summing two different orthogonal complements

Question

Let $V$ be an inner product space, not necessarily finite-dimensional, with subspaces $W, X, Y, Z$ such that

• $W$ and $Y$ are orthogonal complements ($W \perp Y$ and $W + Y = V$)
• $X$ and $Z$ are orthogonal complements
• $W \cap Z = \{0\}$.

I want to show that $X + Y = V$.

I think I can prove this in finite dimensions by projecting $v$ into $(X + Y)^\perp = W \cap Z = \{0\}$, but it’s not clear to me in general how to construct the projection without e.g. Gram–Schmidt.

Answer 1

You're right, it is true in finite dimension. No need to project anything: every subspace is closed so $X+Y=\overline{X+Y}=(X+Y)^{\perp\perp}=X^\perp\cap Y^\perp=W\cap Z=V$.

By your observation, the third assumption is equivalent to $(X+Y)^{\perp\perp}=\overline{X+Y}=V$. So just take, in an infinite dimensional Hilbert space, $X$ and $Y$ closed such that $X+Y$ is not closed. This will give you a counterexample, with $W=Y^\perp$ and $Z=X^\perp$.