ab) Exercises In all these exercises, assume A and B are square matrices, and I is the identity matrix of the same size 1 Is det(ab) = det(ba)? Justify your answer 2 Show that det(ab 1 ) = det A det B 3 Show that det(a T B T ) = det(a) det(b T ) 4 Show that if AB = I, then det A 0 and det B 0 5 (a) Show that if A = A 1, then det A = ±1 (b) If A T = A 1, then what is det A? 6 If A is invertible and A 2 = A, what is det A? 7 If A 7 = I, what is det A? 8 Show that if A is an n n matrix with n odd, and if A is skew-symmetric (ie A T = A), then det(a) = 0