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Can there be two matrices that have the same eigenvalues and corresponding eigenvectors? Lets say there is a matrix with eigenvalues 1 and 2 and eigenvectors $(1,2)^T$ and $(3,4)^T$, will there be another matrix with these? I want to intuitively say no, but I'm not positive.

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    $\begingroup$ The two $2 \times 2$ diagonal matrices $diag(1,2)$ and $diag(2,1)$ have the same eigenvalues and eigenvectors, just in a different order. $\endgroup$ – Simon S Apr 9 '15 at 14:26
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It is possible for multiple matrices to have the same eigenvalues and eigenvectors as long as the diagonals have the same entries, but in a different order.

If the elements of the trace are equal, and the traces are equivalent, then the eigenvalues and eigenvectors will be the same.

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  • $\begingroup$ I tried this with my example, by solving $BDB^{-1}$ where B has the eigenvectors as columns and D with the eigenvalues as corresponding diagonal values. I then did the same with the eigenvalues and eigenvectors switched. Both gave me the same matrix, which I believe would make it unique. Or is this just a special case? $\endgroup$ – TheStrangeQuark Apr 9 '15 at 14:43
  • $\begingroup$ It would indeed be unique. @TheStrangeQuark $\endgroup$ – Chan Hunt Apr 9 '15 at 15:33
  • $\begingroup$ Okay! I figured it out by making a generic matrix and solving by definition to find it. But I understand what you were saying. $\endgroup$ – TheStrangeQuark Apr 9 '15 at 16:05

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