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If the matrix $M\in M_5(\mathbb C)$ has the property $M^2=M^4$ but $M\neq M^2$, are there any special implications regarding the characteristic and minimal polynomials? I suppose then that the eigenvalues are s.t. $\lambda^2=\lambda^4$...(right?) but what else is there?

Thank you.

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Since $M^2=M^4$, then $M^4-M^2 = 0$. That means that the minimal polynomial divides $t^4-t^2 = t^2(t^2-1) = t^2(t-1)(t+1)$. So the characteristic polynomial must be of the form $t^{\alpha}(t-1)^{\beta}(t+1)^{\gamma}$, with $\alpha+\beta+\gamma = 5$.

Since $M\neq M^2$, that means that the minimal polynomial does not divide $t^2-t = t(t-1)$; So the minimal polynomial must be $t^2$, $t^2(t-1)$, $(t+1)$, $t(t+1)$, $t^2(t+1)$, $(t-1)(t+2)$, $t(t-1)(t+1)$ or $t^2(t-1)(t+1)$.

So the minimal polynomial divides $t^2(t-1)(t+1)$, but is not $t$, $t-1$, or $t(t-1)$.

That is all you can deduce. For each of the possibilities above you can find a matrix that satisfies $M^4-M^2=0$ and $M^2\neq M$: it can be $-I$; a diagonal matrix with $0$ and $-1$s in the diagonal, at least one of each; a diagonal matrix with $1$s and $-1$s in the diagonal, at least one of each; a diagonal matrix with $0$s, $1$s, and $-1$s in the diagonal, at least one of each;a matrix with one or two $2\times 2$ Jordan blocks associated to $0$, and the rest of the matrix all $0$s; a matrix with at least one $2\times 2$ Jordan block associated to $0$, at least one $-1$ in the diagonal; etc.

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Thanks, Arturo! –  Fred Nov 20 '11 at 8:56

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