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 $A=\begin{bmatrix}
1 &amp; 1\\
0 &amp; 1
\end{bmatrix}$, 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 $\begin{bmatrix}
0\\
a
\end{bmatrix}$.
Though $A$ and $A^T$ have the same eigenvalues, they may have different eigenvectors for that particular eigenvalue. This is because, if we assume the eigenvector (lets say $v$) to be non zero, then 
$$Av = (eigenvalue(A).v)$$
and 
$$A^Tv = eigenvalue(A^T).v$$
so $$Av - A^Tv = 0$$
so $$(A - A^T).v = 0$$

Now since we assumed $v$ to be non zero,  this will happen only if $(A-A^T)$ has 
a valid non zero null space.

So $A$ and $A^T$ will not have same eigenvectors for a given eigenvalue, unless the above condition holds true.