# Hilbert Spaces and linear subspaces

Let X and Y be linear subspaces of a Hilbert space $$\mathcal{H}$$. $$\\$$

Recall that $$\\$$

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

Show that $$\\$$

$$(X+Y)^\bot$$ = $$X^\bot\cap Y^\bot$$ $$\\$$

I tried solving the problem as follows;$$\\$$

Let $$z\in(X+Y)^\bot$$, if$$\;$$ $$\forall$$ $$x+y\in(X+Y)$$ we have $$ = 0$$. And if $$x+y\neq 0$$, this implies $$z=0$$.Hence $$(X+Y)^\bot=z=0$$. $$\\$$

Again from $$ = 0$$, we have $$ = +=0$$. Which implies $$z\in X^\bot$$ and $$z\in Y^\bot$$. Thus $$X^\bot\cap Y^\bot=z=0$$. $$\\$$

Therefore, $$(X+Y)^\bot$$ = $$X^\bot\cap Y^\bot$$. $$\\$$

I don't know if my working is logically correct. Your comments and hints, would be greatly appreciated. Thanks.

• Your claim that $<z,x+y> = <z,x>+<z,y>=0$ implies $x \in X^{\perp}$ is not correct. You need to show that $<z,x>=0$. Also $<z,x+y>=0$ and $x+y \neq 0$ does not necessarily mean that $z=0$. May 15 '19 at 18:06

Let $$z \in (X+Y)^{\perp}$$. This means $$\langle z,u \rangle =0$$ for all $$u \in X+Y$$. In particular, take $$u=x+0$$, where $$x \in X$$ and $$0 \in Y$$ (since $$Y$$ is a subspace so $$0 \in Y$$). Thus $$\langle z,x \rangle =0$$. Consequently $$z \in X^{\perp}$$.

Now you can proceed similarly to show that $$z \in Y^{\perp}$$. This would prove that $$(X+Y)^{\perp} \subseteq X^{\perp} \cap Y^{\perp}$$. You will now have to show the other containment.

• Thanks Anurag. But how do I show the other containment. Hint please. May 15 '19 at 18:19
• To start: consider an element $z$ in $X^\perp\cap Y^\perp$, and try to show $z\perp(X+Y)$. May 15 '19 at 19:28
• @Berci Thanks very much. May 16 '19 at 9:13

With

$$z \in (X + Y)^\bot \tag 1$$

we have by definition

$$\langle z, x + y \rangle = 0, \; \forall x + y \in X + Y; \tag 2$$

so in particular if we take

$$y = 0, \tag 3$$

then

$$\forall x \in X, \; \langle z, x \rangle = 0 \Longrightarrow z \in X^\bot; \tag 4$$

likewise with

$$x = 0 \tag 5$$

we obtain

$$\forall y \in Y, \; \langle z, y \rangle = 0 \Longrightarrow z \in Y^\bot; \tag 6$$

combining (4) and (6) yields

$$z \in X^\bot \cap Y^\bot, \tag 7$$

and thus we have

$$(X + Y)^\bot \subset X^\bot \cap Y^\bot. \tag 8$$

Going the other way,

$$z \in X^\bot \cap Y^\bot \Longrightarrow z \in X^\bot, \; z \in Y^\bot$$ $$\Longrightarrow \forall (x, y) \in X \times Y, \; \langle z, x \rangle = \langle z, y \rangle = 0, \tag 9$$

whence

$$\forall x + y \in X + Y, \; \langle z, x + y \rangle = \langle z, x \rangle + \langle z, y \rangle = 0 + 0 = 0, \tag{10}$$

whence

$$X^\bot \cap Y^\bot \subset (X + Y)^\bot. \tag{11}$$

Now by virtue of (8) and (11) we conclude that

$$(X + Y)^\bot = X^\bot \cap Y^\bot. \tag{12}$$

$$OE\Delta.$$

• Thanks alot. You've made all simple and clear. Thanks. May 16 '19 at 9:19
• You are most welcome, sir. Cheers! May 16 '19 at 16:10