Can the same vector be an eigenvector of both $A$ and $A^T$?

Question

It is proven that $A$ and $A^T$ have the same eigenvalues. I want to study what stands for eigenvectors. Let me make a try. Given:

$$Ax=\lambda x$$ we know that $x\in C(A)$ for $\lambda \neq 0$. Suppose that for $A^T$ we have the same eigenvectors $x$:

$$A^Tx=\lambda x$$ but now we have that $x\in C(A^T)$. Based on this, eigenvector's $x$ belong both in column and row space which is impossible. So, $A$ and $A^T$ have different eigenvectors.

Am I right about this deduction? In any case, could you please suggest a different way if possible?

Thanks.

^{PS: After @G Tony Jacobs comments I made some changes hopping that I have less mistakes.}

Answer 1

For matrices with distinct eigenvalues, (same eigenvalues) + (same eigenvectors) = (same matrix).

Therefore any asymmetric $A$ with distinct eigenvalues is an example where $A$ and $A^T$ have different eigenvectors.

To write down such an example, take any upper triangular matrix with distinct entries on the diagonal.

Answer 2

A matrix A is diagonalizable into a diagonal matrix D (D's diagonal components are eigenvalues). The transformation matrix from the original basis to the diagonal basis of A is P. In other words, P is the eigenvectors of A expressed in the original basis (or the columns of P are the eigenvectors).

This is equivalent to: There exists a diagonal matrix D and an invertible matrix P such that: $D=P^{-1}AP$

Tranposing the above eqt gives: $D=P^{T}A^TP^{-T}$

We conclude that the matrix $A^T$ is diagonalizable into the same diagonal matrix D, thus its eigenvalues are identical to those of A. Additionally, the transformation matrix from the original basis to the diagonal matrix of $A^T$ is $P^{-T}$. In other words, $P^{-T}$ is the eigenvectors of $A^T$ expressed in the the original basis. Generally, $P^{-T}\ne P$.

If P is orthogonal i.e eigenvectors of A are orthonormal, we have $P^{-T}= P$, which means that eigenvectors of A and $A^T$ are identical. This result is logic because if P is orthogonal, A is symmetric hence $A=A^T$.