How would I show that if the spectral radius of a matrix $M$ is less than $1$ then the matrix $I - M$ is invertible?
I believe that I'm supposed to work towards a contradiction to show that if the matrix was singular then 1 would be an eigenvalue, but I'm not sure how to do this.
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 $\rho(M)\ge 1$.