Show that $A$ and $A^T$ do not have the same eigenvectors in general I understood that $A$ and $A^T$ have the same eigenvalues, since
$$\det(A - \lambda I)= \det(A^T  - \lambda I) = \det(A - \lambda I)^T$$
The problem is to show that $A$ and $A^T$ do not have the same eigenvectors. I have seen around some posts, but I have not understood yet why. 
Could you please provide an exhaustive explanation of why in general $A$ and $A^T$ do not have the same eigenvectors?

The matrix $A=\begin{bmatrix}
1 & 1\\
0 & 1
\end{bmatrix}$, and its transpose $A^T$, have only one eigenvalue, namely $1$. However, the eigenvectors of $A$ are of the form $\begin{bmatrix}
a\\
0
\end{bmatrix}$, whereas the eigenvectors of $A^T$ are of the form $\begin{bmatrix}
0\\
a
\end{bmatrix}$.
