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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 $<z,x+y> = 0$. And if $x+y\neq 0$, this implies $z=0$.Hence $(X+Y)^\bot=z=0$. $\\$

Again from $<z,x+y> = 0$, we have $<z,x+y> = <z,x>+<z,y>=0$. 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.

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  • $\begingroup$ 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 $z=0$. $\endgroup$
    – Anurag A
    May 15 '19 at 18:06
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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.

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

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

$OE\Delta.$

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

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