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Given that $P$ is a projection linear map, s.t. $P^2=P$. I can show that for some vector space $V$ this implies that $V=\text{Ran}(P) \bigoplus \text{Ker}(P)$. Basically, $v=(I-P)v+Pv$ and $P(I-P)v=0$. Then $\text{Ker}(P)\cap \text{Ran}(P) = \{0\}$

I don't know whether I am failing to take into account the infinite dimensional $V$. Does this have any effect? How might I take into account of this?

Thanks in advance!

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Doesn't your argument show that any vector of $V$ can be represented as a sum of something from $\mathrm{Im}(P)$ plus something from $\ker(P)$? And also that $\mathrm{Im}(P)\cap\ker(P)=\{0\}$? Looks like a direct sum decomposition to me! – Jyrki Lahtonen Sep 11 '11 at 19:17
up vote 2 down vote accepted

In order to show that $V = W\oplus Z$, you need to show that $W\cap Z=\{\mathbf{0}\}$ and that $V=W+Z$. The problem of finite dimension does not arise, unless you attempt to show one of the two steps by using dimension arguments.

Implicit in your argument is that $\mathrm{Ran}(I-P) = \mathrm{Ker}(P)$. This does not require $V$ to be finite dimensional: for $P(I-P)(v) = (P-P^2)(v) = Pv - P^v = Pv-Pv = \mathbf{0}$; and if $w\in\mathrm{Ker}(P)$, then $w = w-\mathbf{0} = w-P(w) = (I-P)(w)$. No dimension argument used.

Your first step, showing that $v=(I-P)v+Pv$ shows that $V\subseteq \mathrm{Ker}(P)+\mathrm{Im}(P)$; the other inclusion is trivial. No dimension arguments used.

And your second step is noting that $\mathrm{Ran}(I-P)\cap\mathrm{Ran}(P)=\{\mathbf{0}\}$; again, no dimension arguments are used: if $v$ lies in the intersection, then $v=Pw$ for some $w$, and $v=(I-P)z$ for some $z$. Then $w = Pw = P^2w = P(Pw) = Pv = P(I-P)z = \mathbf{0}$, so $v=P\mathbf{0}=\mathbf{0}$.

Since nowhere are any assumptions made on the dimension of $V$, the argument holds for any vector space, whether finite dimensional or infinite dimensional.

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