27. Further properties of determinants

27. Further properties of determinants

Recall that (abcd) is invertible if and only if its determinant ad-bc is nonzero. We'll prove the analogue of this for n -by-n matrices.

Theorem:

Let A be an n -by-n matrix. Then A is invertible if and only if det(A)0 .

We'll also prove the following:

Theorem:

det(AB)=det(A)det(B) .

Proof of the invertibility criterion

Recall that A is invertible if and only if its reduced echelon form (say Ared ) is the identity matrix. To get from A to Ared , we used some row operations. The effect of these on the determinant is:

  • (type I) determinant unchanged,

  • (type II) determinant rescaled by a nonzero factor,

  • (type III) determinant switches sign.

Therefore det(A) differs from det(Ared) by a nonzero factor. In particular, det(A)0 if and only if det(Ared)0 .

If A is invertible, det(Ared)=det(I)=1 , so det(A)0 .

If A is not invertible, Ared has a row of zeros (Ared is a square matrix in reduced echelon form, so if every row is nonzero then the leading entries have to go along the diagonal, and you get Ared=I ; but we're assuming A is not invertible). Since Ared has a row of zeros, det(Ared)=0 , so det(A)=0 too.

Proof of multiplicativity of the determinant

We first prove det(AB)=det(A)det(B) in the case where A is an elementary matrix.

  • If A=Eij(λ) then AB is obtained from B by RiRi+λRj , so det(AB)=det(B) (such a row operation doesn't change the determinant). We also have det(A)=1 , so we can verify the formula det(AB)=det(A)det(B) in this case just by calculating both sides.

  • If A=Ei(λ) then AB is obtained from B by RiλRi , so det(AB)=λdet(B) . We also have det(A)=λ , so again we can verify the formula just by computing both sides.

This shows that det(AB)=det(A)det(B) whenever A is an elementary matrix.

By induction, we can now show that det(AB)=det(A)det(B) whenever A is a product of elementary matrices, in other words whenever A is an invertible matrix.

If A is not invertible, then det(A)=0 (by the previous theorem) so to verify the formula in this case, we need to prove that det(AB)=0 . Suppose for a contradiction that det(AB)0 . Then AB is invertible. But if AB is invertible then B(AB)-1 is an inverse for A : A(B(AB)-1)=AB(AB)-1=I. This gives a contradiction, so we deduce that det(AB)=0 .

We have now checked the formula for all possible matrices A , which proves the theorem.