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I've tried solving this exercise for quiet a while now but can't get a satisfying solution:

Let $V$ be a finite dimensional vector space and $f: V \to V$ an endomorphism. Show that $$V = \text{Im } f + \text{Ker } f \Longleftrightarrow V = \text{Im } f \oplus \text{Ker } f$$

The $"\Longleftarrow"$-direction is trivial.
For $"\Longrightarrow"$ here's my attempt: We assume $V = \text{Im } f + \text{Ker }f$ and have to show that $\text{Im } f \cap \text{Ker }f = \{0\}$. Suppose $x\in \text{Im } f \cap \text{Ker }f$. We want to show that $x = 0$. Since $x\in \text{Im }f$ and $\text{Im }f \subseteq V$ there is $v\in \text{Im }f, u \in \text{Ker }f$ with $x = u + v$. Since also $x\in \text{Ker }f$ we have that $$0 = f(x) = f(v+ u) = f(v) + f(u) = f(v),$$ hence $f(v) = 0 \implies v\in \text{Ker } f$. Now we have $v\in \text{Ker }f$ and $v \in \text{Im }f$ but I don't know how to proceed (in fact I don't even know if my proof goes in the right direction, but I wanted to give you my thoughts). Any help appreciated.

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    The most-satisfying solution to a not-always-true exercise is a counterexample. ​ ​ – 2017-02-11
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    I assumed that if the exercise says "Show..." that the statement is correct. Is there a counterexample? If yes, could you give me one? – 2017-02-11
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    It presumably means the writer _believed_ the statement was correct, which is one reason why giving a counterexample is satisfying. ​ ​ – 2017-02-11
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    That is indeed true. However I cannot come up with a counter example... – 2017-02-11
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    After your edit, I also can't. ​ ​ – 2017-02-11
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    I'm really sorry I forgot that, I assume that makes quite a big difference. Are you now able to help me with my proof however? :) – 2017-02-11

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The dimensions theorem tells us that

$$n=\dim V=\dim\ker T+\dim\text{ Im}\,V$$

But we're also given $\;V=\ker T+\text{ Im}\,T\;$ , so:

$$n=\dim V=\dim\left(\ker T+\text{ Im}\,T\right)=\overbrace{\dim\ker T+\text{ Im}\,T}^{=n}-\dim\left(\ker T\cap\text{ Im}\,T\right)\implies$$

$$\dim\left(\ker T\cap\text{ Im}\,T\right)=0\implies \ker T\cap\text{ Im}\,T=\{0\}\implies$$

$$\ker T+\text{ Im}\,T=\ker T\oplus\text{ Im}\,T$$