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# canonical basis for symmetric bilinear forms

If $B:V\times V\rightarrow K$ is a symmetric bilinear form over a finite-dimensional vector space, where the characteristic of the field is not 2, then we may prove that there is an orthogonal basis such that $B$ is represented by

$\bordermatrix{&\cr&a_{{1}}&0&\ldots&0\cr&0&a_{{2}}&\ldots&0\cr&\vdots&\vdots&% \ddots&\vdots\cr&0&0&\ldots&a_{{n}}\cr}$ |

Recall that a bilinear form has a well-defined rank, and denote this by $r$.

If $K=\mathbb{R}$ we may choose a basis such that $a_{1}=\cdots=a_{t}=1$,
$a_{{t+1}}=\cdots=a_{{t+p}}=-1$ and $a_{{t+p+j}}=0$, for some integers $p$ and $t$,
where $1\leq j\leq n-t-p$.
Furthermore, these integers are *invariants* of the bilinear form.
This is known as *Sylvester’s Law of Inertia*.
$B$ is *positive definite* if and only if
$t=n$, $p=0$. Such a form constitutes a *real inner product space*.

If $K=\mathbb{C}$ we may go further and choose a basis such that $a_{1}=\cdots=a_{r}=1$ and $a_{{r+j}}=0$, where $1\leq j\leq n-r$.

If $K=F_{p}$ we may choose a basis such that $a_{1}=\cdots=a_{{r-1}}=1$,

## Mathematics Subject Classification

47A07*no label found*11E39

*no label found*15A63

*no label found*

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## Corrections

capitalization by igor ✓

capitalization by igor ✓

finite field case. by Algeboy ✓