# Prove that linear mapping is an isomorphism [closed]

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.

## closed as off-topic by YuiTo Cheng, GNUSupporter 8964民主女神 地下教會, Wrzlprmft, José Carlos Santos, Eevee TrainerMay 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 $$Rank(f)=Rank(f^2)$$
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.