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$A$ is element of $M_n (C)$ and $A^n=I$ and $A^k \neq I$ , $1\leq k <n$. Show that $A$ can diagonalization

I try to use Jordan decomposition to solve this. There is $P$ such that $P^{-1}AP=J_1 + ... + J_k$ when $J_i$ is jordan matrix and I got $P^{-1}A^nP=(J_1)^n + ... + (J_k)^n = I$ and $(J_i)^n= I_i$ . can I take the conclusion that $J_i$ is diagonal matrix?how to prove that?

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As $A^n-I=0$, the minimal polynomial of $A$ must be a factor of $x^n-1$ which has distinct roots. So the matrix must be diagonalizable.

NOTE: You have assumed the size of the matrix ($n$ from $n\times n$) and the $n$ appearing in the index $A^n$ are the same. The theorem is true even when they are unrelated.

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  • $\begingroup$ why must the minimal polynomial have distinct roots ? $\endgroup$ – Gabriel Romon Nov 8 '17 at 9:50
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You can compute powers of Jordan blocks; see here for one example. If a Jordan block is not $1 \times 1$, then its $n$th power will not be diagonal (never mind the identity matrix), so it must be that every Jordan block is $1 \times 1$, i.e. the Jordan matrix is diagonal.

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Let $A = D + N$ where $D$ is a diagonal matrix and $N$ is a nilpotent matrix, then $D$ and $N$ are uniquely determined (by Jordan decomposition) and $DN=ND$ (see Prove that $ND = DN$ where $D$ is a diagonalizable and $N$ is a nilpotent matrix.).

Thus $$ I = A ^n = (D + N)^n = D ^n + N(D ^{n-1} + \dots + N ^{n-1})$$ because the diagonal and nilpotent part in the decomposition are unique, we have $D ^ n = I$ and $N(D ^{n-1} + \dots + N ^{n-1}) = 0$. Now $(D ^{n-1} + \dots + N ^{n-1})$ is invertible (because $D$ has only non zero entries, otherwise $A^n \neq I$ for all $n$), yielding $N = 0$.

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