# Can we show $(U\oplus V)^\perp=U^\perp\oplus V^\perp$?

Let $$H$$ be a complex Hilbert space, $$T\in\mathfrak L(H)$$ and $$U,V$$ be subspaces of $$H$$. Let $$W:=U\oplus V$$ denote the orthogonal direct sum of $$U$$ and $$V$$. Suppose we know $$TU^\perp\subseteq U^\perp$$ and $$TV^\perp\subseteq V^\perp$$.

Are we able to conclude $$TW^\perp\subseteq W^\perp$$? Maybe this is a consequence of $$W^\perp=W^\perp\oplus V^\perp$$. Can we show this?

I've tried to show $$W^\perp=W^\perp\oplus V^\perp$$, but neither of the inclusions is obvious to me. For example, if $$x\in U^\perp\oplus V^\perp$$, then there are $$(y,z)\in U^\perp\times V^\perp$$ with $$y\perp z$$ and $$x=y+z$$. Analogously, if $$w\in U\oplus V$$, there are $$(u,v)\in U\times V$$ with $$u\perp v$$ and $$w=u+v$$. We need to show that $$\langle x,w\rangle_H=0$$. Now, clearly, $$\langle x,w\rangle_H=\langle y,v\rangle_H+\langle z,u\rangle_H$$ ... Now I guess, if the claim holds, we somehow need to "add a $$0$$" in a clever way.

$$W=U+ V$$ implies $$W^\perp = U^\perp\cap V^\perp$$. That is, a vector is orthogonal to all $$w\in W$$ if and only if it is orthogonal to all $$u\in U$$ and $$v\in V$$. And, purley set-theoreticaly, $$f(X\cap Y)\subseteq f(X)\cap f(Y)$$. Thus
$$TU^\perp\subseteq U^\perp, TV^\perp\subseteq V^\perp \implies$$ $$TW^\perp=T(U^\perp\cap V^\perp)\subseteq TU^\perp\cap TV^\perp\subseteq U^\perp\cap V^\perp=W.$$
• When you write $W=U+V$, do you understand this sum to be orthogonal? – 0xbadf00d Feb 6 at 7:19
• So $U\perp V$ is not important here, right? What's definition of a direct sum in this context? – 0xbadf00d Feb 6 at 7:29
• $U+V:=\{u+v\mid u\in U,v\in V\}$, where $U,V$ are any two subsets of $H$. It's just a sum, being direct isn't important. – runway44 Feb 6 at 7:36
• I see, $U$ and $V$ don't even need to be subspaces, if I'm not missing something. – 0xbadf00d Feb 6 at 7:46