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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?

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    $\begingroup$ You just need on counterexample, and they are not difficult to find. For example, $\begin{pmatrix}1\\ -1\end{pmatrix}$ is an eigenvector of $A=\begin{pmatrix} 0 & -1 \\ 2 & 3\end{pmatrix}$ but not of $A^T$. $\endgroup$ – Dylan Jun 6 '15 at 21:05
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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 $\begin{bmatrix} 0\\ a \end{bmatrix}$.

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    $\begingroup$ I think it must be the other way around. The eigenvectors of $A$ are $\begin{bmatrix}a\\0\end{bmatrix}$ and the of $A^T$ are $\begin{bmatrix}0\\a\end{bmatrix}$. $\endgroup$ – philmcole Apr 10 '18 at 9:48
  • $\begingroup$ Is there a condition that will assure that $A$ and $A^T$ have the same eigenvector? Or are they always different? $\endgroup$ – Fareed Abi Farraj Nov 21 '19 at 22:42
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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.

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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 $(0,1)$.

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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

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  • $\begingroup$ $A\neq A^{T}$, then $\vec{v}=\vec{w}=\vec{0}$ is incorrect statement. $\endgroup$ – Prashant Singh Feb 13 '19 at 19:24
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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.

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  • $\begingroup$ Hi Tyler, you can typeset the mathematics nicely using Mathjax, tutorial here - math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Calvin Khor Oct 30 '19 at 11:12
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    $\begingroup$ I think somone could misunderstand this to say that if $A-A^T$ has a nontrivial null space then $A$ and $A^T$ have a common eigenvector. This is not true. For example, if $n$ is odd and $A$ is $n \times n$, then $A-A^T$ must have a nontrivial nullspace. However, $A$ and $A^T$ still need not have a common eigenvector. $\endgroup$ – Eric Oct 30 '19 at 16:22

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