Prove that if $W \subseteq V$ is a subspace s.t. $W \cap W^\perp = \{0\}$, then $V = W\oplus W^\perp$

Question

EDIT: Please do not waste your time on this question! I missed a hypothesis (namely that $\beta$ is non-degenerate) which makes this whole question trivial.

Let $\beta:V \times V \rightarrow \mathbb{F}$ be a symmetric bilinear form.

In class we've defined:

• If $W \subseteq V$ is a subspace, then $W^\perp := \{v \in V \mid \beta(v,w)=0 \text{ for all }w \in W\}$.
• The form $\beta$ is non-degenerate if $V^\perp = \{0\}$

We've also proved the proposition

• If $\beta$ is non-degenerate and $W\subseteq V$ is a subspace, then $\dim V = \dim W + \dim W^\perp$.

How can I prove the following corollary?

Prove that if $W \subseteq V$ is a subspace s.t. $W \cap W^\perp = \{0\}$, then $V = W\oplus W^\perp$.

This is neither a homework question nor an exercise, but was stated in the lecture notes as a proof. I'm presuming that this should be something easy, but I can't see how to use the proposition to get this.

Answer 1

$$\dim(W + W^\perp)=\dim W +\dim W^\perp -\dim (W\cap W^\perp)=\dots=\dim V$$ and so $$W+W^\perp=V.$$