Bilinear form explained

In mathematics, a bilinear form is a bilinear map on a vector space (the elements of which are called vectors) over a field K (the elements of which are called scalars). In other words, a bilinear form is a function that is linear in each argument separately:

• and
• and

The dot product on

\R ^{n} is an example of a bilinear form.^[1]

The definition of a bilinear form can be extended to include modules over a ring, with linear maps replaced by module homomorphisms.

When is the field of complex numbers, one is often more interested in sesquilinear forms, which are similar to bilinear forms but are conjugate linear in one argument.

Coordinate representation

Let be an -dimensional vector space with basis .

The matrix A, defined by is called the matrix of the bilinear form on the basis .

If the matrix represents a vector with respect to this basis, and similarly, the matrix represents another vector, then:$$B(\backslash mathbf,\; \backslash mathbf)\; =\; \backslash mathbf^\backslash textsf\; A\backslash mathbf\; =\; \backslash sum\_^n\; x\_i\; A\_\; y\_j.$$

A bilinear form has different matrices on different bases. However, the matrices of a bilinear form on different bases are all congruent. More precisely, if is another basis of, then$$\backslash mathbf\_j=\backslash sum\_^n\; S\_\backslash mathbf\_i,$$where the

S _{i,j} form an invertible matrix . Then, the matrix of the bilinear form on the new basis is .

Properties

Non-degenerate bilinear forms

Every bilinear form on defines a pair of linear maps from to its dual space . Define byThis is often denoted aswhere the dot (⋅) indicates the slot into which the argument for the resulting linear functional is to be placed (see Currying).

For a finite-dimensional vector space, if either of or is an isomorphism, then both are, and the bilinear form is said to be nondegenerate. More concretely, for a finite-dimensional vector space, non-degenerate means that every non-zero element pairs non-trivially with some other element:

B(x,y) =0 for all y \in V implies that andB(x,y) =0 for all x \in V implies that .

The corresponding notion for a module over a commutative ring is that a bilinear form is if is an isomorphism. Given a finitely generated module over a commutative ring, the pairing may be injective (hence "nondegenerate" in the above sense) but not unimodular. For example, over the integers, the pairing is nondegenerate but not unimodular, as the induced map from to is multiplication by 2.

If is finite-dimensional then one can identify with its double dual . One can then show that is the transpose of the linear map (if is infinite-dimensional then is the transpose of restricted to the image of in). Given one can define the transpose of to be the bilinear form given by

The left radical and right radical of the form are the kernels of and respectively; they are the vectors orthogonal to the whole space on the left and on the right.

If is finite-dimensional then the rank of is equal to the rank of . If this number is equal to then and are linear isomorphisms from to . In this case is nondegenerate. By the rank–nullity theorem, this is equivalent to the condition that the left and equivalently right radicals be trivial. For finite-dimensional spaces, this is often taken as the definition of nondegeneracy:

Given any linear map one can obtain a bilinear form B on V via

This form will be nondegenerate if and only if is an isomorphism.

If is finite-dimensional then, relative to some basis for, a bilinear form is degenerate if and only if the determinant of the associated matrix is zero. Likewise, a nondegenerate form is one for which the determinant of the associated matrix is non-zero (the matrix is non-singular). These statements are independent of the chosen basis. For a module over a commutative ring, a unimodular form is one for which the determinant of the associate matrix is a unit (for example 1), hence the term; note that a form whose matrix determinant is non-zero but not a unit will be nondegenerate but not unimodular, for example over the integers.

Symmetric, skew-symmetric, and alternating forms

We define a bilinear form to be

• symmetric if for all, in ;
• alternating if for all in ;
• or if for all, in ;

Proposition: Every alternating form is skew-symmetric.

Proof: This can be seen by expanding .

If the characteristic of is not 2 then the converse is also true: every skew-symmetric form is alternating. However, if then a skew-symmetric form is the same as a symmetric form and there exist symmetric/skew-symmetric forms that are not alternating.

A bilinear form is symmetric (respectively skew-symmetric) if and only if its coordinate matrix (relative to any basis) is symmetric (respectively skew-symmetric). A bilinear form is alternating if and only if its coordinate matrix is skew-symmetric and the diagonal entries are all zero (which follows from skew-symmetry when).

A bilinear form is symmetric if and only if the maps are equal, and skew-symmetric if and only if they are negatives of one another. If then one can decompose a bilinear form into a symmetric and a skew-symmetric part as follows$$B^\; =\; \backslash tfrac\; (B\; +\; ^B)\; \backslash qquad\; B^\; =\; \backslash tfrac\; (B\; -\; ^B),$$where is the transpose of (defined above).

