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I have a matrix $A$ which has distinct, real eigenvalues. Will $A$ and $A^T$ have the same eigenvectors. I thought this result was false, but I am reading a paper which uses this as a fact when $A$ is a $2\times 2$ real matrix.

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    $\begingroup$ It is indeed false. Check $A = \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}$ for example. $\endgroup$ Sep 9, 2020 at 21:29
  • $\begingroup$ Is the paper taking the transpose of the eigenvectors as well as the transpose of the matrix? $\endgroup$ Sep 9, 2020 at 21:31
  • $\begingroup$ Oops, I misread the question. Deleting my previous comments because they are misleading. $\endgroup$ Sep 9, 2020 at 21:31
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    $\begingroup$ This seems strange, but eigenvalues are indeed equal, maybe authors just mixed it somehow? For eigenvectors, right eigenvectors of $A^t$ are left eigenvectors of $A$. $\endgroup$ Sep 9, 2020 at 21:54
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    $\begingroup$ @PeterFranek it turns out the author stated that they were left eigenvectors and I just read that they were eigenvectors $\endgroup$ Sep 10, 2020 at 18:22

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With the hypothesis that the eigenvalues of $A$ are distinct, $A$ and $A^T$ have the same eigenvectors iff $A$ is normal ($A^T A = A A^T$), and with the further hypothesis that the eigenvalues of $A$ are real, $A$ and $A^T$ have the same eigenvalues iff $A$ is symmetric ($A^T = A$). So any non-symmetric matrix with real distinct eigenvalues is a counterexample; in the $2 \times 2$ case we can take, for example,

$$A = \left[ \begin{array}{cc} 0 & -2 \\ 1 & 3 \end{array} \right].$$

$A$ has eigenvectors $v_1 = \left[ \begin{array}{c} -1 \\ 1 \end{array} \right]$ and $v_2 = \left[ \begin{array}{c} -2 \\ 1 \end{array} \right]$ with eigenvalues $\lambda_1 = 2$ and $\lambda_2 = 1$. Its transpose

$$A^T = \left[ \begin{array}{cc} 0 & 1 \\ -2 & 3 \end{array} \right]$$

has the same eigenvalues but eigenvectors $w_1 = \left[ \begin{array}{c} 1 \\ 2 \end{array} \right], w_2 = \left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$.

You'll notice in the above example that $v_1$ is orthogonal to $w_2$ and vice versa. This generalizes: if $v_i$ is an eigenvector of $A$ with eigenvalue $\lambda_i$ and $w_j$ is an eigenvector of $A^T$ with eigenvalue $\lambda_j \neq \lambda_i$, then

$$w_j^T A v_i = w_j^T (\lambda v_i) = \lambda_i w_j^T v_i$$

but we also have

$$w_j^T A v_i = (A w_j)^T v_i = (\lambda_j w_j)^T v_i = \lambda_j w_j^T v_i$$

which gives $w_j^T v_i = 0$. This says that if $A$ has distinct real eigenvalues then the eigenvectors of $A$ and $A^T$ are dual bases with respect to the inner product (up to scale), which among other things is one way to prove the above result: if the eigenvectors of $A$ aren't orthogonal then they can't be their own dual basis.

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  • $\begingroup$ Is it true that: "$A$ and $B$ (with distinct real eigenvalues same as $A$) have the same eigenvectors iff $B A = A B$"? I think the the last equality holds iff $B=A^k$ for some $k$. (so in case $B=A^T$ it should be $B=A^k$. isn't?) $\endgroup$
    – C.F.G
    Sep 10, 2020 at 4:57
  • $\begingroup$ The first statement is true but the second isn't. Any polynomial in $A$ commutes with $A$. $\endgroup$ Sep 10, 2020 at 5:20
  • $\begingroup$ I actually meant that. $P(A)=I+aA+...$. So $A^T$ is a polynomial of $A$. $\endgroup$
    – C.F.G
    Sep 10, 2020 at 5:23
  • $\begingroup$ Yes, for a matrix with distinct eigenvalues that's equivalent, but normality is much easier to check! $\endgroup$ Sep 10, 2020 at 5:36

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