An interesting linear algebra question Let $A$ and $u$ be $n\times n$ matrix and $n\times 1$ vector of $\mathbb{C}$.
Denote $\overline{A}$ is the matrix $(\overline{A})_{ij}=A_{ij}^*$, the conjugate number; ($\overline{A}$ is not the conjugate transpose matrix) and similarly $\overline{u}$.
Prove that if $\lambda$ is a nonnegative eigenvalue of $A\overline{A}$, ie $\exists v\ne 0:A\overline{A}v=\lambda v$, then $\exists u\ne 0$ such that:
$$A\overline{u}=\sqrt{\lambda}u$$

Actually I have found a simple way to directly point out the vector $u$:


*

*$u=\sqrt{\lambda}A\overline{v}+\lambda v$ if $\sqrt{\lambda}A\overline{v}+\lambda v\ne0$

*If $\sqrt{\lambda}A\overline{v}+\lambda v=0$ then $A\overline{v}=-\sqrt{\lambda}v$ then we choose $u=iv$
Also thanks for the elegant answer from User1551.
Thanks to Leeuwen's comment I think the more appropriate choice is $u=A\overline{v}+\sqrt{\lambda}v$
