Prove matrix AB is invertible. 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? 

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. 
