# Symmetric bilinear form

Symmetric bilinear forms on finite-dimensional vector spaces precisely correspond to symmetric matrices given a basis for *V*. Among bilinear forms, the symmetric ones are important because they are the ones for which the vector space admits a particularly simple kind of basis known as an orthogonal basis (at least when the characteristic of the field is not 2).

Given a symmetric bilinear form *B*, the function *q*(*x*) = *B*(*x*, *x*) is the associated quadratic form on the vector space. Moreover, if the characteristic of the field is not 2, *B* is the unique symmetric bilinear form associated with *q*.

The last two axioms only establish linearity in the first argument, but the first axiom (symmetry) then immediately implies linearity in the second argument as well.

Let *V* = **R**^{n}, the *n* dimensional real vector space. Then the standard dot product is a symmetric bilinear form, *B*(*x*, *y*) = *x* ⋅ *y*. The matrix corresponding to this bilinear form (see below) on a standard basis is the identity matrix.

Let *V* be any vector space (including possibly infinite-dimensional), and assume *T* is a linear function from *V* to the field. Then the function defined by *B*(*x*, *y*) = *T*(*x*)*T*(*y*) is a symmetric bilinear form.

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

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*).

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.

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

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.

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.

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:

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.