Existence of Matrix inverses depending on the existence of the inverse of the others.. Let $A_{m\times n}$ and $B_{n\times m}$ be two matrices with real entries. Prove that $I-AB$ is invertible iff $I-BA$ is invertible.

This follows from the fact that $\det(I - AB) = \det(I - BA)$; this is called the Sylvester's determinant theorem. A neat explanation is given at the beginning of this beautiful blog post by Terry Tao.
Actually, this is a results that holds in any ring with $1$, I first discovered it in an exercise in Jacobson's Basic Algebra I.
Here's a proof that in any ring $A$ with $1$, we have that $1-ab$ is invertible if and only if $1-ba$ is. Assume $1-ab$ invertible. Compute
$$
ba 
= 
b (1-ab)(1-ab)^{-1} a
=
(b-bab)(1-ab)^{-1} a
=
(1 - ba) b (1-ab)^{-1} a.
$$
Thus
$$
1 - ba = 1 - (1 - ba) b (1-ab)^{-1} a.
$$
Taking the second term on the RHS to the LHS,
$$
(1-ba) \cdot (1 + b(1-ab)^{-1} a) = 1.
$$
Of course one has to check that  $(1 + b(1-ab)^{-1} a) \cdot (1-ba) = 1$ also holds, and then one has $$(1-ba)^{-1} = 1 + b(1-ab)^{-1} a.$$
