Proofs That Det(A T)

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 − ×   a11 a12 a1k  a a ··· a   21 22 2k  A =  . . ···.. .   . . . .  ak1 ak2 akk ··· so that   a11 a21 ak1  a a ··· a  t  12 22 k2  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.