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.

  • $\begingroup$ the spectral radius of the zero matrix is less than one. $\endgroup$ – Thoth Mar 3 '16 at 1:51
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    $\begingroup$ 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. $\endgroup$ – Robert Israel Mar 3 '16 at 2:12
  • $\begingroup$ Sorry, you are correct. How would I go about solving this? $\endgroup$ – 100001 Mar 3 '16 at 2:36

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 $\rho(M)\ge 1$.


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