Decomposing a vector parallel and perpendicular to a fixed vector

[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.