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?

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
