Let $W \subset V$ vector spaces, let $W'$ be the orthogonal complement of $W$. Show $\dim(W)+\dim(W')=\dim V$

Question

As title says: Let $W \subset V$ vector spaces, let $W'$ be the orthogonal complement of $W$. Show $\dim(W)+\dim(W')=\dim V$.

We are given that $W \subset V$ finite vector spaces, symmetric bilinear mapping $B: V\times V \to \mathbb R$, $Q$ is the quadratic form that corresponds to $B$ (meaning $Q(x)=B(x,x)$)

$W'=\{u\in V| B(u,x)=0 \forall x \in W\}$

We are given also that $Q$ is nondegenerate (meaning, it has no zero eigenvalue. the number of positive eigenvalues+the number of negative eigenvalues = dimV)

Show that $\dim W+\dim W'=\dim V$

I nearly got the solution, just need one last push

What I did:

I wanted to show that $V$ is a direct sum of $W$ and $W'$:

Lemma: if Q is nondegenerate then $Q(x)=0$ implies $x=0$

proof: Q is nondegenerate so there is no $0$ eigenvalue, so the only vector that $Q$ sends to $0$ has to be the $0$ vector.

let $x \in W\cap W'$, then $B(x \in W',x \in W)=B(x,x)=0$ which means $Q(x)=0$, from the lemma, it follows that $x=0$. So it is indeed a direct sum.

Now we need to prove that $V\subseteq W\cup W'$ and $W\cup W' \subseteq V$:

$W\cup W' \subseteq V$:

let $x\in W\cup W'$, if $x\in W$ then $x \in V$ because $W \subset V$. If $x\in W'$ then for the same reason $x \in V$, because $W' \subset V$.

Now we just need to show that $V\subseteq W\cup W'$:

let $x \in V$. if $x \in W$ then it is trivial. So assume $x \in V$ and $x \notin W$

Why is $x\in W'$ in that case?

Answer 1

By nondegeneracy and symmetry, $B(\cdot,\cdot)$ is an inner product on $V$. Let ${x_1, \ldots, x_m}$ be a basis for $W$. After modification by the Gram-Schmidt process, if necessary, you can assume that the basis is orthogonal with respect to $B(\cdot, \cdot)$, that is $B(x_i, x_j) = 0$ when $i \neq j$. Now complete the basis such that ${x_1, \ldots, x_m, y_1, \ldots, y_n}$ is a basis for $V$. Again, let's say it is orthogonal with respect to $B(\cdot,\cdot)$. Of course, $m + n = \mathrm{dim}V$. Moreover, you can check that $W'$ is the span of the $y_i$'s. So $\mathrm{dim}W = m$ and $\mathrm{dim}W' = n$.

I hope this provides some intuition at least, if not the expected answer.