Orthogonal basis

The concept of an orthogonal basis is applicable to a vector space ${\displaystyle V}$ (over any field) equipped with a symmetric bilinear form ${\displaystyle \langle \cdot ,\cdot \rangle ,}$ where orthogonality of two vectors ${\displaystyle v}$ and ${\displaystyle w}$ means ${\displaystyle \langle v,w\rangle =0.}$ For an orthogonal basis ${\displaystyle \left\{e_{k}\right\}:}$

$${\displaystyle \langle e_{j},e_{k}\rangle ={\begin{cases}q(e_{k})&j=k\\0&j\neq k,\end{cases}}}$$

where ${\displaystyle q}$ is a quadratic form associated with ${\displaystyle \langle \cdot ,\cdot \rangle$ :} ${\displaystyle q(v)=\langle v,v\rangle }$ (in an inner product space, ${\displaystyle q(v)=\|v\|^{2}.}$).

Hence for an orthogonal basis ${\displaystyle \left\{e_{k}\right\},}$

$${\displaystyle \langle v,w\rangle =\sum _{k}q(e_{k})v^{k}w^{k},}$$

where ${\displaystyle v_{k}}$ and ${\displaystyle w_{k}}$ are components of ${\displaystyle v}$ and ${\displaystyle w}$ in the basis.

The concept of orthogonality may be extended to a vector space (over any field) equipped with a quadratic form ${\displaystyle q(v)}$. Starting from the observation that, when the characteristic of the underlying field is not 2, the associated symmetric bilinear form ${\displaystyle \langle v,w\rangle ={\tfrac {1}{2}}(q(v+w)-q(v)-q(w))}$ allows vectors ${\displaystyle v}$ and ${\displaystyle w}$ to be defined as being orthogonal with respect to ${\displaystyle q}$ when ${\displaystyle q(v+w)-q(v)-q(w)=0.}$