# Eigenvector of unitary matrix

We are given that, $\;\;\;\;\;\large{Ax=\lambda x}$ $\;\;\;$eqn- $(1)$

Let $\lambda$ be the eigenvalue of a unitary matrix such that $A{\overline{A}}^{T}=I$.

Then we know for a unitary matrix $|\lambda|=1\implies\lambda\overline{\lambda}=1$.

We also have, $\;\;\;\large{A^{T}x=\lambda x}$.

Taking conjugate both sides, $\;\;\;\;{\overline{A}}^{T}\overline{x}=\overline{\lambda}\overline{x}$

We can then have,

$A^{-1}\overline{x}=\overline{\lambda}\overline{x}$

Rearranging this,

$\large{A\overline{x}=\frac{1}{\overline{\lambda}}\overline{x}}$

Using property of eigenvalue for unitary matrix mentioned above.

$\large{A\overline{x}=\lambda\overline{x}}$

But comparing this and eqn $(1)$ gives weird result. Am i doing something wrong, or am I not able to understand something? Please help.

#### Solutions Collecting From Web of "Eigenvector of unitary matrix"

If $\lambda$ is an eigenvalue of an unitary matrix then $| \lambda |^2=1$ where $| \lambda |^2=\lambda {\overline {\lambda}}$ Indeed if $A$ is unitary matrix and $v$ is an eigenvector of $A$ then $$Av=\lambda v$$ and taking the transpose and the conjugate both sides $$^t{\overline{v}} ^t{\overline{A}}={\overline {\lambda}} ^t{\overline{v}}$$ and multiplying both sides for $Av$ we obtain that $$^t{\overline{v}}^t{\overline{A}} Av={\overline {\lambda}} ^t{\overline{v}} Av=^t{\overline{v}} v=\lambda {\overline {\lambda}} ^t{\overline{v}} v$$ from here $$\lambda {\overline {\lambda}} =1$$