Diagonalizable matrix

Question

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 $A^{4}$.

Answer 1

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 $\mathbf R$.

Answer 2

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$