If $A$ is an involutory matrix, i.e. $A^2=I$, then is $A$ diagonalizable?

  • 3
    $\begingroup$ Using Representation theory one can show that every complex matrix of finite order is diagonalisable. $\endgroup$
    – user38268
    Jan 17 '12 at 17:38
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    $\begingroup$ @Benjamin: Or just using the same argument: if $A^n=I$, then $A$ satisfies $t^n-1$, which has $n$ distinct roots over $\mathbb{C}$; hence the minimal polynomial of $A$ over $\mathbb{C}$ has no repeated roots, and hence $A$ is diagonalizable. $\endgroup$ Jan 17 '12 at 17:46
  • $\begingroup$ @ArturoMagidin You said it slicker than me. I was thinking Maschke's Theorem. $\endgroup$
    – user38268
    Jan 17 '12 at 17:46

Since $A^2=I$, then $A$ satisfies the polynomial $t^2-1 = (t-1)(t+1)$. Hence, the minimal polynomial of $A$ divides $(t-1)(t+1)$; so the minimal polynomial of $A$ splits and has distinct roots, so $A$ is diagonalizable.

As N.S. points out in the comments, the above fails if you are working in characteristic 2. There, the matrix $$A=\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)$$ has minimal and characteristic polynomials $t^2+1 = (t+1)^2$, and it is not diagonalizable (the eigenspace of $1$ has dimension $1$). But if $1\neq -1$, you are set.

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    $\begingroup$ Unless you are in characteristic 2 ;) $\endgroup$
    – N. S.
    Jan 17 '12 at 17:03
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    $\begingroup$ Thanks! Nice and simple answer. $\endgroup$
    – Godisemo
    Jan 17 '12 at 17:09
  • $\begingroup$ @N.S. Ah, true. $\endgroup$ Jan 17 '12 at 17:21

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