How would I show that if the spectral radius of a matrix $M$ is less than $1$ then the matrix $I - M$ is invertible?
• It helps to get the question right. I suspect it should be: if the spectral radius of $M$ is less than $1$, then $I - M$ is invertible. – Robert Israel Mar 3 '16 at 2:12
Suppose that $I-M$ is not invertible, then its kernel is not trivial so $\exists v\ne 0$ such that $$(I-M)v=0$$ i.e. $Mv=v$. This shows that 1 is an eigenvalue of $M$ so $1\in \sigma(M)$, thus $\rho(M),$ the spectral radius of $M$, satisfies $\rho(M)\ge 1$.