matrix&eigenvalues

Question

Let $A \in M_n(C)$ and $A^{-1}=A^*$. Prove that the eigenvalues of A have the modul equal with one.

P.S. I know that this is a property well-known, but I couldn't find a demonstration for it.

Answer 1

The eigenvalues of $A^{-1}$ and $A^*$ are $1\over\lambda$ and $\lambda^*$ respectively. So we can write: ${1\over\lambda}=\lambda^*$ or $\lambda=e^{i\theta}$

Answer 2

Let $\lambda$ be an eigenvalue of $A$. Then $\lambda\neq0$ because $A$ is invertible. Let $v$ be an eigenvector with eigenvalue $\lambda$. Then $A.v=\lambda v$. Therefore$$A^{-1}.v=A^{-1}.\left(\frac1\lambda.\lambda.v\right)=\frac1\lambda.(A^{-1}.(\lambda v))=\frac1\lambda v.$$But $A^{-1}=A^*$ and $A^*.v=\overline\lambda.v$. Therefore…

Answer 3

If $v$ is an eigenvector of $A$, then $$ \|Av\|^2 = (Av)^*(Av) = (\lambda v)^*(\lambda v) = |\lambda|^2 \|v\|^2 $$ However, in this case we can also write $$ \|Av\|^2 = v^*A^*Av = v^*v = \|v\|^2 $$