# Show I - AB is invertible related [duplicate]

Let $A$ and $B$ be $n$ x $n$ matrices over reals. Show that $I - BA$ is invertible if $I - AB$ is invertible. Deduce that $AB$ and $BA$ have the same eigenvalues.

I know how to prove that $AB$ and $BA$ have same eigen values, when either $A$ or $B$ is non-singular.

But here, that condition is not mentioned.

## marked as duplicate by Robert Israel matrices StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 6 '17 at 18:20
Note that $$B^{-1}(I-BA)B=I-AB.$$ Hence $\det(I-BA)=\det(I-AB)$, so that $I-BA$ is invertible if and only if $I-AB$ is invertible. Actually, the question has been answered here already. The second part is a duplicate as well, see here, and here.