Why is the orthognal complment of $W_1 + W_2$ equal to the intersection of the orthogonal complement of $W_1$ with the orthogonal complement of $W_2$?

Question

Suppose that $W_1, W_2$ are subspaces of $V$. Show that

$$(W_1 + W_2)^⊥ = W_1^⊥ ∩ W_2^⊥.$$

Answer 1

By definition, $$ \left( W_1+W_2 \right)^\perp = \left\{ \ x \in V \ \colon \ \langle x, v \rangle = 0 \mbox{ for all } v \in W_1 + W_2 \ \right\} $$

As $W_1 \subset W_1 + W_2$ and $W_2 \subset W_1 + W_2$, so we can also conclude that any $x \in \left( W_1 + W_2 \right)^\perp$ also is in $W_1^\perp \cap W_2^\perp$, and so $$ \left( W_1 + W_2 \right)^\perp \subset W_1^\perp \cap W_2^\perp. \tag{1} $$

Now let $x \in W_1^\perp \cap W_2^\perp$. Then $x \in W_1^\perp$ and $x \in W_2^\perp$. Therefore $$ \left\langle x, w_1 \right\rangle = 0 \ \mbox{ for all } w_1 \in W_1 $$ and $$ \left\langle x, w_2 \right\rangle = 0 \ \mbox{ for all } w_2 \in W_2. $$ So for any element $v \in W_1 + W_2$, we have $$ v = w_1 + w_2 $$ for some $w_1 \in W_1$ and $w_2 \in W_2$. Therefore $$ \begin{align} \langle x, v \rangle &= \left\langle x, w_1 + w_2 \right\rangle \\ &= \left\langle x, w_1 \right\rangle + \left\langle x, w_2 \right\rangle \\ &= 0 + 0 \\ &= 0, \end{align} $$ thus showing that $x \in \left( W_1 + W_2 \right)^\perp$. Thus $$ W_1^\perp \cap W_2^\perp \subset \left( W_1 + W_2 \right)^\perp. \tag{2} $$

From (1) and (2) above, we obtain $$ \left( W_1 + W_2 \right)^\perp = W_1^\perp \cap W_2^\perp. $$

Hope this helps.

Answer 2

Hints.

1. If $U_1$ and $U_2$ are subspaces with $U_1\subseteq U_2$, then $U_1^\perp\supseteq U_2^\perp$

2. As a consequence, $(W_1+W_2)^\perp\subseteq W_1^\perp\cap W_2^\perp$.

3. Now prove that $W_1^\perp\cap W_2^\perp\subseteq(W_1+W_2)^\perp$; what are the elements of $W_1+W_2$?