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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

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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, $x \in W_1 \cap W_2$.

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  • $\begingroup$ your answer looks more natural to me, with all respect to @Ben West beautiful solution which was very insightful as well... Thanks ! $\endgroup$ – the8thone Aug 16 '14 at 20:13
  • $\begingroup$ Oops, I did not mean to usurp Ben's answer, it was just an additional approach. $\endgroup$ – copper.hat Aug 16 '14 at 20:16
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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 $$ (W_1\cap W_2)^\perp=((W_1^\perp)^\perp\cap (W_2^\perp)^\perp)^\perp=((W_1^\perp+W_2^\perp)^\perp)^\perp=W_1^\perp+W_2^\perp. $$

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    $\begingroup$ Wow !!! Thank you so much ! $\endgroup$ – the8thone Aug 16 '14 at 3:18
  • $\begingroup$ @the8thone No problem. $\endgroup$ – BW. Aug 16 '14 at 3:19
  • $\begingroup$ 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. $\endgroup$ – InfiniteElementMethod Aug 16 '14 at 4:43
  • $\begingroup$ @InfiniteDifferenceMethod I suppose a proof can be found in some other posts on this site, for example: math.stackexchange.com/questions/246967/… $\endgroup$ – Martin Sleziak Aug 16 '14 at 5:37

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