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Is there an example such that $A$ and $B$, three by three, that have the same eigenvectors, but different eigenvalues?

What would be the eigenvectors and eigenvalues if it exists because I'm stuck on this practice problem.

I know that if matrices $A$ and $B$ can be written such that $AB=BA$, they share the same eigenvectors, but what about their eigenvalues? precisely if they're squared matrices ($3\times 3$ case)

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How about $I$ and $-I$? Then all $x\neq0$ are eigenvectors. But the eigenvalues are $1$ and $-1$ respectively.

For a less trivial example, how about $\begin{pmatrix}1&0&0\\0&2&1\\0&0&2\end{pmatrix}$ and $\begin{pmatrix}4&0&0\\0&3&1\\0&0&3\end{pmatrix}$?

They share eigenvectors $(1,0,0)$ and $(0,1,0)$, but for different eigenvalues.

As to your question about commuting matrices, you could take any two diagonal matrices, with different entries on the respective diagonals.

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    $\begingroup$ It only works for the identity matrices? $\endgroup$ – xim Jan 29 at 3:04
  • $\begingroup$ Hmm. I'll have to think about it. It looks like multiples of identity matrices will work. How about a permutation matrix? $\endgroup$ – Chris Custer Jan 29 at 3:33
  • $\begingroup$ It would work if you put different values on the diagonal. Maybe using Jordan normal form we can make examples. $\endgroup$ – Chris Custer Jan 29 at 3:55

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