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Suppose $A$ and $B$ are $n \times n$ matrices such that $||I-AB||$ is less than $1$, then show that $AB$ is invertible.

I started to prove this by contradiction. Assuming it wasn't invertible so $(I-AB)x=0$ and $||x|| = 1$ thus $x=ABx$. I tried to find a contradiction but didn't have any luck. Any suggestions?

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Why the two matrices $A$ and $B$? Why not just say "If $||I-A|| < 1$, show that $A$ is invertible"? – TonyK Apr 9 '12 at 17:52
It's not $I-AB$ that isn't invertible. – Robert Israel Apr 9 '12 at 17:53
up vote 7 down vote accepted

Given any matrix $M$, it is easy to see that if $\lambda$ is an eigenvalue, then $||M|| \geq |\lambda|$.

Your proof by contradiction should start by assuming that $AB$ is not invertible, so there exists a unit norm $x$ such that $ABx = 0$. From this it follows that the matrix $I-AB$ has an eigenvalue of $1$, which leads to a contradiction.

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Show that $I - C$ is invertible when $||C|| < 1$ by using the geometric series $\Sigma_{k = 0}^{\infty}C^k$. Then choosing $C = I - AB$ gives you the result.

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