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?
 A: The thing is that, unless the matrix is symmetric, $A$ and $A^T$ represent different systems of equations. Try with a simple example. 
When calculating the eigenvectors you solve the equations $(A-\lambda I)v =0$ and $(A^T-\lambda I)w=0$, which again are different systems.
A: Since there isn't an answer with everyone's favourite counterexample yet, here's one: consider the matrix
$$ A=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}. $$
Then $A$ has only one eigenvector, namely $(1,0)$, with eigenvalue $0$. Meanwhile,
$$ A^T=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} $$
also has only one eigenvector, namely $(0,1)$.
A: 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 $\begin{bmatrix}
0\\
a
\end{bmatrix}$.
A: Think of it like this: the eigenvector $\vec{v}$ is the vector that satisfies $$A\vec{v}=\lambda\vec{v}$$ when $A$ is an $n*n$ matrix and $\lambda$ is an eigenvalue of $A$.
We know that $A$ and $A^{T}$ have the same eigenvalues, but $A=A^{T}$ is only true if $A$ is a symmetric matrix.
Let's assume $A$ is an $n*n$ matrix but not a symmetric matrix.
Therefore $A\neq A^{T}$. Let's also assume that $A$ has at least one real eigenvalue, $\lambda_{1}$. Therefore, $A^{T}$ has at least one real eigenvalue, $\lambda_{1}$. Therefore  $$A\vec{v}=\lambda_{1}\vec{v} $$ and $$A^{T}\vec{w}=\lambda_{1}\vec{w}$$
Let's assume that the eigenvectors $\vec{v}=\vec{w}$
$$\vec{v}=\vec{w} \implies\lambda_{1}\vec{v}=\lambda_{1}\vec{w}\implies A\vec{v}=A^{T}\vec{w}$$
but if $\vec{v}=\vec{w}, $ and $A\neq A^{T}$, then $\vec{v}=\vec{w}=\vec{0}$ which is definitionally not an eigenvector. Therefore, if $\vec{v}$ and $\vec{w}$ are eigenvectors of $A$ and $A^{T}$ respectively, $\vec{v}\neq \vec{w}$. Therefore $A$ and $A^{T}$ have at least one eigenvector not in common
A: 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.
