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Let $W_1$ and $W_2$ be subspaces of a finite dimensional inner product space space. Prove that $$(W_1 \cap W_2)^\perp=W_1^\perp + W_2^\perp $$

My Try

One direction is easy : Let $\alpha \neq 0$ and $\alpha \in W_1^\perp + W_2^\perp$, i.e. $\alpha$ can be written as $\alpha = \beta + \gamma$ such that $\beta \in W_1^\perp$ and $\gamma \in W_2^\perp$, hence $(\beta|\eta)=0$ for all $\eta \in W_1 $ and $(\gamma|\delta)=0$ for all $\delta \in W_2 $. Now for all $\eta \in W_1 \cap W_2$ it is clear that $(\alpha | \eta) =0$ . hence $$(W_1 \cap W_2)^\perp\supset W_1^\perp + W_2^\perp $$ For proving the other containment, let $\alpha \neq 0$ and $\alpha \in (W_1 \cap W_2)^\perp$, it means that for all $\beta \in W_1 \cap W_2$, $(\alpha|\beta)=0$. Hence $\alpha \in V \setminus (W_1 \cap W_2) = V \setminus (W_1) \cup V \setminus (W_2)$. Hence $\alpha \in W_1^c$ or $\alpha \in W_2^c$. WLOG suppose $\alpha \in W_1^c$. We also have that $$V=(W_1 ) \oplus (W_1 )^\perp$$therefore $\alpha = \eta + \delta $ where $\eta \in W_1$ and $\delta \in W_1^\perp$ and indeed $\delta \neq 0$. From here I have to somehow show that $\eta=0$, but I am stuck in here...

I already appreciate any help

Asaf Karagila
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the8thone
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  • See also: http://math.stackexchange.com/questions/408973/w-1-perp-w-2-perp-w-1-cap-w-2-perp-can-a-set-be-a-function-can-tw – Martin Sleziak Aug 16 '14 at 05:35
  • An example showing that this does not hold in infinite-dimensional spaces can be found here: http://math.stackexchange.com/questions/899273/does-s-bott-bot-s-cap-t-bot-hold-in-infinite-dimensional-spaces – Martin Sleziak Aug 19 '14 at 13:55

2 Answers2

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It is not hard to see that $(W_1+W_2)^\perp=W_1^\perp\cap W_2^\perp$. Try to prove that if you haven't already. A proof can be found here.

Since the spaces are finite dimensional, $(W_i^\perp)^\perp=W_i$ for $i=1,2$. Then using these two facts, observe $$ (W_1\cap W_2)^\perp=((W_1^\perp)^\perp\cap (W_2^\perp)^\perp)^\perp=((W_1^\perp+W_2^\perp)^\perp)^\perp=W_1^\perp+W_2^\perp. $$

Community
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Ben West
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    Wow !!! Thank you so much ! – the8thone Aug 16 '14 at 03:18
  • @the8thone No problem. – Ben West Aug 16 '14 at 03:19
  • Let $z\in W_1 \cap W_2$, $x\in W_1^\perp$ , $y\in W_2^\perp$. Since $z\in W_1$, $\langle z, x \rangle =0$. Since $z\in W_2$, $\langle z, y \rangle =0$. Let $u=x+y\in W_1^\perp+ W_2^\perp $. Combining these, we see

    $$ \langle z, u \rangle= \langle z, x+y \rangle=\langle z, x \rangle+\langle z, y \rangle = 0+0$$

    So $u\in (W_1 \cap W_2)^\perp$ and thus $W_1^\perp+ W_2^\perp \subset (W_1 \cap W_2)^\perp$

    I am also lost on the reverse inclusion.

     – InfiniteElementMethod Aug 16 '14 at 04:43
  • @InfiniteDifferenceMethod I suppose a proof can be found in some other posts on this site, for example: http://math.stackexchange.com/questions/246967/show-w-1-w-2-perp-w-1-perp-cap-w-2-perp-and-w-1-perp-w-2-perp – Martin Sleziak Aug 16 '14 at 05:37
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Some relevant facts are (1) that if $W$ is a closed subspace, then $W^{\bot \bot} = W$, and (2) if $W \subset X$, then $X^\bot \subset W^\bot$, and (3) finite dimensional subspaces are always closed (and so (1) applies).

You have shown one direction, you wish to show that $(W_1 \cap W_2)^\bot\subset W_1^\bot + W_2^\bot$. Because of the above facts, this is equivalent to showing $W_1 \cap W_2 \supset (W_1^\bot + W_2^\bot)^\bot$.

So, suppose that $x \in (W_1^\bot + W_2^\bot)^\bot$. This means that $\langle x,w_1'+w_2'\rangle = 0$ whenever $w_k' \in W_k^\bot$, $k=1,2$.

In particular, we have $\langle x,w_1'\rangle = 0$ for all $w_1' \in W_1^\bot$, and so $x \in W_1^{\bot\bot} = W_1$. Similarly, we have $x \in W_2$. And so, $x \in W_1 \cap W_2$.

copper.hat
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  • your answer looks more natural to me, with all respect to @Ben West beautiful solution which was very insightful as well... Thanks ! – the8thone Aug 16 '14 at 20:13
  • Oops, I did not mean to usurp Ben's answer, it was just an additional approach. – copper.hat Aug 16 '14 at 20:16