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 {K}

. A application

B: V \ times V \ rightarrow \ mathbb {K}: (U, v) \ mapsto B (U, v)

is a symmetrical bilinear form on space if:

$\ forall (U, v) \ in V^2 \ quad B (U, v) =B (v, U) \$
$\ forall (U, v, W) \ in V^3 \ quad B (u+v, W) =B (U, W) +B (v, W) \$
$\ forall \ lambda \ in K, \ forall (v, W) \ in V^2 \ quad B (\ lambda v, W) = \ lambda B (v, W) \$

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 $C= (e_ {1}, \ ldots, e_ {N})$ a base of a vector space $V$ . Let us define the matrix order $n$ $A$ by $a_ {ij} =B (e_ {I}, e_ {J})$ . The matrix $A$ is symmetrical according to the symmetry of the bilinear form. If the matrix of the type $(N, 1)$ $x$ 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 $B (v, W)$ is equal to:

${} ^t X has y= {} ^t has x.$

there

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

$A' = {} ^t S.A.S.$

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 $B (v, W) =0$ , which, thanks to reflexivity, is equivalent to $B (W, v) =0$ .

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

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

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 $W^ {\ perp}$ , 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 $W^ {\ perp} =n- \ dim (W)$ .

Orthogonal bases

A base

C= (e_ {1}, \ ldots, e_ {N})

is othogonale for

B

if:

$\ forall I \ neq J, B (e_ {I}, e_ {J}) =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 {K}

, 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= (e_ {1}, \ ldots, e_ {N})

an orthogonal base.

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

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

\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 $C= (e_ {1}, \ ldots, e_ {N})$ an orthogonal base.

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

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

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

\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.

See too

bilinear Form
square Form

References

Law of inertia of Sylvester
traditional quadratic Forms and groups of Rene Deheuvels. Editions puf.

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