# Show that the direct sum of a kernel of a projection and its image create the originating vector space.

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?

• 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 $P$? – user29743 Dec 18 '12 at 21:16
• (by 1 i mean the identity) – user29743 Dec 18 '12 at 21:17
• $1 - P$ 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? – iblue Dec 18 '12 at 21:22
• sorry I know it is off-topic, but am i the only person for which mathjax does not render latex on stackexchange ? – nicolas Mar 1 '14 at 13:48
• @nicolas: you should post on meta describing your problem. This is not the place for this kind of question. – 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 $$V=\operatorname{Im}P \oplus \operatorname{Ker}P$$.
• Thank you! I don't understand the last step. Why $\operatorname{Im} P \cap \operatorname{Ker} P = 0 \Rightarrow V=\operatorname{Im}P \oplus \operatorname{Ker}P$? – iblue Dec 18 '12 at 22:27
• 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 $V=Im(P)\oplus Ker(P)$. – Manos Dec 18 '12 at 23:29
• We should be writing $\operatorname{im} P \cap \operatorname{ker} P = \{0\}$ rather than just $0$. – 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 $v$?