# Symmetric bilinear form

A symmetric bilinear form is a bilinear form on a vector space that is symmetric. More simply, it is a function that maps a pair of elements of the vector space to its underlying field in such a way that the order of the elements into the function does not affect the element of the field to which it maps. Symmetric bilinear forms are of great importance in the study of orthogonal polarity and quadrics.

They are also more briefly referred to as just symmetric forms when "bilinear" is understood. They are closely related to quadratic forms; for the details of the distinction between the two, see ε-quadratic forms.

## Definition

Let V be a vector space of dimension n over a field K. A map ${\displaystyle B:V\times V\rightarrow K:(u,v)\mapsto B(u,v)}$ is a symmetric bilinear form on the space if:

The last two axioms only imply linearity in the first argument, but the first immediately implies linearity in the second argument then too.

## Matrix representation

Let ${\displaystyle C=\{e_{1},\ldots ,e_{n}\}}$ be a basis for V. Define the n×n matrix A by ${\displaystyle A_{ij}=B(e_{i},e_{j})}$. The matrix A is a symmetric matrix exactly due to symmetry of the bilinear form. If the n×1 matrix x represents a vector v with respect to this basis, and analogously, y represents w, then ${\displaystyle B(v,w)}$ is given by :

${\displaystyle x^{\mathsf {T}}Ay=y^{\mathsf {T}}Ax.}$

Suppose C' is another basis for V, with : ${\displaystyle {\begin{bmatrix}e'_{1}&\cdots &e'_{n}\end{bmatrix}}={\begin{bmatrix}e_{1}&\cdots &e_{n}\end{bmatrix}}S}$ with S an invertible n×n matrix. Now the new matrix representation for the symmetric bilinear form is given by

${\displaystyle A'=S^{\mathsf {T}}AS.}$

## Orthogonality and singularity

A symmetric bilinear form is always reflexive. Two vectors v and w are defined to be orthogonal with respect to the bilinear form B if B(v, w) = 0, which is, due to reflexivity, equivalent to B(w, v) = 0.

The radical of a bilinear form B is the set of vectors orthogonal with every vector in V. That this is a subspace of V follows from the linearity of B in each of its arguments. When working with a matrix representation A with respect to a certain basis, v, represented by x, is in the radical if and only if

${\displaystyle Ax=0\Longleftrightarrow x^{\mathsf {T}}A=0.}$

The matrix A is singular if and only if the radical is nontrivial.

If W is a subset of V, then its orthogonal complement W is the set of all vectors in V that are orthogonal to every vector in W; it is a subspace of V. When B is non-degenerate, the radical of B is trivial and the dimension of W is dim(W) = dim(V) − dim(W).

## Orthogonal basis

A basis ${\displaystyle C=\{e_{1},\ldots ,e_{n}\}}$ is orthogonal with respect to B if and only if :

${\displaystyle B(e_{i},e_{j})=0\ \forall i\neq j.}$

When the characteristic of the field is not two, V always has an orthogonal basis. This can be proven by induction.

A basis C is orthogonal if and only if the matrix representation A is a diagonal matrix.

### Signature and Sylvester's law of inertia

In its most general form, Sylvester's law of inertia says that, when working over an ordered field, the numbers of diagonal elements which are positive, zero and negative respectively are independent of the chosen orthogonal basis. These three numbers form the signature of the bilinear form.

### Real case

When working in a space over the reals, one can go a bit a further. Let ${\displaystyle C=\{e_{1},\ldots ,e_{n}\}}$ be an orthogonal basis.

We define a new basis ${\displaystyle C'=\{e'_{1},\ldots ,e'_{n}\}}$

${\displaystyle e'_{i}={\begin{cases}e_{i}&{\text{if }}B(e_{i},e_{i})=0\\{\frac {e_{i}}{\sqrt {B(e_{i},e_{i})}}}&{\text{if }}B(e_{i},e_{i})>0\\{\frac {e_{i}}{\sqrt {-B(e_{i},e_{i})}}}&{\text{if }}B(e_{i},e_{i})<0\end{cases}}}$

Now, the new matrix representation A will be a diagonal matrix with only 0, 1 and −1 on the diagonal. Zeroes will appear if and only if the radical is nontrivial.

### Complex case

When working in a space over the complex numbers, one can go further as well and it is even easier. Let ${\displaystyle C=\{e_{1},\ldots ,e_{n}\}}$ be an orthogonal basis.

We define a new basis ${\displaystyle C'=\{e'_{1},\ldots ,e'_{n}\}}$ :

${\displaystyle e'_{i}={\begin{cases}e_{i}&{\text{if }}\;B(e_{i},e_{i})=0\\e_{i}/{\sqrt {B(e_{i},e_{i})}}&{\text{if }}\;B(e_{i},e_{i})\neq 0\\\end{cases}}}$

Now the new matrix representation A will be a diagonal matrix with only 0 and 1 on the diagonal. Zeroes will appear if and only if the radical is nontrivial.

## Orthogonal polarities

Let B be a symmetric bilinear form with a trivial radical on the space V over the field K with characteristic not 2. One can now define a map from D(V), the set of all subspaces of V, to itself:

${\displaystyle \alpha :D(V)\rightarrow D(V):W\mapsto W^{\perp }.}$

This map is an orthogonal polarity on the projective space PG(W). Conversely, one can prove all orthogonal polarities are induced in this way, and that two symmetric bilinear forms with trivial radical induce the same polarity if and only if they are equal up to scalar multiplication.

## References

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