$$im(f^2)=im(f) \implies V = ker(f) + im(f)$$
What I tried: take $v \in V$. Then $v = v - f(v) + f(v)$
$f(v) \in im(f)$ trivial. So i tried to show $v - f(v) \in ker(f)$. $f(v - f(v)) = f(v)-f(f(v))$
and that's where I'm stuck. Can someone help me?
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2Please use MathJax. Here is a tutorial: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Mark Feb 11 '23 at 21:55
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5Anyway, you are pretty close. By assumption we have $f(v)=f(f(w))$ for some $w$. So now try to find a very similar decomposition of $v$ to the one you tried, but using $w$. – Mark Feb 11 '23 at 21:59
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thanks, I got it! – Cake Feb 11 '23 at 22:04
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Follow from the Rank-Nullity Theorem this. – Arturo Magidin Feb 12 '23 at 01:04