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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.

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Hint:$(I-BA)^{-1}=X$ (say), Now expand left side. we get $$X=I+BA+ (BA)(BA)+(BA)(BA)(BA)+\dots$$ $$AXB=AB+(AB)^2+(AB)^3+(AB)^4+\dots$$ $$I+AXB=I+(AB)+(AB)^2+\dots+(AB)^n+\dots=(I-AB)^{-1}$$

Check yourself: $(I+AXB)(I-AB)=I$, $(I-AB)(I+AXB)=I$

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You're writing a series $\sum_{i=1}^{\infty} (AB)^{i}$, but this in general won't converge. – Andreas Caranti Feb 8 '13 at 11:24

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 $$(1-ba)^{-1} = 1 + b(1-ab)^{-1} a.$$

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Hint: Assume $m<n$, use the fact that $p_{BA}(t)=t^{n-m}p_{AB}(t)$, where $p_{AB}$ is the characteristic polynomial of $AB$. Similar for $m>n$.

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