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Let $E$ be a vector space over some field $K$ and $f\in L(E)$ such that $\ker f \oplus \operatorname{Im}f=E$.

What can be said about $f$ in general? What if $E$ is finite-dimensional ?

It is well-known that whenever $f$ is a projection (ie $f\circ f=f$), we have $\ker f \oplus \operatorname{Im}f=E$.

It seems to me that $\ker f \oplus \operatorname{Im}f=E$ does not imply $f\circ f=f$, even in a finite-dimensional setting. Indeed, if $x=x_1+x_2$ where $x_1\in \ker f$ and $x_2 \in \operatorname{Im}f$, then $f(x)=f(x_2)$ and $f(f(x))=f(f(x_2))=f(f(f(x_3)))$ for some $x_3\in E$. I can't find any reason why $f(f(x))$ should be equal to $f(x)$...

Am I missing something ? Can something be said about $f$ in terms of eigenvalues for example ?

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    $f\lvert_{\operatorname{im} f}$ is an automorphism of $\operatorname{im} f$. – 2017-01-08
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    It means that $f$ induces an automorphism on $E / \ker f$ by $$x+ \ker f \mapsto f(x)+ \ker f$$ using the first isomorphism theorem ($\mathrm{Im} f \cong E / \ker f$), it is equivalent on saying that $f$ induces an automorphism on $\mathrm{Im} f$. – 2017-01-08
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    $\ker f \oplus \operatorname{Im}f=E$ is always true if $f$ is diagonalizable (because in this case the kernel is the eigenspace for $0$ and the image of $f$ is the sum of the other eigenspaces). – 2017-01-08

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