# Symmetrical bilinear form

A symmetrical bilinear form is the name given to a bilinear Forme on a vector Space which is symmetrical. The symmetrical bilinear forms play a big role in the study of the quadric .

## Definition

That is to say $V$ a vector space of dimension $n$ on a commutative body $\ mathbb \left\{K\right\}$. A application $B: V \ times V \ rightarrow \ mathbb \left\{K\right\}: \left(U, v\right) \ mapsto B \left(U, v\right)$ is a symmetrical bilinear form on space if:
• $\ forall \left(U, v\right) \ in V^2 \ quad B \left(U, v\right) =B \left(v, U\right) \$
• $\ forall \left(U, v, W\right) \ in V^3 \ quad B \left(u+v, W\right) =B \left(U, W\right) +B \left(v, W\right) \$
• $\ forall \ lambda \ in K, \ forall \left(v, W\right) \ in V^2 \ quad B \left(\ lambda v, W\right) = \ lambda B \left(v, W\right) \$

Remarque: The last two axioms imply only the linearity compared to the “first variable” but the first makes it possible to deduce from it the linearity compared to the “second variable”.

## Matric representation

That is to say $C= \left(e_ \left\{1\right\}, \ ldots, e_ \left\{N\right\}\right)$ a base of a vector space $V$. Let us define the matrix order $n$ $A$ by $a_ \left\{ij\right\} =B \left(e_ \left\{I\right\}, e_ \left\{J\right\}\right)$. The matrix $A$ is symmetrical according to the symmetry of the bilinear form. If the matrix of the type $\left(N, 1\right)$ $x$ represents the coordinates of a vector $v$ compared to this base, and in a similar way $y$ represents the coordinates of a vector $w$, then $B \left(v, W\right)$ is equal to:

$\left\{\right\} ^t X has y= \left\{\right\} ^t has x.$

there

Let us suppose that $C\text{'}= \left(e\text{'}_ \left\{1\right\}, \ ldots, e\text{'}_ \left\{N\right\}\right)$ is another base of $V$, consider the Matrice of passage (invertible) $S$ of order $n$ of the base $C$ at the base $C\text{'}$. Maintaining in this new base the matric representation of the symmetrical bilinear form is given by

$A\text{'} = \left\{\right\} ^t S.A.S.$

## Orthogonality and singularity

A symmetrical bilinear form is always reflexive. By definition, two vectors $v$ and $w$ are orthogonal for the bilinear form $B$ if $B \left(v, W\right) =0$, which, thanks to reflexivity, is equivalent to $B \left(W, v\right) =0$.

The core of a bilinear form $B$ is the whole of the orthogonal vectors to any other vector of $V$. It is easy to check that it is a subspace of $V$. When we work with a matric representation of $A$ relative at a certain base, a vector $v$ represented by its matrix column of the coordinates $x$, belongs to the core if and only if

$A x=0 \ Longleftrightarrow \left\{\right\} ^t X A=0.$

The matrix $A$ is noninvertible or singular if and only if the core of $B$ is nonreduced to the singleton null vector, i.e. noncommonplace.

If $W$ is a vectorial subspace of $V$, then $W^ \left\{\ perp\right\}$, the whole of all the orthogonal vectors to any vector of $W$ is also a subspace of $V$. When the core of $B$ is commonplace, the dimension of $W^ \left\{\ perp\right\} =n- \ dim \left(W\right)$.

## Orthogonal bases

A base $C= \left(e_ \left\{1\right\}, \ ldots, e_ \left\{N\right\}\right)$ is othogonale for $B$ if:

$\ forall I \ neq J, B \left(e_ \left\{I\right\}, e_ \left\{J\right\}\right) =0 \$.

When the characteristic of the body is different from two, there exists always an orthogonal base. That can be shown by recurrence.

A base $C$ is othogonale if and only if the matrix $A$ representative $B$ in this base is a diagonal Matrice.

### Signature and law of inertia of Sylvester

In its most general form, the Loi of inertia of Sylvester affirms, that while working on a Corps ordered $\ mathbb \left\{K\right\}$, the number of diagonal elements null, or strictly positive, or strictly negative, is independent of the selected orthogonal base. These three numbers constitute the signature bilinear form.

### Real case

While working on the body of realities, it is possible to go a little further. That is to say $C= \left(e_ \left\{1\right\}, \ ldots, e_ \left\{N\right\}\right)$ an orthogonal base.

Let us define new bases $C\text{'}= \left(e\text{'}_ \left\{1\right\}, \ ldots, e\text{'}_ \left\{N\right\}\right)$ by

$e\text{'}_ \left\{I\right\} = \ left \ \left\{$

\begin{matrix} e_ {I} & \ mbox {if} B (e_ {I}, e_ {I}) =0 \ \ \ frac {e_ {I}} {\ sqrt {B (e_ {I}, e_ {I})}} & \ mbox {if} B (e_ {I}, e_ {I}) >0 \ \ \ frac {e_ {I}} {\ sqrt {- B (e_ {I}, e_ {I})}} & \ mbox {if} B (e_ {I}, e_ {I}) <0 \end{matrix}\right.

Now, the matrix $A$ representing the symmetrical bilinear form, in this new base, is a diagonal matrix having of the 0 or the 1 only on its diagonal. From the zeros appear on the diagonal if and only if the core is noncommonplace.

### Complex case

While working on the body of the complex numbers, one can establish a result similar to that of the real case.

That is to say $C= \left(e_ \left\{1\right\}, \ ldots, e_ \left\{N\right\}\right)$ an orthogonal base.

For all $i \ in \ \left\{1, \ ldots, N \\right\}$ such as $B \left(e_ \left\{I\right\}, e_ \left\{I\right\}\right) \ neq 0$, let us note $r_i$ one of the square roots of $B \left(e_ \left\{I\right\}, e_ \left\{I\right\}\right)$.

Let us define a new base $C\text{'}= \left(e\text{'}_ \left\{1\right\}, \ ldots, e\text{'}_ \left\{N\right\}\right)$ by

$e\text{'}_ \left\{I\right\} = \ left \ \left\{$

\begin{matrix} e_ {I} & \ mbox {if} \; B (e_ {I}, e_ {I}) =0 \ \ e_ {I} /r_i & \ mbox {if} \; B (e_ {I}, e_ {I}) \ neq 0 \ \ \end{matrix}\right.

Now, the matrix $A$ in the new base is a diagonal matrix having only of the 0 or 1 on the diagonal. From the zeros appear if and only if the core is noncommonplace.

## References

• Law of inertia of Sylvester