# Orthogonality in subspaces of Hilbert Spaces

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 $$=0=+$$

How can I show $$0==$$ 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 $$a\in Y^\perp$$, $$a\in X^\perp\cap Y^\perp$$.