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Prove that if $T$ and $U$ are simultaneously diagonalizable linear operators on a finite-dimensional vector space $V$, then the matrices $[T]_\beta$ and $[U]_\beta$ are simultaneously diagonalizable for any ordered basis $\beta$.

Does that mean I only need to find an invertible matrix $Q$ such that $Q^{-1} [T]_\beta Q$ and $Q^{-1} [U]_\beta Q$ are diagonalizable? I am totally confused!

Can I prove the "there exists" statement by just finding an invertible identity matrix such that $I^{-1} [T]_\beta I$ and $I^{-1} [U]_\beta I$ are diagonalizable? thanks for helping.

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Suppose that $T$ and $U$ are simultaneously diagonalized in basis $\beta$ (i.e. $[T]_\beta$ and $[U]_\beta$ are both diagonal). If we switch to another basis, say $\gamma$, what is the relationship between $[T]_\beta$ and $[T]_\gamma$?

Answer: It's the change of basis matrix $P=[I]_\beta^\gamma$. We have $[T]_\beta = [I]_\gamma^\beta [T]_\gamma [I]_\beta^\gamma = P^{-1} [T]_\gamma P$. The same is true for $U$. Thus $P$ simultaneously diagonalizes both $[T]_\gamma$ and $[U]_\gamma$.

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  • $\begingroup$ I am still confused.. aren't we trying to prove that there exists an invertible matrix Q such that Q^-1 [T]_beta Q and Q^-1 [U]_beta Q are diagonal matrix? since we already know [T]_beta and [U]_beta are diagonal matrix, is it necessary to find [T]_gamma and [U]_gamma? $\endgroup$ – Gigi Oct 23 '11 at 17:58
  • $\begingroup$ @Gigi: Yes. That matrix is $P$. $\endgroup$ – joriki Oct 23 '11 at 18:00
  • $\begingroup$ Oh I see it now, thank you. $\endgroup$ – Gigi Oct 23 '11 at 18:04

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