# Is there any relation between an eigenvector of $A$ and the eigenvector of $A^T$ with the same eigenvalue?

Let $$A$$ be a square matrix over $$\mathbb C$$, and let $$A^T$$ denote its transpose.

It is not hard to see that $$A$$ and $$A^T$$ have the same set of eigenvalues, so given $$Ax=\lambda x$$ for some vector $$x\in V$$ and eigenvalue $$\lambda\in\mathbb C$$, we know that there must always be some other $$y\in V$$ such that also $$A^T y=\lambda y.$$

We also know that, if $$Ax=\lambda x$$ and $$A^T y=\mu y$$ with $$\lambda\neq \mu$$, then $$\langle y^*,x\rangle=0$$, where $$y^*$$ denotes the vector whose elements are complex conjugate of those of $$y$$, as follows from $$\langle y^*,Ax\rangle=\lambda \langle y^*,x\rangle=\mu\langle y^*,x\rangle.$$

The same argument, however, does not provide any information for the case $$\mu=\lambda$$. Is there relation holding in general for such a case?

More precisely, given $$Ax=\lambda x$$ and $$A^T y=\lambda y$$, is there any general relation between $$x$$ and $$y$$?

This is a classical result that any matrix $$A$$ is similar to its transpose.
Then, write $$A^T=PAP^{-1}$$, you have that if $$A^Tx=\lambda x$$, then $$A(P^{-1}x)=\lambda (P^{-1}x)$$
So that $$E_\lambda(A^T)=P\cdot E_\lambda(A).$$
The matrix $$P$$ isn't always easy to find, and indeed the "classical result" is not that easy to show. This can be done with a density argument when the field is for example $$\mathbb C$$. In a more general context, you can use Frobenius matrix reduction to see it.