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Suppose $V$ is a finite dimensional vector space, $A_1, \dots, A_k : V \to V$ are simultaneously diagonalizable and $E$ is a $A_i$ invariant subspace for $i = 1, \dots, k$. I know that $A_i|_E$ is also diagonalizable for each $i$, but is the collection $\{A_i|_E\}$ simultaneously diagonalizable?

In the proof that $A_i|_E$ is diagonalizable, we use the fact that its minimal polynomial $m_{A_i|E}$ divides $m_{A_i}$. Since the roots of these minimum polynomials are eigenvalues for their respective matrices, I was thinking that we can take the associated eigenvectors to form an eigenbasis. Then I think the question would come down to showing that $m_{A_i|_E} = m_{A_j|_E}$ for each $i,j \in \{1,\dots, m\}$. Would this work?

Note: I'm trying to prove this so that I can show a collection of diagonalizable matrices are simultaneously diagonalizable iff they commute.

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Each $A_i$ is diagonalizable and simultaneous diagonalizability is equivalent to commutativity, ie, $A_i$ and $A_j$ are simultaneously diagonalizable iff $A_iA_j = A_jA_i$.

For every $v\in E$, we have $A_iA_j(v) = A_jA_i (v)\in E$. Then the commutativity is preserved under the restriction to $E$. Therefore the $\left. A_i \right|_E$ are simultaneously diagonalizable.

EDIT: From the later edit by the OP, commutativity must not be used. I'd rather point to this answer that contains a neat digression on the subject. I hope it will be helpful.

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  • $\begingroup$ Thank you for your answer. I should have made this clear from the start, but I want to use this result to prove that simultaneous diagonalizability is equivalent to commutativity. $\endgroup$ – matt stokes Sep 17 '18 at 23:16
  • $\begingroup$ Oh OK. I'll edit my answer then... $\endgroup$ – Alan Muniz Sep 17 '18 at 23:22

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