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If so, is there something that can be said about a relationship between the set of eigenvectors of the first matrix and the set of eigenvectors of the second matrix? For example, if the answer to the above question is yes, is it true that the pairwise angles between eigenvectors remains fixed for every matrix with the same set of unique eigenvalues?

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    $\begingroup$ Just pick two disjoint orthonormal bases of $\mathbb{R}^n$ $\endgroup$ – Max Dec 1 '17 at 20:58
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    $\begingroup$ Recall the notion of similarity. Similar matrices will have the same eigenvalues, but different eigenvectors. Adjusting the matrix by which you conjugate allows you to choose these eienvectors freely. $\endgroup$ – Theo Bendit Dec 1 '17 at 21:07
  • $\begingroup$ Oh yeah of course, so given a matrix with a set of unique eigenvalues, with a change of basis it is possible to give it any arbitrary set of eigenvectors as long as they form a basis. $\endgroup$ – user023049 Dec 1 '17 at 21:16
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    $\begingroup$ $\begin{bmatrix}2&1\\0&2\end{bmatrix}$ and $\begin{bmatrix}2&0\\0&2\end{bmatrix}$ have the same set of eigenvalues (with the same multiplicity) but they don't even have the same number of linearly independent eigenvectors. $\endgroup$ – Rahul Dec 1 '17 at 21:41
  • $\begingroup$ @Rahul yes but that is precisely why I specified unique eigenvalues. $\endgroup$ – user023049 Dec 1 '17 at 22:07
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Yes it is possible.

Let's consider a n-by-n matrix $A$ and $P$ a n-by-n invertible matrix $P$

thus $$B=P^{-1}AP$$

has the same set of eigenvalus but diffirent eigenvectors.

Infact if $Av = \lambda v$

$$BP^{-1}= P^{-1}A$$

then $$BP^{-1}v =P^{-1}Av =\lambda P^{-1}v \ \implies \ B(P^{-1}v) = \lambda (P^{-1}v)$$

A and B are defined as similar matrices.

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    $\begingroup$ In particular, when $P$ is orthogonal, angles between the eigenvectors are preserved. $\endgroup$ – Theo Bendit Dec 1 '17 at 21:16
  • $\begingroup$ oh yes of course I've swapped the P's, I correct the answer. Thanks! $\endgroup$ – gimusi Dec 1 '17 at 21:53
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    $\begingroup$ Thanks taking the time to write this out. I had this in the back of my mind somewhere but for some reason I wasn't thinking about how a change of basis leaves the eigenvalues unchanged and obviously changes the eigenvectors until @TheoBendit mentioned it. $\endgroup$ – user023049 Dec 1 '17 at 22:20

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