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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.

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$W=U+ V$ 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

$$ TU^\perp\subseteq U^\perp, TV^\perp\subseteq V^\perp \implies $$ $$ TW^\perp=T(U^\perp\cap V^\perp)\subseteq TU^\perp\cap TV^\perp\subseteq U^\perp\cap V^\perp=W. $$

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  • $\begingroup$ When you write $W=U+V$, do you understand this sum to be orthogonal? $\endgroup$ – 0xbadf00d Feb 6 at 7:19
  • $\begingroup$ It doesn't actually matter if it's orthogonal, or even direct. $\endgroup$ – runway44 Feb 6 at 7:27
  • $\begingroup$ So $U\perp V$ is not important here, right? What's definition of a direct sum in this context? $\endgroup$ – 0xbadf00d Feb 6 at 7:29
  • $\begingroup$ $U+V:=\{u+v\mid u\in U,v\in V\}$, where $U,V$ are any two subsets of $H$. It's just a sum, being direct isn't important. $\endgroup$ – runway44 Feb 6 at 7:36
  • $\begingroup$ I see, $U$ and $V$ don't even need to be subspaces, if I'm not missing something. $\endgroup$ – 0xbadf00d Feb 6 at 7:46

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