Prove $(V_1 \cap V_2)^{\perp_L} = V_1^{\perp_L} + V_2^{\perp_L}$ if $f$ is non-degenerate

Question

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?

Answer 1

$\newcommand{\lbot}{\perp_L}$ $\newcommand{\rbot}{\perp_R}$

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 $(f(\alpha_i, \alpha_j))$.)