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Prove given subspaces $\mathcal{R},\mathcal{S}\subseteq \mathbb{R}^n$, that $(\mathcal{R}\cap \mathcal{S})^\perp = \mathcal{R}^\perp + \mathcal{S}^\perp$.

Here is my attempt:

We will show "$\supseteq$" first. Note part (c) of this post is the same, but here is the succinct version. Now, $\mathscr{L}$et $m\in \mathcal{R}^\perp $ and $n \in \mathcal{S}^\perp$. Then set $x = m + n \in \mathcal{R}^\perp + \mathcal{S}^\perp$. Let $y \in \mathcal{R} \cap \mathcal{S}$ be arbitrary $y$. Observe, $$0 = \langle y,m\rangle + \langle y, n \rangle = \langle y, m+ n \rangle =\langle y, x \rangle.$$ Thus, $x \in (\mathcal{R} \cap \mathcal{S})^\perp$.

For "$\subseteq$" Here is my attempt: Let $x = x_1 + x_2\in(\mathcal{R} \cap \mathcal{S})^\perp$ for $x_i \in (\mathcal{R} \cap \mathcal{S})^\perp$ for $i \in\{1,2\}$. Let $y \in\mathcal{R} \cap \mathcal{S}$. We have $$0 = \langle x, y\rangle = \langle x_1 + x_2,y\rangle = \langle x_1,y\rangle + \langle x_2,y\rangle.$$ Then $x_i \in\mathcal{R}^\perp \cap\mathcal{S}^\perp$ for $i\in\{1,2\}$. Now let $v = v_1+v_2 \in \mathcal{R} + \mathcal{S}$ where $v_1 \in \mathcal{R}$ and $v_2\in\mathcal{S}$. Then $$\begin{equation}\begin{split} \langle x, v\rangle &= \langle x, v_1 + v_2\rangle \\ &= \langle x, v_1\rangle + \langle x, v_2\rangle \\ &= \langle x_1 + x_2, v_1\rangle + \langle x_1 + x_2, v_1\rangle\\ &= \langle x_1, v_1\rangle + \langle x_2, v_1\rangle + \langle x_1, v_2\rangle + \langle x_2, v_2\rangle\\ &= 0 + 0 + 0 + 0\\ &= 0\end{split} \end{equation}$$ Finally, $x = x_1 + x_2 \in\mathcal{R}^\perp + \mathcal{S}^\perp$ where $x_1 \in\mathcal{R}^\perp$ and $x_2 \in\mathcal{S}^\perp$.

Please let me know if I did that second one correctly. BTW Here is a post where someones hints at the second part that I have attempted but not actually given full answer.

Owen Murphy
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  • The beginning of "$\subseteq$" is wrong: $\langle x_1,y\rangle + \langle x_2,y\rangle=0$ for all $y \in\mathcal{R} \cap \mathcal{S}$ doesn't mean that $x_1,x_2 \in\mathcal{R}^\perp \cap\mathcal{S}^\perp$. – azif00 Sep 22 '21 at 23:52
  • Once you establish the $\supseteq$ part, my initial inclination would be to try to finish the argument by comparing the dimensions of the two parts. (And as a random guess, it's possible part of the argument might involve using $\mathcal{R}^\perp \cap \mathcal{S}^\perp \subseteq (\mathcal{R} + \mathcal{S})^\perp$ to establish an inequality of dimensions.) – Daniel Schepler Sep 23 '21 at 00:09
  • @azif00 is it because I need to show $\langle x_1, y\rangle = \langle x_2, y \rangle = 0$? I presume this is where I have made an erroneous assumption. – Owen Murphy Sep 23 '21 at 00:13

1 Answers1

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Given that for subspaces $A,B\subseteq \mathbb{R}^n$, we know $$(A + B)^\perp = A^\perp \cap B^\perp.$$ This is easily proven. Then just choose $A = \mathcal R^\perp$ and $B = \mathcal S^\perp$. Plug these in and take the perp of the equation and observe: $$ \begin{equation}\begin{split}(\mathcal R^\perp + \mathcal S^\perp)^\perp &= (\mathcal R^\perp)^\perp \cap (\mathcal S^\perp)^\perp \\ \left((\mathcal R^\perp + \mathcal S^\perp)^\perp\right)^\perp &= \left((\mathcal R^\perp)^\perp \cap (\mathcal S^\perp)^\perp\right)^\perp \\ \mathcal R^\perp + \mathcal S^\perp &= (\mathcal R \cap \mathcal S)^\perp. \end{split}\end{equation} $$

Owen Murphy
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  • Your very first first equality asumes what you are trying to prove: that the "perp" of a sum is the intersection of the "perps". So you are using what you are supposed to be proving in the first place. If you first equality is true, the last one is true. But how do you know the first equation is true? Starting from something false gives you absolutely no warrant for determining the validity of your endpoint. – Arturo Magidin Oct 06 '21 at 20:13
  • @ArturoMagidin, if by the "first equality" you mean that "$(A+B)^\perp = A^\perp \cap B^\perp$" then please read the original question. We need to prove the bottom but simply used the fact that $(A+B)^\perp = A^\perp \cap B^\perp$ from a previous proof to come up with a more cute proof. – Owen Murphy Oct 06 '21 at 20:18
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    Did I get turned around along the way? If so, sorry. – Arturo Magidin Oct 06 '21 at 20:20
  • This is nice. Show that $(R \cap S)^{\bot} \subseteq (R^{\bot}+S^{\bot})$ by showing that $(R^{\bot}+S^{\bot})^{\bot} \subseteq (R \cap S)$. And then taking the $\bot$ of both sides and reversing the binary relation. – Mike Oct 06 '21 at 23:13
  • The first equality in this answer by the way is straightforward to show in fact. – Mike Oct 06 '21 at 23:21