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I found this problem, with $f\in\text{End}_\Bbb{C}(V)$ such that ${f}^{m}= Id_V$ for some $m$. I know that $f$ is diagonalizable because it can't be nilpotent because of the hypothesis. The problem is that I don't know how to find how many eigenvalues does $f$ have.

The problem:

Suppose that $V$ is a finite dimensional vector space over $\Bbb{C}$ (the field of complex numbers), $m$ is an integer and $f$ ∈ $\text{End}_\Bbb{C}(V )$ satisfies ${f}^{m} = Id_V$ for some integer $m$. Show that $f$ is diagonalizable. (How many eigenvalues does f have?)

Thank you

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1 Answer 1

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The mth power of $f$ is associated to the composition of the application with itself, repeated m times, whose associated matrix is $A^m$ if $A$ is the matrix for $f$.

Therefore if $m = 2$ and $\vec{v}$ is an eigenvector and $\lambda$ its eigenvalue it will happen that $f \circ f$ has as eigenvalue $\lambda^2$ as $f(f(\vec{v})) = f(\lambda\vec{v}) = \lambda(f(\vec{v})) = \lambda(\lambda\vec{v})= \lambda^2\vec{v}$. This can be generalized. If $m$ is generic then $f(f(f(....f(\vec{v}) ... ))) = \lambda^m\vec{v}$.

If the composition results in the identity matrix then $f(f( .... f(\vec{v}) ... )) = 1\vec{v} = \lambda^m\vec{v}$

All the complex solutions for $\lambda^m = 1$ are possible eigenvalues.

I could be wrong cause I did this long time ago, so double check everything.

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