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Can someone please clarify for me these two statements.

Let $A$ of order ($n\times n$) be a matrix over $\Bbb R$. $A^{4} = I$. so:

  1. $A$ diagonalizable over $\Bbb R$.
  2. $A$ diagonalizable over $\Bbb C$.

The only statement I know is concerning eigenvalues: if $a$ is an eigenvalue of $A$ then $a^{4}$ is an eigenvalue of $A^{4}$.

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    $\begingroup$ A matrix with $A^4$ is not necessarily diagonalizable over $\mathbb{R}$. e.g. $[[0,1],[-1,0]]$ $\endgroup$ Commented Feb 10, 2017 at 9:34
  • $\begingroup$ Well, if it is diagonalizable over $\Bbb R$ it should be diagonalizable over $\Bbb C$. However, I don't believe that such a matrix is always diagonalizable over $\Bbb R$: math.stackexchange.com/questions/1197797/… $\endgroup$
    – Surb
    Commented Feb 10, 2017 at 9:34
  • $\begingroup$ you are both correct, The matrix is not necessarily diagonalizable over R. how do I know for sure that it is diagonalizable over C ? $\endgroup$
    – Tom.A
    Commented Feb 10, 2017 at 9:36
  • $\begingroup$ Can you use Jordan form? If you can, for $\mathbb{C}$ you just have to prove $A$ doesn't have Jordan blocks of size greater than $1$. $\endgroup$ Commented Feb 10, 2017 at 9:46

2 Answers 2

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Note that $A$ is annihilated by the polynomial $p(X) = X^4 - 1$. Therefore, $A$'s minimal polynomial $\mu_A$ is a divisor of $p$, that is $$ \mu_A(X) \in \{X - 1, X + 1, X^2 - 1, X^2 + 1, X^4- 1 \} $$ Hence, in each case, the minimal polynomial of $A$ has distinct linear factors and therefore $A$ is diagonizable over $\mathbf C$.

As matrices with minimal polynomial $X^2 + 1$ are not diagonizable over $\mathbf R$, $A$ need not to be diagonizable over $\mathbf R$.

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From $A^4=I$ we get

$(A-I)(A+I)(A-iI)(A+iI)=0$, hence

$\mathbb C^n=ker(A-I) \oplus ker(A+I) \oplus ker(A-iI) \oplus ker(A+iI)$.

This shows that $A$ diagonalizable over $ \mathbb C$

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