# matrix&eigenvalues

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.

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 $$\|Av\|^2 = v^*A^*Av = v^*v = \|v\|^2$$
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…
• how you obtain this: $A^*.v=\overline\lambda.v$? – Cosmin Cretu Jan 19 '18 at 15:59
• @CosminCretu How do you define $A^*$? – José Carlos Santos Jan 19 '18 at 16:01
• @CosminCretu So, it is the adjugate matrix of $A$! I am sorry; I thought that it was the adjoint matrix. But then the statement is false. If$$A=\begin{pmatrix}1&2\\3&7\end{pmatrix},$$then $A^{-1}=A^*$, but the eigenvalues of $A$ are $4\pm\sqrt{15}$. – José Carlos Santos Jan 19 '18 at 16:10
The eigenvalues of $A^{-1}$ and $A^*$ are $1\over\lambda$ and $\lambda^*$ respectively. So we can write: ${1\over\lambda}=\lambda^*$ or $\lambda=e^{i\theta}$