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Show that every idempotent matrix is diagonalizable. What can you say if $A$ is tripotent ($A^3=A?$) What if $A^k=A?$

The first two cases is obvious since we can find the minimal polynomial to be $t(t-1)$ or $t(t-1)(t+1)$, which are products of distinct linear factors. However, what can we say about the general case $A^k=A?$ Any solutions, hints, or suggestions would be appreciated.

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In general, we can factor $$ x^k - x = x(x^{k-1} - 1) $$ the roots of $x^{k-1} - 1$ are the $k-1$ roots of unity, given by $$ x = e^{2 \pi ij/(k-1)} = \cos(2 \pi j/(k-1)) + i\sin(2 \pi j/(k-1)), \quad j = 0,\dots,k-2 $$ Thus, for all $k$, the polynomial $x^k - x$ is the product of distinct linear factors.

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