Actually, I saw normalizer of diagonal matrices are permutation matrices. I read the answer but I don't know how to prove that conjugation preserves the spectrum. Actually I do some proof on 2x2 matrices, but had no idea on expanding this proof to n by n matrices.


Hint: If $Av=\lambda v$, we have $(S A S^{-1})(Sv)=\lambda (Sv)$.

  • $\begingroup$ Oh thanks, I will try it! $\endgroup$ – user122655 Jan 21 '14 at 9:19
  • $\begingroup$ Oh I see why the spectrum is preserved. Thank you! $\endgroup$ – user122655 Jan 21 '14 at 9:22
  • $\begingroup$ Just Brilliant. $\endgroup$ – MR_BD Apr 7 '17 at 9:09

From a more abstract perspective, conjugation of a matrix $A$ by another matrix $B$ simply gives you the same linear transformation $A$, but in a different basis. Fortunately, eigenvectors and eigenvalues don't depend on a choice of basis, so they will be the same for a matrix and one of its conjugates.

  • $\begingroup$ But in order to make the connection you are using, one exactly has to prove what is asked here! Hence: Your argument is circular, it is just a different perspective. $\endgroup$ – Martin Brandenburg Jan 21 '14 at 9:30
  • $\begingroup$ @MartinBrandenburg The part about conjugation being a change of basis or eigenvalues not depending on a choice of basis? By the second part I mean that eigenvalues are a property of an abstract linear transformation, and have little to do with a basis. $\endgroup$ – Dylan Yott Jan 21 '14 at 9:33
  • $\begingroup$ I think your perspective makes me see the problem differently. However, I needed to why eigenvalues do not depend on a choice of basis. Thanks! $\endgroup$ – user122655 Jan 21 '14 at 9:36
  • $\begingroup$ @MartinBrandenburg Okay, I see what you're saying now. I hope this is still helpful for intuition even if it isn't a proof. $\endgroup$ – Dylan Yott Jan 21 '14 at 9:38
  • $\begingroup$ @DylanYott Yes of course it is helpful for me. Thanks Dylan! $\endgroup$ – user122655 Jan 23 '14 at 5:48

Another alternative proof:

Let $\lambda$ be an eigenvalue of $A$.
Then $|A-\lambda I_n|=0$.
Next, $|TAT^{-1}-\lambda I_n|=|TAT^{-1}-\lambda TT^{-1}|=|T||A-\lambda I_n||T|^{-1}=|A-\lambda I_n|=0$
Hence $\lambda$ is also an eigenvalue for $TAT^{-1}$.


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