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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.

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    If $V$ is finite-dimensional, we have $W\cap W^{\perp} = \{0\} \implies V = W \oplus W^{\perp}$ also if $V^{\perp} \neq \{0\}$. – 2017-02-18

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

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    Sorry, I did not pay enough attention. – 2017-02-18
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    If $V$ is finite-dimensional, then we always have $\dim W + \dim W^{\perp} \geqslant \dim V$. With $\varphi \colon V \to V^{\ast};\, w \mapsto \beta(w,\,\cdot\,)$, $W^{\perp}$ is the annihilator of $\varphi(W)$, so $\dim W^{\perp} = \dim V^{\ast} - \dim \varphi(W) \geqslant \dim V - \dim W$ since $\dim V^{\ast} = \dim V$ and $\dim \varphi(W) \leqslant \dim W$. – 2017-02-18