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Let $A\in \mathbb{C}^{n^2}$ such that $A^m=I_n$, for some $m,n\in \mathbb{N}$.

Please prove that $A$ is diagonalizable.

Now let $B\in \mathbb{C}^{n^2}$ such that $B^m=B$, for some $m\in \mathbb{N}$ such that $m>1$.

Please prove that $B$ is diagonalizable.

1 Answers 1

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Hints:
(1) A matrix $M$ is diagonalizable if and only if it's minimal polynomial $m_M(x)$ factors completely over the field (here $\mathbb{C}$) into distinct linear factors.The relevant theorem is here
(2) If $p(x)$ is a polynomial such that $p(M)=0$ then $m_M(x)|p(x)$.
(3) If we have two polynomials $f(x),g(x)$, $f(x)|g(x)$ and $g(x)$ factors completely into distinct linear factors, then so does $f(x)$.
(4) Observe that if $p(x)=x^m-1$ and $q(x)=x^m-x$ then $p(A)=0=q(B)$

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    I noticed what you said is basically the same. The fact that $J_A^m=I_n$ implies $J_A$ diagonal follows from this identity easily: http://en.wikipedia.org/wiki/Jordan_Normal_Form#Powers Thanks.2012-12-15