This exercise comes from a university exam (http://www.ubacs.com.ar/foro/viewtopic.php?f=67&t=3079, link in spanish). I'll copy it in english for everyone. It's #3:
We define in $C^{n×n}$ the canonical inner product $<A, B> = tr(AB^*)$. Let $P \in GL(n,\mathbb{C})$ be and $f \in End_\mathbb{C}(\mathbb{C^{n×n}})$, defined as $f(A)=P^{-1} A P$.
(a) Calculate $f^*$ and prove that if P is hermitic (i.e. self-adjoint), then f is self-adjoint.
(b) If P is hermitic, prove that if v,w $\in \mathbb{C^n}$ are column vectores eigenvectors of $P$ then $ v\cdot w^* \in \mathbb{C^{n×n}}$ is an eigenvector of $f$. ¿With what eigenvalue?
--
(a) is easy; I have problems with (b).
(I'm assuming that v and w are linearly independent, I know they could be linearly dependent).
I thought that if I define $B= \{v,w,z_3,...z_n\}$ a base of eigenvectors then $(v)_B=e_1$ and $(w)_B=e_2$. If I work in that base then the exercise is easy and the eigenvector will be $\frac{\lambda_v}{\lambda_w}$ ($Pv = \lambda_v v$; $Pw = \lambda_w w$). But...I don't think that's right. Does anywone have any idea? Thanks!
We have $$w^*P=(w^*P)^{**}= (P^*w)^*=(Pw)^*=(\lambda_ww)^*= \overline{\lambda_w}w^*.$$ Since $Pv=\lambda_vv$, $v=P^{-1}Pv=P^{-1}\lambda_vv=\lambda_v P^{-1}v$ and $P^{-1}v=\lambda_v^{-1}v$.
Finally $$f(v\cdot w^*)=P^{-1} v\cdot w^*P=\overline{\lambda_w} P^{-1}v\cdot w^*= \overline{\lambda_w} \lambda_v^{-1}v\cdot w^*.$$