Hermitian adjoint - can I apply star to eigenvalue? $T(v)=kv$ $ >> $ $T^*(v)=k^*v$

Question

Is what I wrote true? can I apply Star ( $^*$) like that and get the equality?
Sorry if you can find online what I said ( my native is not English and I didn't know how to write - as you see I wrote star - $^*$ )

Answer 1

You seem to be assuming that an eigenvector of $T$ implies it's also an eigenvector of $T^*$. This is not true even in the real case. Counterexample: $$A =\begin{pmatrix}1 & 0\\2 &-1\end{pmatrix}$$

check that $v=(1,1)$ is an eigenvector. However $$A^T =\begin{pmatrix}1 & 2\\0 &-1\end{pmatrix}$$

and $v=(1,1)$ is not an eigenvector.