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We have a square matrix A, and i need to prove that it's eigenvectors are the same as the one of $A^T$.

I was thinking to an example; but i know it's not a proof, Let A:

\begin{bmatrix}1 & 4\\2 & 3\end{bmatrix}

It's eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 5$, and after some calculation we can prove that it's eigenvectors are for $\lambda_1=$ \begin{bmatrix}1\\1\over 2\end{bmatrix} and \begin{bmatrix}1\\1\end{bmatrix} for $\lambda_2$.

If we now take $A^T$ and we make the same calculations we can see that the Eigenvector of $A$ are the not the same as the one of $A^T$, but how to prove this without an example?

Edit: Eigen vector of \begin{bmatrix}1 & 4\\2 & 3\end{bmatrix}= $v_1=(1,1);v_2=(-2,1)$

Eigen vector of \begin{bmatrix}1 & 2\\4 & 3\end{bmatrix}= $v_1=(1,2);v_2=(-1,1)$

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    $\begingroup$ After the edit the statement is still false in general, just take any matrix that equals its transpose. $\endgroup$ – Git Gud May 9 '15 at 11:59
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    $\begingroup$ I just said that it is false. $\endgroup$ – Git Gud May 9 '15 at 12:03
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    $\begingroup$ Would it help to mention that $A$ and $A^T$ share their eigenvalues? In other words, we have $A v_1 = \lambda_1 v_1$ and $A^T w_1 = \lambda_1 w_1$. If I understand your question correctly, you now have to prove that $v_1$ and $w_1$ are not always parallel. $\endgroup$ – Ailurus May 9 '15 at 12:17
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    $\begingroup$ @BioShock You have indeed proved that the statement “For every square matrix $A$, the eigenvectors of $A$ and $A^T$ are equal” is false: an example suffices. Note that you can't prove the (different) statement “For every square matrix $A$, $A$ and $A^T$ share no eigenvector”, because it's false as well (think to a symmetric matrix). $\endgroup$ – egreg May 9 '15 at 12:20
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    $\begingroup$ Now you've returned to the original version of the statement which I've said in a now deleted comment that it is false as per your example. $\endgroup$ – Git Gud May 9 '15 at 12:22
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It seems that your problem is one of logic, not of linear algebra.

There are matrices $A$ such that $A$ and $A^\top$ have the same eigenvectors. Here is an example: $$A=\left[\matrix{1&2\cr 2&1\cr}\right]\ .$$ There are matrices $A$ such that $A$ and $A^\top$ do not have the same eigenvectors. Here is an example: $$A=\left[\matrix{1&1\cr 0&1\cr}\right]\ .$$ Therefore the statements (a) "For any square matrix $A$, the matrices $A$ and $A^\top$ have the same eigenvectors" and (b) "For any square matrix $A$, the matrices $A$ and $A^\top$ do not have the same eigenvectors" are both false.

Note that any $\forall$-statement, when wrong, is falsified with a single counterexample.

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