-1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $$ \|Av\|^2 = v^*A^*Av = v^*v = \|v\|^2 $$

$\endgroup$
1
$\begingroup$

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…

$\endgroup$
  • $\begingroup$ how you obtain this: $A^*.v=\overline\lambda.v$? $\endgroup$ – Cosmin Cretu Jan 19 '18 at 15:59
  • $\begingroup$ @CosminCretu How do you define $A^*$? $\endgroup$ – José Carlos Santos Jan 19 '18 at 16:01
  • $\begingroup$ the transpose of its cofactor matrix $\endgroup$ – Cosmin Cretu Jan 19 '18 at 16:03
  • 1
    $\begingroup$ @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 $4\pm\sqrt{15}$. $\endgroup$ – José Carlos Santos Jan 19 '18 at 16:10
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.