27. Further properties of determinants

27. Further properties of determinants

Recall that $\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ 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 $\det(A)\neq 0$ .

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 $A_{red}$ ) is the identity matrix. To get from $A$ to $A_{red}$ , 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 $\det(A_{red})$ by a nonzero factor. In particular, $\det(A)\neq 0$ if and only if $\det(A_{red})\neq 0$ .

If $A$ is invertible, $\det(A_{red})=\det(I)=1$ , so $\det(A)\neq 0$ .

If $A$ is not invertible, $A_{red}$ has a row of zeros ($A_{red}$ 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 $A_{red}=I$ ; but we're assuming $A$ is not invertible). Since $A_{red}$ has a row of zeros, $\det(A_{red})=0$ , 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 $A=E_{ij}(\lambda)$ then $AB$ is obtained from $B$ by $R_{i}\mapsto R_{i}+\lambda R_{j}$ , 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 $A=E_{i}(\lambda)$ then $AB$ is obtained from $B$ by $R_{i}\mapsto\lambda R_{i}$ , so $\det(AB)=\lambda\det(B)$ . We also have $\det(A)=\lambda$ , 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 $\det(AB)\neq 0$ . 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.