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

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)

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• I can't reach the book but there is something weird, $I_m$ is a matrix and $Az$ is a vector, can you clarify? – Cherny Mar 2 '17 at 15:10
• @Cherny sorry I should have mentioned that $z$ is a scalar. – tosik Mar 2 '17 at 15:11
• Are you familiar with the fact that eigenvalues are roots of the characteristic polynomial of the matrix? If so, no further demonstration is needed, because this result is an immediate and obvious consequence. If not, then consult a linear algebra text or en.wikipedia.org/wiki/Characteristic_polynomial. – whuber Mar 2 '17 at 15:18

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.

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 $\frac 1 z , |\frac 1 z|>1$

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

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

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