Does $Ax=x$ imply $A^* x=x$, if $A^*$ is the conjugate transpose of $A$?

I have a fairly simple question. If $$A$$ is a matrix and $$A^*$$ denotes its conjugate transpose, is it true that if $$Ax = x$$, then $$A^*x = x$$?

The matrix $$A^*$$ will certainly have $$1$$ as an eigenvalue, but will it be with the same eigenvector? And if not, what is the relation between the eigenvector of $$A$$ and the one of $$A^*$$?

• I take it you mean that $A(x) = x$ for some particular $x$ (that is, not for any $x$)? Feb 10 '13 at 16:44
• One simple thing that should be clarified here: if one distinguishes left and right eigenvectors then one CAN say something, ie Ax = kx is the same as xA = kx. Sep 22 '13 at 10:07

$$\begin{pmatrix} 1 & 1\\ 0 & -1\end{pmatrix}$$
which has $x=(1,0)^T$ as an eigenvector with eigenvalue 1. Yet $A^*x=(1,1)^T\neq x$.
The easiest example would be to consider the rank one matrix $$A = xy^\top$$ Then $$Ax = (xy^\top) x= x(y^\top x) = x\lambda = \lambda x$$ and $$A^* x = (\bar{y} \bar{x}^\top) x =\bar{y} (\bar{x}^\top x )= \bar{y} k = k\bar{y}$$