Can the same vector be an eigenvector of both $A$ and $A^T$? 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.
 A: 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.
A: Let $n\geq 2$ and $Z_n=\{A\in M_n(\mathbb{C}); A,A^T \text{have at least one common eigenvector}\}$.
Proposition. $M_n(\mathbb{C})\setminus Z_n$ is a Zariski open dense subset. That implies (for example) that if you randomly choose the $(a_{j,k})=(\alpha_j+i\beta_k)$ according to a normal law, then $A,A^T$ have no common eigenvector, with probability $1$.
Proof. According to Shemesh, cf. Remark 3.1:
$A\in Z_n$ IFF $(*)$ the matrix $[[A,A^T]^T,\cdots,[A^k,{A^l}^T]^T,\cdots,[A^{n-1},{A^{n-1}}^T]^T]$ has rank $<n$. Since $(*)$ can be written as a complex algebraic system of relations, $Z_n$ is a Zariski closed subset. It remains to show that, for every $n$, $Z_n$ is not $M_n(\mathbb{C})$.
Choose $a_{j,j}=j$ and if $j<k$ then $a_{j,k}=1$, the other $a_{i,j}$ being $0$.
A: 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 $A=A^T$.
