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Let $\mathcal{A}$ be the algebra generated by a set $\mathcal{M}$ of commuting matrices over complex field. What can I say about the eigenvalues of the members of $\mathcal{A}$ in terms of the eigenvalues of the members of $\mathcal{M}$?

If all members of $\mathcal{M}$ are diagonalizable, I think there is a simple relationship. Each member of $\mathcal{A}$ is a form of $f(M_1, \cdots, M_n)$ where $f$ is a polynomial and $M_1, \cdots$ are member of $\mathcal{M}$. Then for each eigenvalue (counting the multiplicity) of $f(M_1, \cdots, M_n)$ there is an eigenvector shared by all members of $\mathcal{A}$. The eigenvalue must be $f(\lambda_1, \cdots, \lambda_n)$ where $\lambda_i$ is the eigenvalue of $M_i$ corresponding to the common eigenvector.

What relations are there if other conditions are assumed instead of the diagonalizability? I know that the quation is quite broad. But any relations between the eigenvalues may be helpful for me. Thank you.

TameBlie
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  • Have you googled with the main keywords ? There are many reference on the web. For example (https://www.researchgate.net/post/What_is_the_relation_between_eigenvalues_of_two_commuting_matrices). See as well the answer of Jose Brox in (http://math.stackexchange.com/q/1227031) – Jean Marie Sep 20 '16 at 10:51
  • Do you know the McCoy theorem? You might look it up. There are also a bunch of other relations bound with it, like semi commuting matrices and other relations. – Patrick Abraham Sep 20 '16 at 10:51
  • @JeanMarie, Of course I have googled, but didn't get much relevant results. I've read your first link, but the second one is new for me. Thank you. – TameBlie Sep 20 '16 at 10:59
  • @Abraham, I didn't aware of the theorem. Thank you. – TameBlie Sep 20 '16 at 11:00
  • Your line of reasoning applies even if the matrices are not diagonalizable; any family of commuting matrices has a common eigenvector. Moreover, the entire commuting family can be simultaneously brought into upper-triangular form with a unitary similarity. – Ben Grossmann Sep 20 '16 at 11:57
  • With your definition, does the algebra generated by a set include the identity? – Ben Grossmann Sep 20 '16 at 11:58
  • @Omnomnomno, Thank you for pointing the existence of a common eigenvector. I have no condition on the identity. It can be included or not. – TameBlie Sep 20 '16 at 15:53
  • @TameBlie the only change that results from not including the identity is that if all matrices in $\mathcal M$ are not invertible, then it's possible that every element of $\mathcal A$ has $0$ has an eigenvalue. – Ben Grossmann Sep 20 '16 at 15:54
  • Notably, $\mathcal A$ contains the identity if and only if it contains any invertible matrix. – Ben Grossmann Sep 20 '16 at 15:56

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