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I know this is true for complex matrices and have been very useful in proving a series of statement regarding diagonalizability of complex matrices (idempotent matrices are diagonalizable in complex). But is this true for real matrices with real field--I suppose it is false. Is there anyway we have say additional conditions to derive a analogous result to diagnose diagonalizability of real matrices?

Daniel Li
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    In general, a matrix over a field $\mathbb F$ is diagonalizable iff its minimal polynomial has no repeated roots and splits over $\mathbb F$. See https://math.stackexchange.com/questions/56745/minimal-polynomial-and-diagonalizable-matrix for a proof. – boink Sep 08 '19 at 00:19
  • @boink I have seen the word "splits". Can you explain what it means in simple terms so that I can apply directly to show say real idempotent matrix is diagonalizable. – Daniel Li Sep 08 '19 at 00:20
  • It basically means that all the roots of that polynomial are in that field. The idea is that it "splits" into linear factors over $\mathbb F$. – boink Sep 08 '19 at 00:22
  • @boink Thank you very much! Just to double check. "$A^4=I \implies A$ is diagonalizable" is true for complex but not necesarilly true for real because $z^4=1$ does not split in real, only in complex. But "$A^2=I \implies A$ is diagonalizable" is true in real because it splits in real as well. Am I right? – Daniel Li Sep 08 '19 at 00:37
  • As it stands, the statement in the title of the question is incorrect. If $L: \mathbb{R}^2 \to \mathbb{R}^2$ is a rotation by 1/4 of a turn, then the minimal polynomial of $L$ is $x^2 + 1$, which has no repeated roots. But $L$ is not diagonalisable over $\mathbb{R}$. – Joppy Sep 08 '19 at 03:31
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    @DanielLi Yep, that seems right! :) – boink Sep 08 '19 at 04:16
  • @Joppy You should see I wasn't very certain of what I was saying. But these discussions in comment sections have been very helpful. – Daniel Li Sep 09 '19 at 01:45

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