Reflexive bilinear forms and orthogonal vectors

A bilinear form is reflexive if and only if it is either symmetric or alternating. In the absence of reflexivity we have to distinguish left and right orthogonality. In a reflexive space the left and right radicals agree and are termed the kernel or the radical of the bilinear form: the subspace of all vectors orthogonal with every other vector. A vector, with matrix representation, is in the radical of a bilinear form with matrix representation, if and only if . The radical is always a subspace of . It is trivial if and only if the matrix is nonsingular, and thus if and only if the bilinear form is nondegenerate.

Suppose is a subspace. Define the orthogonal complement$$W^\; =\; \backslash left\backslash \; .$$

For a non-degenerate form on a finite-dimensional space, the map is bijective, and the dimension of is .

Bounded and elliptic bilinear forms

Definition: A bilinear form on a normed vector space is bounded, if there is a constant such that for all,$$B\; (\backslash mathbf,\; \backslash mathbf)\; \backslash le\; C\; \backslash left\backslash |\; \backslash mathbf\; \backslash right\backslash |\; \backslash left\backslash |\backslash mathbf\; \backslash right\backslash |\; .$$

Definition: A bilinear form on a normed vector space is elliptic, or coercive, if there is a constant such that for all,$$B\; (\backslash mathbf,\; \backslash mathbf)\; \backslash ge\; c\; \backslash left\backslash |\; \backslash mathbf\; \backslash right\backslash |\; ^2\; .$$

Associated quadratic form

For any bilinear form, there exists an associated quadratic form defined by .

When, the quadratic form Q is determined by the symmetric part of the bilinear form B and is independent of the antisymmetric part. In this case there is a one-to-one correspondence between the symmetric part of the bilinear form and the quadratic form, and it makes sense to speak of the symmetric bilinear form associated with a quadratic form.

When and, this correspondence between quadratic forms and symmetric bilinear forms breaks down.

Relation to tensor products

By the universal property of the tensor product, there is a canonical correspondence between bilinear forms on and linear maps . If is a bilinear form on the corresponding linear map is given byIn the other direction, if is a linear map the corresponding bilinear form is given by composing F with the bilinear map that sends to .

The set of all linear maps is the dual space of, so bilinear forms may be thought of as elements of which (when is finite-dimensional) is canonically isomorphic to .

Likewise, symmetric bilinear forms may be thought of as elements of (dual of the second symmetric power of) and alternating bilinear forms as elements of (the second exterior power of). If, .

Generalizations

Pairs of distinct vector spaces

Much of the theory is available for a bilinear mapping from two vector spaces over the same base field to that field

Here we still have induced linear mappings from to, and from to . It may happen that these mappings are isomorphisms; assuming finite dimensions, if one is an isomorphism, the other must be. When this occurs, B is said to be a perfect pairing.

In finite dimensions, this is equivalent to the pairing being nondegenerate (the spaces necessarily having the same dimensions). For modules (instead of vector spaces), just as how a nondegenerate form is weaker than a unimodular form, a nondegenerate pairing is a weaker notion than a perfect pairing. A pairing can be nondegenerate without being a perfect pairing, for instance via is nondegenerate, but induces multiplication by 2 on the map .

Terminology varies in coverage of bilinear forms. For example, F. Reese Harvey discusses "eight types of inner product". To define them he uses diagonal matrices A_ij having only +1 or −1 for non-zero elements. Some of the "inner products" are symplectic forms and some are sesquilinear forms or Hermitian forms. Rather than a general field, the instances with real numbers, complex numbers, and quaternions are spelled out. The bilinear form$$\backslash sum\_^p\; x\_k\; y\_k\; -\; \backslash sum\_^n\; x\_k\; y\_k$$is called the real symmetric case and labeled, where . Then he articulates the connection to traditional terminology:

General modules

Given a ring and a right -module and its dual module, a mapping is called a bilinear form iffor all, all and all .

The mapping is known as the natural pairing, also called the canonical bilinear form on .

A linear map induces the bilinear form, and a linear map induces the bilinear form .

Conversely, a bilinear form induces the R-linear maps and . Here, denotes the double dual of .

References

• • • • • • • . Also:

Notes and References

1. Web site: 2021-01-16. Chapter 3. Bilinear forms — Lecture notes for MA1212.