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Let $W\subset V$ be vector spaces. Is there exist a canonical way to construct a projector $\pi: V \to W$?

The same question for W is the kernel or image of some map.

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It should be doable if you have an inner product. – Dylan Moreland Jul 20 '12 at 23:23
Are they finite dimensional? – copper.hat Jul 20 '12 at 23:31
@copper.hat may be infinite – Aspirin Jul 20 '12 at 23:34
Any continuity requirements? – copper.hat Jul 20 '12 at 23:44
$W$ is always a kernel of some map [like $V \to V/W$], and the same goes for the image [$W \to V$]. – Dylan Moreland Jul 20 '12 at 23:59

Here is an example. $\mathbb R$ is a vector space over the field $\mathbb Q$. Also $\mathbb Q \subseteq \mathbb R$ is a (one-dimensional) subspace. I defy you to "canonically" write down a projection!

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For general linear space I don't think there exist a canonical construction. But anyway you can proceed as follows. Take Hamel basis $\{e_\alpha:\alpha\in A_0\}$ of $V$. Complete it up to Hamel basis $\{e_\alpha:\alpha\in A\}$ of $W$, where $A_0\subseteq A$. Now define $$ \pi: W\to V: \sum\limits_{\alpha\in A}\lambda_\alpha e_\alpha\mapsto \sum\limits_{\alpha\in A_0}\lambda_\alpha e_\alpha $$ This construction is not canonical because it relies on axiom of choice. Because of the axiom of choice completion of basis is quite "random", and not canonical.

But if $W$ is a Hilbert space, there is a good procedure to get canonical projection. Consider orthogonal complement $V^\perp$ of $V$, then from theorem about orthogonal projection for each $w\in W$ we have unique $w_1\in V$, $w_2\in V^\perp$ such that $w=w_1+w_2$. So we define $$ \pi: W\to V: w\mapsto w_1 $$ Note that "hilbertablity" is necessary, there exist inner product spaces for which you can't construct orthogonal projections on particular subspaces. Here you can find a counterexample, but anyway you can apply non-canonical construction (involving Hamel basis) to get some, not necessary orthogonal projection $\pi$.

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If I grant/believe for the moment the questioner's claim that there are no continuity requirements, although this is not the only point, the answer is "no", in the following sense. First, as remarked upon in other answers and comments, to have a "projection" is exactly to have a "complementary subspace". If we have a vector space $V$ and for every subspace $W$ we have a "canonically" associated complementary subspace, this entails quite a few things.

If/when the vector space is over $\mathbb R$, this allows us to choose/find an "orthonormal basis", and, in doing so, define an inner product.

Stopping short of seeing what else is entailed... I myself am content to note that we might as well have an inner product to begin, in which case _of_course_ we have canonical complementary subspaces.

Over fields other than $\mathbb R$ or $\mathbb C$, and in general Banach spaces (now starting to refuse to ignore topology) or more general TVS's, certainly things are less clear. A thing that has always been disturbing to me is Serre's (by now 50-years-old) discussion of Banach spaces over $p$-adic fields... so, conceivably, depending upon the questioner's ulterior purposes, there might be something non-trivial there, of interest.

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Dear @paul garrett, Could you elaborate on your comment about Serre's discussion of $p$-adic Banach spaces? I'm not familiar with what you're referring to. – Keenan Kidwell Jul 21 '12 at 5:45
Motivated by looking at p-adic interpolation of zetas and p-adic modular forms, in the 1970s Serre proved some fairly surprising results about Banach spaces over p-adic fields, completely continuous operators thereupon. For example, I seem to recall that he showed that the strong dual of $\ell^1$ is again of the form $\ell^1$, but this may be wrong. A paper by Coleman in Inv. Math. 1997 cites two of the papers of Serre I'd vaguely recalled. – paul garrett Jul 21 '12 at 14:54
Thanks for the reference! – Keenan Kidwell Jul 21 '12 at 19:34

What does canonical mean? Unless that's some sort of invariant there are obviously many ways to project a space to a subspace. For example, given $W = \mathbb{R} \subset V = \mathbb{R}^2 $, the projection matrix:

$P_a= \pmatrix{1&a\cr 0&0\cr}$ fixes the basis vector $e_1$, and maps $e_2$ to an arbitrary vector $a e_1 \in W$.

Which $P_a$ is canonical?

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