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.
Best Answer
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: $$ \begin{equation}\begin{split}(\mathcal R^\perp + \mathcal S^\perp)^\perp &= (\mathcal R^\perp)^\perp \cap (\mathcal S^\perp)^\perp \\ \left((\mathcal R^\perp + \mathcal S^\perp)^\perp\right)^\perp &= \left((\mathcal R^\perp)^\perp \cap (\mathcal S^\perp)^\perp\right)^\perp \\ \mathcal R^\perp + \mathcal S^\perp &= (\mathcal R \cap \mathcal S)^\perp. \end{split}\end{equation} $$