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?

  • $\begingroup$ very light hint: think about the transformation $1 - P$. What happens when you square it? What happens when you multiply it by $P$? What happens when you add $P$? $\endgroup$ – user29743 Dec 18 '12 at 21:16
  • $\begingroup$ (by 1 i mean the identity) $\endgroup$ – user29743 Dec 18 '12 at 21:17
  • $\begingroup$ $1 - P$ seems to give $\operatorname{ker} P$, $1 - P \circ 1 - P = 1 - P$. And when I add $P$ to $1 - P$, I get $1$, but how does that help me? $\endgroup$ – iblue Dec 18 '12 at 21:22
  • $\begingroup$ sorry I know it is off-topic, but am i the only person for which mathjax does not render latex on stackexchange ? $\endgroup$ – nicolas Mar 1 '14 at 13:48
  • $\begingroup$ @nicolas: you should post on meta describing your problem. This is not the place for this kind of question. $\endgroup$ – robjohn Mar 2 '14 at 4:44

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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  • $\begingroup$ Thank you! I don't understand the last step. Why $\operatorname{Im} P \cap \operatorname{Ker} P = 0 \Rightarrow V=\operatorname{Im}P \oplus \operatorname{Ker}P$? $\endgroup$ – iblue Dec 18 '12 at 22:27
  • $\begingroup$ Because this is the definition of a vector space being the direct sum of two subspaces. The subspaces must span the whole space and have zero intersection. $\endgroup$ – Manos Dec 18 '12 at 22:28
  • $\begingroup$ I need to clarify that strictly speaking what you have written is not true. What is true is $Im(P) \cap Ker(P)=0$ and $V=Im(P)+Ker(P)$ imply $V=Im(P)\oplus Ker(P)$. $\endgroup$ – Manos Dec 18 '12 at 23:29
  • $\begingroup$ We should be writing $\operatorname{im} P \cap \operatorname{ker} P = \{0\}$ rather than just $0$. $\endgroup$ – Jon Warneke Oct 5 '16 at 19:18

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 $v$?

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