Prove the following statement or give a counterexample if it is false.

If $A$ is an $m\times n$ matrix and $B$ is an $n\times m$ matrix then $AB$ is invertible if and only if $BA$ is invertible.

What i tried:

I mentioned that it is true

To prove the forward implication

$\det(AB)=\det(A)\det(B)\neq 0$

Then $\det(A)\det(B)=\det(B)\det(A)=\det(BA)\neq 0$

Hence $BA$ is invertible

We do the same to prove the backward implication

$\det(BA)=\det(B)\det(A)\neq 0$

Then $\det(B)\det(A)=\det(A)\det(B)=\det(AB)\neq 0$

Hence $AB$ is invertible

Is my proof correct? Could anyone explain. Thanks


It's not correct. You cannot speak of the determinants of $A$ and $B$ when they are not square.

  • $\begingroup$ But since $AB$ and $BA$ are defined, both $A$ and $B$ are square matrices. $\endgroup$ Sep 19 '15 at 14:46
  • $\begingroup$ That means for a non square matrix, the determinant does not exists? $\endgroup$
    – ys wong
    Sep 19 '15 at 14:47
  • $\begingroup$ @Scientifica No. E.g. $A=(1,0),\ B=\pmatrix{1\\ 0}$ give $AB=1,\ BA=\pmatrix{1&0\\ 0&0}$. $\endgroup$
    – user1551
    Sep 19 '15 at 14:48
  • 1
    $\begingroup$ @Scientifica No. If $A$ is $n\times m$ and $B$ is $m\times n$ then both $AB$ and $BA$ are defined (and square) even if $n\ne m$. $\endgroup$ Sep 19 '15 at 14:49
  • 1
    $\begingroup$ @yswong Determinant is defined for square matrices only. Your proof works if $m=n$, but this is not a given condition. $\endgroup$
    – user1551
    Sep 19 '15 at 14:49

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