The proof said the following, let $A \in M_{n \times n}(\mathbb{C})$ such that $A^r=I_{n \times n}$ for some $r \in \mathbb{N},r>0$ then $A$ is diagonalizable. Mi attempt is the next, since $A$ is equal to the identity matrix, then we can approach the jordan normal form of the $A$, but, i can´t see, the zise of the jordan blocks. someone who give me a hint, or in her case the proof
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1Why do you say that $A$ is equal to the identity matrix? – José Carlos Santos Jun 26 '20 at 17:50
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$A$ not is necessary the identity matrix, the hipotesis is $A^r$ is the idenity matrix for some $r \in \mathbb{N}$$ – Jun 26 '20 at 17:52
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But you wrote that $A$ is equal to the identity matrix! – José Carlos Santos Jun 26 '20 at 17:53