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Let $V$ be a finite-dimensional vector space over field $F$ and let $f:V \rightarrow V$ be an endomorphism. Using the rank-nullity theorem show that if ker$(f \circ f) = $ker$(f)$ then $f:$ im $(f)\rightarrow $im$(f \circ f)$ is an isomorphism.

I have no idea how to even approach this question.

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closed as off-topic by YuiTo Cheng, GNUSupporter 8964民主女神 地下教會, Wrzlprmft, José Carlos Santos, Eevee Trainer May 7 at 6:47

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Apply rank-nullity theorem to $f$ and $f^2$. Let $V$ has the dimension $n$.

Then $Rank(f)+Nullity(f)=n$ and $rank(f^2)+Nullity(f^2)=n$

since it is given to us that $Ker(f)=Ker(f^2)$ we can conclude that $Rank(f)=Rank(f^2)$

Can you proceed from here?

Added later: Now show that $f:im(f)\rightarrow im(f^2) $ is injective. Since both domain and range have same dimension, it follows that map is onto.

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