Let $X$ and $Y$ are subspaces of the Hilbert space $H$

If $X+Y=\{x+y : x\in X , y\in Y\}$

Show that $(X+Y)^{\perp}=X^{\perp} \cap Y^{\perp}$

My tend:

I have tried to prove coverings from bothside. Namely,$(X+Y)^{\perp} \subseteq X^{\perp} \cap Y^{\perp}$ and $(X+Y)^{\perp} \supseteq X^{\perp} \cap Y^{\perp}$

I have shown $(X+Y)^{\perp} \supseteq X^{\perp} \cap Y^{\perp}$ side but I cannot show the second covering. I have written belows and I’m stuck.

Let $ a \in (X+Y)^{\perp} $. We get $<a,x+y>=0=<a,x>+<a,y>$

How can I show $0=<a,x>=<a,y>$ namely $ a \in X^{\perp} \cap Y^{\perp}$??

It will probably so clear but I cannot see in no way. Thanks in advance...


Suppose that $a\in (X+Y)^\perp$. Then $\langle a, x+y\rangle = 0$ for all $x\in X$ and $y\in Y$. However, $Y$ is a subspace, and so $0\in Y$. As such, for any $x\in X$, $\langle a, x+0\rangle=\langle a,x\rangle=0$ and so $a\in X^\perp$. On the other hand, $0$ is also an element of $X$, and so $\langle a, 0+y\rangle =\langle a,y\rangle=0$. Therefore, $a\in Y^\perp$. As $a\in X^\perp$ and $a\in Y^\perp$, $a\in X^\perp\cap Y^\perp$.

  • $\begingroup$ Very nice viewpoint thanks a lot :)) $\endgroup$
    – user519955
    Oct 29 '18 at 23:56

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