Every orthogonal space is a direct sum of a hyperbolic and an anisotropic space

Nondegenerate version
Suppose $$K$$ is a field of characteristic not equal to two, and $$b$$ is a nondegenerate symmetric bilinear form on a finite-dimensional vector space $$V$$ over $$K$$. Then, we can write an internal direct sum decomposition of $$V$$:

$$V = W \oplus W^\perp$$

where the following are true:


 * $$W$$ and $$W^\perp$$ are orthogonal complements with respect to $$b$$ in $$V$$, i.e., $$b(w,w') = 0 \forall w \in W$$ if and only if $$w' \in W^\perp$$, and $$b(w,w') = 0$$ for all $$w' \in W^\perp$$ if and only if $$w \in W$$.
 * The restriction of $$b$$ to $$W$$ makes $$W$$ a hyperbolic space, i.e., a direct sum of finitely many hyperbolic planes. In other words, we can choose a basis for $$W$$ in which the matrix for $$b$$ is a diagonal block matrix with blocks $$\begin{pmatrix} 0 & 1 \\ 1 & 0 \\\end{pmatrix}$$.
 * The restriction of $$b$$ to $$W^\perp$$ makes $$W^\perp$$ an anisotropic space, i.e., for any nonzero vector $$v \in W^\perp$$, $$b(v,v) \ne 0$$.