Determinants multiply

Determinants multiply Let A and B be two n × n matrices. The point of this note is to prove that det(AB) = det(A) det(B). The textbook gives an algebraic proof in Theorem 6.2.6 and a geometric proof in Section 6.3. Our proof, like that in Theorem 6.2.6, relies on properties of row reduction. I’ll write ∆(w ~ 1, w ~ 2, . . . , w ~ n ) for the determinant of the n × n matrix with rows w ~ 1, w ~ 2, . . . , w ~ n. We begin by checking some special cases: Special Case 1: The statement is true if det B = 0. Proof: If det B = 0 then B is not injective; say B~v = ~0 with ~v 6= ~0. Then AB~v = ~0 so AB is not injective and det(AB) = 0. Special Case 2: The statement is true if det A = 0. Proof: If det A = 0 then A is not surjective. Suppose that w ~ is not in the image of A. Then w ~ is also not in the image of AB, so AB is not surjective and det(AB) = 0. Special Case 3: A diagonal matrix. Let D be a matrix with d1 , d2 , . . . , dn on the diagonal and zeroes off the diagonal. So det(D) = d1 d2 · · · dn . I claim that det(DB) = d1 d2 · · · dn det B. Proof: Let the rows of B be w ~ 1, w ~ 2, . . . , w ~ n . Then the rows of DB are d1 w ~ 1 , d2 w ~ 2 , . . . , dn w ~ n . So we want to show that ∆(d1 w ~ 1 , d2 w ~ 2 , . . . , dn w ~ n ) = d1 d2 · · · dn ∆(w ~ 1, . . . , w ~ n ). This follows from the basic properties of determinant (specifically, Theorem 6.2.3.(a) in the book). Special Case 4: A matrix with one off-diagonal entry. Let E be a matrix which looks like   1 0 0 0 0 0 0 1 0 0 0 0    0 0 1 0 0 0    0 k 0 1 0 0   0 0 0 0 1 0  0 0 0 0 0 1 with the k entry in position (i, j). (So the example is i = 4, j = 2.)Then det E = 1. I claim that det(EB) = 1 · det(B). Proof: Let the rows of B be w ~ 1, w ~ 2, . . . , w ~j, . . . , w ~ i, . . . w ~ n . Then the rows of EB are w ~ 1, w ~ 2, . . . , w ~j, . . . , w ~ i + kw ~j, . . . w ~ n . We have det(EB) = ∆(w ~ 1, w ~ 2, . . . , w ~j, . . . , w ~ i + kw ~j, . . . w ~ n) = ∆(w ~ 1, w ~ 2, . . . , w ~j, . . . , w ~ i, . . . w ~ n ) + k∆(w ~ 1, w ~ 2, . . . , w ~j, . . . , w ~j, . . . w ~ n) = ∆(w ~ 1, w ~ 2, . . . , w ~j, . . . , w ~ i, . . . w ~ n) + 0 = det(B) Going from the first line to the second line we have used the linearity of determinant; going from the second line to the third line we have used that the row w ~ j occurs twice. Special Case 5: A matrix which switches two  1 0 0 0 0 0 0 0  0 0 1 0  0 0 0 1  0 1 0 0 0 0 0 0 rows. Let S be a matrix which looks like  0 0 1 0  0 0  0 0  1 0 0 1 with the off diagonal entries in positions (i, j) and (j, i). (So the example is i = 2, j = 5.) Then det S = −1. I claim that det(SB) = − det(B). Proof: Let the rows of B be w ~ 1, w ~ 2, . . . , w ~j, . . . , w ~ i, . . . w ~ n . Then the rows of SB are w ~ 1, w ~ 2, . . . , w ~ i, . . . , w ~j, . . . w ~ n . Our claim is the standard fact that ∆(w ~ 1, w ~ 2, . . . , w ~j, . . . , w ~ i, . . . w ~ n ) = −∆(w ~ 1, w ~ 2, . . . , w ~ i, . . . , w ~j, . . . w ~ n ). Using row reduction to compute determinants Let A be an invertible matrix. Using the row reduction procedure, I claim that A can be written as a product of matrices which look like the matrices D, E and S discussed in Special Cases 3, 4 and 5. An example is clearer than a proof. Let A = ( 25 36 ). In the left columns below I’ll run row reduction; in the right columns I’ll start building an expression for A as a product of matrices of the various special forms. 2 3 ! 5 6 ! 1 3/2 2 3 5 5 6 6 ! 1 3/2 1 5 0 −3/2 ! 1 3/2 5 6 0 1 ! 1 0 5 6 0 1 5 6 1 6 3/2 ! 0 −3/2 ! 1 3/2 0 1 3/2 ! 2 3 2 3 2 3 ! = 2 0 ! 0 1 ! = 2 0 = 2 0 = 2 0 6 ! 1 5 1 0 1 ! 0 1 ! ! 5 ! 0 1 ! 1 3/2 1 5 ! 0 1 ! 0 1 1 5 0 1 3/2 ! 0 −3/2 ! ! 1 0 1 3/2 0 3/2 ! 1 0 0 3/2 0 ! 1 1 3/2 0 ! 1 1 0 0 1 In short, every time that we add one row to another, we pick up an E. Every time we rescale a row, we pick up a D. If we had switched two rows, we would have gotten an S. At the end of the day, we have A = X1 X2 . . . Xm Id, where each Xi is either a D-matrix, an E-matrix or an S-matrix. Notice that we know the row reduction ends at the identity, since A is invertible. Using special cases 3, 4 and 5, we know that know that det(A) = det(X1 X2 . . . Xm Id) = det(X1 ) det(X2 · · · Xm Id) = det(X1 ) det(X2 ) det(X3 · · · Xm Id) = · · · = det(X1 ) det(X2 ) · · · det(Xm ) det(Id) = det(X1 ) det(X2 ) · · · det(Xm )·1 = det(X1 ) det(X2 ) · · · det(Xm ). For example, we are saying that           2 3 2 0 1 5 1 0 1 3/2 det = det det det det 5 6 0 1 0 1 0 3/2 0 1 or, in other words, that   2 3 = 2 × 1 × (−3/2) × 1. 5 6 Your book has a very similar discussion in Algorithm 6.2. det The proof Our goal is to show that det(AB) = det(A) det(B). If A is not invertible, we already checked this in Special Case 2. Otherwise, write A = X1 X2 · · · Xm Id with each Xi either a D-matrix, an E-matrix or an S-matrix. As we just discussed, det(A) = det(X1 ) det(X2 ) · · · det(Xm ). Using Special Cases 3, 4 and 5 over and over again, det(AB) = det(X1 X2 . . . Xm B) = det(X1 ) det(X2 · · · Xm B) = det(X1 ) det(X2 ) det(X3 · · · Xm B) = · · · = det(X1 ) det(X2 ) · · · det(Xm ) det(B) = det(A) det(B). This is what we wanted to prove. !