Question about inner product spaces

[AxiomOfChoice]

Suppose you have an inner product space V (not necessarily finite dimensional; so it could be an infinite dimensional Hilbert space or something). Fix a vector \Phi in this space. Given an arbitrary vector \Psi \in V, can I write it as

\Psi = \Psi^{\parallel} + \Psi^{\perp},

where \Psi^{\parallel} is parallel to the given \Phi and \Psi^{\perp} is perpendicular to the given \Phi?

[arkajad]

If the inner product is positive definite, then you can. Just write:

\Psi^\parallel = \frac{(\Phi,\Psi)}{(\Phi,\Phi)}\,\Phi

\Psi^\perp=\Psi-\Psi^\parallel

[Landau]

This is just the fact that

V=<\Phi>\oplus <\Phi>^\perp

where <> denotes the span.

[Fredrik]

The more general theorem says that if x is a member of a Hilbert space H, and K is a closed linear subspace of H, there's a unique y in K such that x-y\perp K. If we define x_\parallel=y and x_\perp=x-y, we can write x=x_\parallel+x_\perp. The theorem also says that y is at the minimum distance from x: d(x,y)=d(x,K).

(I'm saying linear subspace to emphasize that it's a subspace of the vector space, not the Hilbert space. A closed linear subspace is a linear subspace that's also a closed set. Some authors use the term "linear subspace" only when the set is closed, and the term "linear manifold" when it may not be closed).

[Landau]

@Fredrik: V is not assumed to be a Hilbert space (i.e. need not be complete) in this topic.