Show that $A$ and $A^T$ do not have the same eigenvectors in general

Question

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?

Answer 1

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.

Answer 2

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 $\begin{bmatrix} 0\\ a \end{bmatrix}$.

Answer 3

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 $(0,1)$.

Answer 4

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