Problem: Given a real subspace $W$ of $\mathbb{R} ^4$, let $W^\perp$ denote its orthogonal complement with respect to the standard inner product of $\mathbb{R} ^4$. Let $W_1 = \operatorname{span}\left\{ (1,0,0,-1) \right\} $ and $W_2 = \operatorname{span}\left\{ 2,1,0,-1 \right\} $ be real subspaces of $\mathbb{R} ^4$. Then the dimension of $W_1^\perp \cap W_2^\perp $ over $\mathbb{R} $ is equal to _______________.
Solution: For any subspace $W \subseteq V$, we have
\[
W \oplus W^\perp = V \implies \dim W + \dim W^\perp = \dim V.
\]
Using the above relation, we have
\[
\dim W_1^\perp = 3 = \dim W_2^\perp .
\]
As $(2,1,0,-1) \notin W_1$, both the subspaces are distinct which implies, $W_1^\perp \neq W_2^\perp $. That is, there exists $v\in W_2^\perp $ such that $v \notin W_1^\perp $. This implies,
\[
\operatorname{span}\left\{ W_1^\perp \cup \{ v \} \right\} = \mathbb{R} ^4 \implies \operatorname{span}\left\{ \right\} \left\{ W_1^\perp \cup W_2^\perp \right\} = \mathbb{R} ^4.
\]
Now recall that for any subspace $W_1, W_2$ of a finite dimensional vector space $V$, we have
\[
\dim (W_1 + W_2) = \dim W_1 + \dim W_2 - \dim (W_1 \cap W_2).
\]
Since $\dim \left( W_1^\perp + W_2^\perp \right) = 4$, we have
\[
4 = 3 + 3 - \dim \left( W_1^\perp \cap W_2^\perp \right) \implies \dim \left( W_1^\perp \cap W_2^\perp \right) = 2.
\]
Hence, we showed that
\[
\textcolor{teal}{\boxed{
\dim \left( W_1^\perp \cap W_2^\perp \right) = 2
}}
\]