Non-degenerate bilinear forms Every bilinear form on defines a pair of linear maps from to its
dual space . Define by This is often denoted as where 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 and :B(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^{+} = \tfrac{1}{2} (B + {}^{\text{t}}B) \qquad B^{-} = \tfrac{1}{2} (B - {}^{\text{t}}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^{\perp} = \left\{\mathbf{v} \mid B(\mathbf{v}, \mathbf{w}) = 0 \text{ for all } \mathbf{w} \in W\right\} . 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 ( \mathbf{u} , \mathbf{v}) \le C \left\| \mathbf{u} \right\| \left\|\mathbf{v} \right\| .
Definition: A bilinear form on a normed vector space is
elliptic, or
coercive, if there is a constant such that for all , B ( \mathbf{u} , \mathbf{u}) \ge c \left\| \mathbf{u} \right\| ^2 . ==Associated quadratic form==