Condition for all eigenvalues to be less than 1 (proof)

Question

From the Lutkepohl (2005) Appendix A.8:

All eigenvalues of the $m\times m$ matrix $A$ have modulus less than $1$ if and only if $\det(I_m - Az) \neq 0$ for $|z| \leq 1$, that is, the polynomial $\det(I_m - Az)$ has no roots in and on the complex unit circle.

I am seeking for the proof of this statement (as simple as possible)

Answer 1

The inequality holds automatically for $z=0$. For $z \neq 0$, $\operatorname{det}(I-Az)=z^m \operatorname{det}(z^{-1}I-A)$, which will be zero if and only if $z^{-1}$ is an eigenvalue of $A$. Now $\{ z^{-1} : |z| \leq 1,z \neq 0 \}=\{ z : |z| \geq 1 \}$, so the result follows.

Answer 2

So here is a proof:

1. $det(I_m-Az) = 0 \Rightarrow \exists x: Ax=\lambda x, |\lambda|>1$:

$det(I_m-Az) = 0 \Rightarrow \exists x: (I_m -Az)x=0 \Rightarrow Ix=Azx\Rightarrow Ax=\frac x z \Rightarrow $ x is an eigen vector of A with eigenvalue $\frac 1 z , |\frac 1 z|>1$

2. $\exists x: Ax=\lambda x, |\lambda|>1 \Rightarrow det(I_m-Az) = 0$:

$\exists x: Ax = \lambda x, |\lambda | >1\Rightarrow(I_m -A\cdot \frac 1 \lambda )x= I_mx - \frac {Ax} \lambda=x- x=0$

$\Rightarrow det(I_m -A\cdot \frac 1\lambda )=0 , |\frac 1\lambda |<1 $