Subspaces $X$ and $Y$ of a Hilbert Space with $\dim X<\infty$ and $\dim X<\dim Y$.

I am working through some problems in the book Analysis Now by Pedersen, and I came across this problem:

Suppose $H$ is a Hilbert space and $X$ and $Y$ are closed subspaces of $H$ with $\dim X<\infty$ and $\dim X<\dim Y$. Prove that $X^{\perp}\cap Y\not=\{0\}$.

I noticed someone asked the same question here: Hilbert space, functional analysis

But, I'm unsure if the answer is correct since it is not clear to me why $Y=X\cap Y+X^{\perp}\cap Y$.

If $X\subset Y$, then I would know how to solve the problem, but I am having trouble with the general case.

• You are right, $Y=(X\cap Y)+(X^\perp \cap Y)$ might not be true. For instance, in the usual euclidean space $\Bbb R^4$ you could have $X=\operatorname{Span}(e_1)$ (hence $X^\perp=\operatorname{Span}(e_2,e_3,e_4)$ ) and $Y=\operatorname{Span}(e_1+e_2, e_3)$. It holds $Y\cap X^\perp =\operatorname{Span}(e_3)$ and $Y\cap X=\{0\}$.
– user228113
Oct 27 '16 at 21:10
• Consider the orthogonal projection $P$ onto $X$. Since $\dim Y > \dim X$, it follows that $P\lvert_Y$ cannot be injective. But $\ker (P\lvert_Y) = Y \cap X^{\perp}$. Oct 27 '16 at 21:45

By changing $Y$ by a subspace of dimension $\dim(X)+1$, and $H$ by $X+Y$, we can assume $H$ is finite dimensional.

Now consider the canonical quotient map $T:Y\to H/X^\perp\simeq X$. By the rank-nullity theorem, $$\operatorname{dim}(Y)=\operatorname{dim}(TY)+\operatorname{dim}(\ker T)\leq\operatorname{dim}(X)+\operatorname{dim}(Y\cap X^\perp)$$ But $\operatorname{dim}(Y)>\operatorname{dim}(X)$, so $$0<\operatorname{dim}(Y\cap X^\perp).$$

• I'm a bit confused about your argument as to why we are allowed to assume $H$ is finite-dimensional. Could you please elaborate on that point a bit more.
– ervx
Oct 27 '16 at 21:26
• @ervx Of couse, if $\tilde{Y}\cap X^\perp\neq 0$ for some subspace $\tilde{Y}\subseteq Y$, then the same is true with $Y$ in place of $\tilde{Y}$. This is the first part (where we assume that $\operatorname{dim}(Y)=\operatorname{dim}(X)+1$). Now instead of looking at all of $H$, we look at the finite-dimensional subspace $\tilde{H}=X+Y$, with the same inner product. If we use subscripts to denote the space we are working in, we have $X^\perp_{\tilde{H}}=X^\perp_H\cap\tilde{H}$. Anyway, the argument shows $0\neq Y\cap X^\perp_{\tilde{H}}\subseteq Y\cap X^\perp_H$. Oct 27 '16 at 21:49

The answer at the linked question is indeed incorrect. We get a counterexample in $\mathbb{R}^3$ (that generalises of course to Hilbert spaces of arbitrary dimension $\geqslant 3$) by taking $X = \operatorname{span} \{ e_1\}$ and $Y = \operatorname{span} \{ e_1 + e_3, e_2\}$. Then $X\cap Y = \{0\}$ and $X^{\perp} \cap Y = \operatorname{span} \{e_2\} \neq Y$.

In a Hilbert space, we can look at the orthogonal projection $P$ onto $X$. Since $\dim Y > \dim X$, it follows that $P\lvert_Y$ cannot be injective, hence

$$Y \cap X^{\perp} = \ker (P\lvert_Y) \neq \{0\}.$$

We can generalise this to complemented subspaces (every finite-dimensional subspace of a normed space is complemented) $X$ of Banach spaces $B$: If $C$ is a complementary (closed) subspace to $X$, then the canonical projection $\pi \colon B \to B/C$ cannot be injective on $Y$ if $\dim Y > \dim B/C = \dim X$, so $Y\cap C \neq \{0\}$.

Of course we can forget about everything topological and use an algebraic complement (which always exists thanks to the axiom of choice) to reach the same conclusion, but in functional analysis, algebraic complements are typically not useful, while topological complements are.