K^n/Bilinear forms/Arbitrary coefficients/Special cases/Example

Let ${}V=K^{n}$, and let ${}a_{ij}\in K$ denote fixed numbers, for ${}1\leq i,j\leq n$. Then, the assignment

$({\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}},{\begin{pmatrix}y_{1}\\\vdots \\y_{n}\end{pmatrix}})\longmapsto \Psi (x_{1},\ldots ,x_{n},y_{1},\ldots ,y_{n})=\sum _{i,j}a_{ij}x_{i}y_{j}$

is a bilinear form. In case

${}a_{ij}=0\,$

for all ${}i,j$, this is the zero form; in case

${}a_{ij}=\delta _{ij}={\begin{cases}1,{\text{ if }}i=j\,,\\0{\text{ else}}\,,\end{cases}}\,$

we have the standard inner product (where the expression makes sense for every field; the property of being positive definite does only make sense for an ordered field). In case ${}n=4$ and

${}\Psi (x_{1},\ldots ,x_{4},y_{1},\ldots ,y_{4})=x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}-x_{4}y_{4}\,,$

we talk about a Minkowski-form. For ${}n=2$ and

${}\Psi (x_{1},x_{2},y_{1},y_{2})=x_{1}y_{2}-x_{2}y_{1}\,,$

this is the determinant in the two-dimensional case.