Exercise 5.1.8

(a)

A matrix $M$ is invertible if and only if $0$ is not an eigenvalue of $M$.

(b)

Let $M$ be an invertible matrix. We have $\lambda$ is an eigenvalue of $M$ if and only if ${\lambda }^{-1}$ is an eigenvalue of $M^{-1}$.

First, if $M$ is invertible, then there’s no vector $v$ such that $\mathit{Mv}=0v=0$. So $0$ is not an eigenvalue of $M$. If $0$ is not an eigenvalue of $M$, then $v=0$ is the only vector such that $\mathit{Mv}=0$. This means that $M$ is injective and so invertible since $M$ is square. Second, it’s enough to prove one side of that statement since $M={(M^{-1})}^{-1}$. And if we have $\mathit{Mv}=\mathit{\lambda v}$, then we have $M^{-1}v={\lambda }^{-1}v$.