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Under what circumstances is any left eigenvector of a matrix also a right eigenvector (and vice versa)? My guess is that this is true if the matrix is symmetric, but is this necessary, and is it sufficient? Actually I never had attended a course in linear algebra, so please be patient with me.

user136457
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    You confuse left/right eigenvector and singular vectors of a matrix (these are different notions). You seem to talk about eigenvector here, but please clarify this point. – Surb Jul 07 '15 at 22:27
  • @Surb: Why do you think so? – joriki Jul 07 '15 at 22:32
  • @joriki Well, what I know, is that $x\neq 0$ is a left eigenvector of $A$ if $A^Tx=\lambda x$, $y\neq 0$ is a right eigenvector of $A$ if $Ay=\lambda y$ and $(u,v)$ is a singular vector (actually it is a family of singular vectors) of $A$ if $Au = \lambda v$ and $A^Tv = \lambda u$... Thus (to the best of my knowledge), there is no notion of left/right singular vector. BTW, why don't you think so ? – Surb Jul 07 '15 at 22:35
  • Thanks, I am indeed confusing the notions, and I aim to talk about eigenvectors of course! – user136457 Jul 07 '15 at 22:37
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    @Surb: Wikipedia speaks of left and right singular vectors of a matrix. – joriki Jul 07 '15 at 22:37
  • @joriki ok... I did not know that... so basically as in the eigenvalue case, the "left" singular vectors are the singular vectors of the transpose matrix. In any case, symmetry is a sufficient condition for both (the eigenvalues and the eigenvectors). I'm not sure if it's necessary though (I would not be surprised that a counter-example could be given with a matrix which is not diagonalisable). – Surb Jul 07 '15 at 22:43
  • @user136457 can you show that it is sufficient ? (using the definitions I recalled in my second comment) – Surb Jul 07 '15 at 22:45
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    $A = \begin{bmatrix}0 & 1 \ -1 & 0\end{bmatrix}$ is a counterexample. It has eigenvalues $\pm i$. Since $A^T = -A$, the eigenvectors of $A$ are the same as the eigenvectors of $A^T$ with switched eigenvalues.

    If we want $A$ and $A^T$ to have the same eigenvectors with the same corresponding eigenvalues, this is a little trickier. There are counterexamples for complex matrices, but perhaps not for real ones.

     – Andrew Dudzik Jul 07 '15 at 23:09
  • @Surb what do you mean by "both" when you say "In any case, symmetry is a sufficient condition for both (the eigenvalues and the eigenvectors)." – ions me Sep 26 '22 at 01:36
  • And what was the verdict? are left and right singular vectors actually different? – ions me Sep 26 '22 at 01:37

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A real matrix $A$ is normal, that is $AA^T = A^TA$, if and only if it is orthogonal diagonalizable. In that case, the left eigenvectors are also right eigenvectors. I suspect it will be necessary too.

user251257
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