Prove $(V_1 \cap V_2)^{\perp_L} = V_1^{\perp_L} + V_2^{\perp_L}$ if $f$ is non-degenerate Let $f(\alpha, \beta)$ be a bilinear form on the $n$-dimensional linear space $V$ over the number field $F$. Prove, if $f(\alpha, \beta)$ is non-degenerate, for any subspaces $V_1$ and $V_2$ of $V$,
then
\begin{align*}
& (V_1 \cap V_2)^{\perp_L} = V_1^{\perp_L} + V_2^{\perp_L}, \\
& (V_1 \cap V_2)^{\perp_R} = V_1^{\perp_R} + V_2^{\perp_R}.
\end{align*}
where for any subspace $W$ of $V$, the left orthogonal group $W^{\perp_L}$ and the right orthogonal group $W^{\perp_R}$ are defined by
\begin{align*}
& W^{\perp_L} = \{\alpha \in V: f(\alpha, \beta) = 0, \forall \beta \in W\}, \\
& W^{\perp_R} = \{\beta \in V: f(\alpha, \beta) = 0, \forall \alpha \in W\}.
\end{align*}
By definition, I am able to show (in this direction, non-degeneracy of $f$ is not needed) that $V_1^{\perp_L} + V_2^{\perp_L} \subseteq (V_1 \cap V_2)^{\perp_L}$. I don't have much thoughts on the other direction, in particular, how the non-degeneracy of $f$ should be applied?
 A: $\newcommand{\lbot}{\perp_L}$
$\newcommand{\rbot}{\perp_R}$
We first prove by definition that
\begin{align*}
& (V_1 + V_2)^{\lbot} = V_1^{\lbot} \cap V_2^{\lbot}; \tag{1} \\ 
& (V_1 + V_2)^{\rbot} = V_1^{\rbot} \cap V_2^{\rbot}. \tag{2}
\end{align*}
Let $\alpha \in (V_1 + V_2)^{\lbot}$, then for any $\beta_1 \in V_1, \beta_2 \in V_2$, we have
\begin{align*}
& f(\alpha, \beta_1 + \beta_2) = f(\alpha, \beta_1) + f(\alpha, \beta_2) = 0, \\
& f(\alpha, \beta_1 - \beta_2) = f(\alpha, \beta_1) - f(\alpha, \beta_2) = 0.
\end{align*}
Hence $f(\alpha, \beta_1) = f(\alpha, \beta_2) = 0$, i.e., $\alpha \in V_1^{\lbot} \cap V_2^{\lbot}$. Conversely, if $\alpha \in V_1^{\lbot} \cap V_2^{\lbot}$, then for any $\beta = \beta_1 + \beta_2 \in V_1 + V_2$, where $\beta_1 \in V_1, \beta_2 \in V_2$, we have
$$f(\alpha, \beta_1 + \beta_2) = f(\alpha, \beta_1) + f(\alpha, \beta_2) = 0 + 0 = 0,$$
i.e., $\alpha \in (V_1 + V_2)^{\lbot}$. The second equality can be proved similarly.
If $f(\alpha, \beta)$ is non-degenerate, we show that for any subspace $W$ of $V$, $W = (W^{\lbot})^{\rbot}$. By definition, $W \subset (W^{\lbot})^{\rbot}$. To show the other direction, it can be shown by $f$ is non-degenerate that for any subspace $W$,
$$\dim(W^{\lbot}) = \dim(W^{\rbot}) = \dim(V) - \dim(W).$$
It then follows that
\begin{align*}
\dim((W^{\lbot})^{\rbot}) = \dim(V) - \dim(W^{\lbot}) = \dim(V) - (\dim(V) - \dim(W)) = \dim(W). \tag{*}
\end{align*}
This equality and $W \subset (W^{\lbot})^{\rbot}$ imply that $W = (W^{\lbot})^{\rbot}$. Similarly, $W = (W^{\rbot})^{\lbot}$.
Now by $(1)$ and $(2)$, we have
\begin{align*}
(V_1 \cap V_2)^{\lbot} = ((V_1^{\lbot})^{\rbot} \cap (V_2^{\lbot})^{\rbot})^{\lbot} = ((V_1^{\lbot} + V_2^{\lbot})^{\rbot})^{\lbot} = V_1^{\lbot} + V_2^{\lbot}.
\end{align*}
This completes the proof.
(The equality $(*)$ can be established by constructing a map between $W^{\lbot}$ to the solution space of the first $\dim(W)$ columns of the matrix $(f(\alpha_i, \alpha_j))$.)
