t Proofs that det(A ) = det A. Eric O. Korman 1 Proof 1 We consider two cases: det A = 0 and det A = 0. First assume that det A = 0. Then by a theorem in the text, A is not invertible.6 This implies that At is not invertible since we have seen that a matrix is invertible if and only if its trans- t t pose is. Thus det A = 0 so in this case we have det A = det A. Now assume that det A = 0. Then A is invertible and can therefore be written 6 t as a product E1 Ek of elementary matrices. We claim that det E = det E for any elementary··· matrix. This is because if E is of the second or third type of t t elementary matrix then E = E so that det E = det E. If E is of the first type t then so is E . But from the text we know that det E = 1 for all elementary matrices of the first type. This proves our claim. Using properties of the transpose and the multiplicative property of the determinant we have t t det A = det((E1 Ek) ) t ··· t = det(Ek E1) t ··· t = det(Ek) det(E1) ··· = det Ek det E1 ··· = det E1 det Ek ··· = det(E1 Ek) ··· = det A. 2 Proof 2 t We will prove that det A = det A using the fact that the determinant can be computed by cofactor expansion along any row or column (this fact really 1 shouldn’t be assumed...) t We proceed by induction on n. For the base case n = 1 we have A = A so that t det A = det A , as desired. For the inductive step assume the result is true for n = k 1 and let A be a k k matrix. Write − × 0 1 a11 a12 a1k B a a ··· a C B 21 22 2k C A = B . ···.. C @ . A ak1 ak2 akk ··· so that 0 1 a11 a21 ak1 B a a ··· a C t B 12 22 k2 C A = B . ···.. C . @ . A a1k a2k akk ··· Using cofactor expansion along the first column of A we have k+1 det A = a11 det A11 a21 det A21 + + ( 1) ak1 det Ak1 − ··· − where Ai j is the matrix obtained from A by removing the ith row and jth column. Using cofactor expansion along the first row of At we have t t t k+1 t det(A ) = a11 det(A )11 a21 det(A )12 + + ( 1) ak1 det(A )1k. − ··· − t t We see that (A )i j = (Aji) . Since Aji is a (k 1) (k 1) matrix we can use the inductive hypothesis to see that − × − t t det(A )i j = det((Aji) ) = det Aji. t t Making this substitution into the above formula for det(A ) gives det(A ) = det A. 2.