Orthogonal and algebraic compement inclusion

Question

Let $H$ be a Hilbert space. Let $U$ be a subspace and let $E$ be any complement such that

$$ H = U \oplus E$$

I am wondering if it can be said that $U^\bot$ is contained in $E$. If not can it at least be said that $\mathrm{dim}(U^\bot) \le \mathrm{dim}(E)$?

Answer 1

Geometrically the answer can be obvious. Take $H=\Bbb R^2$; writing $H=U\oplus V$ involves finding two one-dimensional subspaces, or equivalently, lines through the origin. The only way one line contains another is if they coincide, so your hypothesis is that there are no nonorthogonal pairs of lines through the origin, which is clearly wrong.

If $U\le H$ is any subspace and $E$ any complement then $E\cong H/U$ (the projection $H\to E$ has kernel $U$), so $\dim E$ does not depend on which complement is chosen.

Answer 2

In general, no. Consider for example $H=\mathbb{R}^2$, $U=\{(x,0)\,:\,x\in\mathbb{R}\}$, $E=\{(x,x)\,:\,x\in\mathbb{R}\}$. Certainly $H=U\oplus E$ but $U^\perp=\{(0,y)\,:\,y\in\mathbb{R}\}$ so we also have $H=U^\perp\oplus E$. Hence $E\not\subseteq U$.

The answer to your other question is yes. Let us assume $E$ is finite dimensional (otherwise it holds trivially). If $U$ is finite dimensional then $\dim(V\oplus W)=\dim V+\dim W$ implies $\dim U^\perp=\dim E$, so assume $U$ is infinite dimensional. First assume $U$ is closed. Then the orthogonal projection $P_{U^\perp}$ onto $U^\perp$ has kernel $U$, so $H/U\cong U^\perp$. Since $H=U\oplus E$, there exists a (not necessarily bounded) projection operator $Q:H\rightarrow E$ such that $x-Qx\in U$ for all $x\in H$. Define $$T(x+U)=Qx$$ We claim this an injective linear map $T:H/U\rightarrow E$. If $x,y\in H$ with $x+U=y+U$ then $x-y\in U$, so $Q(x-y)=0$ This implies $Qx=Qy$, so the map is well-defined. Moreover $T$ is clearly linear and if $Qx=0$ then $x\in U$, so $\ker T=0$. Hence $T$ is injective, completing the proof. This shows $\dim U^\perp=\dim H/U\le\dim E$.

If $U$ is not closed, then let $F$ be the linear span of $\{x\in E\,:\,x\notin\overline{U}\}$. Then the above working shows $\dim U^\perp=\dim\overline{U}^\perp\le\dim F\le\dim E$.