Help with understanding this proposition regarding non-degenerate bilinear forms

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!

• Can you please specify, how does the result depends on $B$? May be I am missing something clear, but imho the proposition concerns only the spaces and not the $B$. Commented Aug 5, 2016 at 12:50
• The orthogonal of $W$ is w.r.t. the bilinear form I suppose? I also presume that it is supposed non-degenerate? Commented Aug 5, 2016 at 12:52
• When you say that $W$ is non-degenerate, do you mean that the restriction of $B$ to $W$ is non-degenerate? Commented Aug 5, 2016 at 12:54
• Just thinking of $B(v,w)=0$ for all $v \in V$ implies $w=0$ (any $w\in V$). Commented Aug 5, 2016 at 12:55
• In other words the matrix is nonsingular, right? And the orthgonality $\perp$ is rather $\perp_B$. Also the reflexivity can be viewed as the matrix being symmetric,right? Commented Aug 5, 2016 at 12:59

Can it be that the implication $$\tilde{v} \in W^{\perp_B} \Rightarrow B(v_i,\tilde{v})=0, i=1,\dotsc ,m$$ is actually equivalency? Then it would solve the problem using the Dimension of Sum and Intersection of Vector Spaces. May be I am missing something, feel free to opose.