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

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 $V=\operatorname{Im}P \oplus \operatorname{Ker}P$.

  • $\begingroup$ 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 $0$. $\endgroup$ 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 $v$?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.