If $f^2=id_V$ then f is diagonizable

linear-algebra

I am trying to prove the following statement, which wasn't proven in my course but seems really important,

Let $f \in End(V)$, if $f^2=id_V \Rightarrow f$ is diagonizable. (the characteristic of the field is not 2)

It seems intuitive to me, but I can't find a way to prove it, the only thing I noticed has to be true is that $f\restriction_{Imf} = id_V$ has to hold, pls help


From $f^2 - \mathrm{id} = 0$ conclude its minimal polynomial divides $x^2-1$. If the characteristic is not two, then the minimal polynomial cannot have repeated roots.

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