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Suppose $A \in \mathbb{R}^{m\times n}$. Then there exists a projection matrix $P$ onto the range of $A$. In other words, there exists a matrix $P \in \mathbb{R}^{m\times m}$ such that $P^2=P$, and $Range(A) = Range(P)$. You can construct $P$ using the SVD, for example.

I am looking for a more general version of this result. In particular, I would like extensions to Banach spaces, as well as topological vector spaces. I know there are many nuances that arise when dealing with infinite-dimensional spaces, so I'm hoping somebody can help clear things up.

I'm hoping for something that looks like: "Suppose $X$ and $Y$ are topological vector spaces, and $A:X\to Y$ is a continuous linear map. There exists a continuous linear map $P:Y\to Y$ such that $P^2=P$ and $Range(A) = Range(P)$."

If this is false, can I modify the statement somehow to make it true? If it's true, how do I prove it, and how do I construct $P$?

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up vote 7 down vote accepted

This is false in general, for (at least) two reasons

  1. The range of a projection operator $P$ is closed, as it is the kernel of $1 - P$.

  2. A closed subspace $F \subset E$ of a topological vector space is said to be complemented if there is a continuous projection onto it. Not every closed subspace is complemented.

To see that the range of an operator is not necessarily closed, simply note that for $p \lt q$ we have a continuous map (inclusion) $\ell^{q} \to \ell^{p}$. Since $\ell^{q} \subsetneqq \ell^{p}$ is dense, the range can't be closed.

The standard example of a closed but not complemented subspace is $c_0 \subset \ell^{\infty}$ (this is called Phillips' lemma). This is not very difficult to prove but not very obvious either.

In fact, a very deep result of Lindenstrauss and Tzafriri characterizes the Banach spaces linearly isomorphic to a Hilbert space as precisely those in which every closed subspace is complemented. (Thanks, Philip, for pointing this out).

Finally, two easy results in a positive direction:

  • If $A: E \to H$ is a map from any vector space to a Hilbert space and the range of $A$ is closed, then you can project orthogonally onto the range.

  • The map $A: E \to F$ between Fréchet-spaces (completely metrizable locally convex topological vector spaces) admits a pseudo-inverse $B: F \to E$, that is a map $B$ such that $ABA = A$ and $BAB = B$ if and only if both the kernel and the range of $A$ are complemented.

To prove this last fact, note first that $1-BA$ and $AB$ are projections onto kernel and range of $A$, so they are complemented. In the other direction, factor $A$ as $E \to E/\operatorname{Ker}{A} \to \operatorname{Im}{A} \to F$. Here, the leftmost map admits a right inverse and the rightmost map a left inverse and the map in the middle is an isomorphism by the open mapping theorem and the hypotheses. Now $B$ can be obtained simply by "going backwards".

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It is probably worth mentioning that the only Banach spaces for which every closed linear subspace is the range of a continuous linear projection are those Banach spaces that are isomorphic to a Hilbert space. – Philip Brooker Jun 2 '11 at 10:34
@Philip, yes of course. I added it in. Thanks! – t.b. Jun 2 '11 at 10:38

In a less general situation, the von Neumann article in wikipedia talks about "subspace that belong" to a von Neumann algebra and that are in one to one correspondance with projections.

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