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I got the following question as my homework.

Given $V$ is a vector space with $P \in \operatorname{End} V$. $P \circ P = P$ ("P is idempotent"). Show that $V = \operatorname{Ker} P \oplus \operatorname{Im} P$.

One $P$ I can imagine is a projection from 3d-space to plane, that just sets some coordinates to zero. For example $\begin{pmatrix} x \\ y \\ z\end{pmatrix} \mapsto \begin{pmatrix}x \\ y \\ 0\end{pmatrix}$. Then $\operatorname{Ker} P$ would give the line $\begin{pmatrix} 0 \\ 0 \\ z\end{pmatrix}$ and $\operatorname{Im} P$ would contain all $\begin{pmatrix}x \\ y \\ 0\end{pmatrix}$. So the result of $\operatorname{Ker} P \oplus \operatorname{Im} P$ is of course $V$.

But how do I prove that in a mathematical way?

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    @nicolas: you should [post on meta](http://meta.math.stackexchange.com/) describing your problem. This is not the place for this kind of question.2014-03-02

2 Answers 2

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Take $x \in V$. Since $P=P^2$ we must have $Px=P^2x$ and so $P(x-Px)=0$. Hence $x-Px=\xi$ for some $\xi \in \operatorname{Ker}P$. Thus $x = Px + \xi$. This shows that $V=\operatorname{Im}P + \operatorname{Ker}P$. Now take $y \in \operatorname{Im}P \cap \operatorname{Ker}P$. Since $y \in \operatorname{Im}P$ we have $y=Pz$ for some $z \in V$. Applying $P$ to both sides we get $Py=P^2z$. But $y \in \operatorname{Ker}P$, hence $0=Py=P^2z=Pz=y$. This shows that $\operatorname{Im}P \cap \operatorname{Ker}P=\{0\}$ and so we have $V=\operatorname{Im}P \oplus \operatorname{Ker}P$.

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    We should be writing $\operatorname{im} P \cap \operatorname{ker} P = \{0\}$ rather than just $0$.2016-10-05
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Hint: $V = \operatorname{Ker}P \oplus \operatorname{Im}P$ iff every $v\in V$ has a unique representation as $v = u+w$ for some $u \in \operatorname{Ker}P, w \in \operatorname{Im}P$ (If you haven't seen that already, it's not too hard to prove.)

How can you find such an expression for general $v$?