I'm reworking a proof of a proposition so that I can be sure I understand it correctly, but I am getting a little stuck.
Consider the following proposition:
$B$ a reflexive bilinear form on a vector space $V$ and $W$ a non-degenerate subspace. Then $ V = W \oplus W^{\bot}$
Proof
By definition, $W \cap W^{\bot} = \{ 0 \}$. It suffices to show that $\dim(V) = \dim(W) + \dim(W^{\bot})$. Let $\dim(W) = m$ and $\dim(V) = n$. Let $\{ v_1,...,v_m \}$ be the basis of $W$, and extend this to a basis of $V$: $\{ v_1,...,v_m \} \cup \{ v_{m+1},...,v_{n} \}$. Let $ \tilde{v} \in W^{\bot}$ be non-zero and write $\tilde{v} = c_1 v_1+...+c_n v_n$. Since $v \in W^{\bot} \implies B(v_i, \tilde{v}) = 0$ for $1 \le i \le m$. Then for each $i \in [1,..,m]$, $\sum_{j=1}^{n} B(v_i, v_j)c_j = 0$. So $\tilde{v}$ is in the kernel of an $m \times n$ matrix $\hat{B}$ where the top $m$ rows of $\hat{B}$ coincide with the matrix corresponding to $B$.
I am ok with this proof until this point, where the claim is made that:
$\dim(W^{\bot}) \ge n-m$.
However, it's not immediately obvious to me why this is so. I seem to have shown that $\dim(W^{\bot}) \subseteq Ker \hat{B}$.
I know by rank nullity that $\dim(V) = \dim(Ker \hat{B}) + \dim(Im \hat{B})$, which would give me $\dim(Ker \hat{B}) = \dim(V) - \dim(Im \hat{B}) = n-m $, but this does not imply $\dim(W^{\bot}) \ge n-m$?
Any enlightenment would be greatly appreciated!
