Bilinear form

In mathematics, a bilinear form on a vector space "V" is a bilinear mapping "V" × "V" → "F", where "F" is the field of scalars. That is, a bilinear form is a function "B": "V" × "V" → "F" which is linear in each argument separately:

:

Any bilinear form on "F"^"n" can be expressed as

:$B(\; extbf\{x\},\; extbf\{y\})\; =\; extbf\{x\}^\{mathrm\{TA\; extbf\{y\}\; =\; sum\_\{i,j=1\}^n\; a\_\{ij\}\; x\_i\; y\_j$

where "A" is an "n" × "n" matrix.

The definition of a bilinear form can easily be extended to include modules over a commutative ring, with linear maps replaced by module homomorphisms. When "F" is the field of complex numbers C, one is often more interested in sesquilinear forms, which are similar to bilinear forms but are conjugate linear in one argument.

Coordinate representation

Let $C=\{e\_\{1\},ldots,e\_\{n\}\}$ be a basis for a finite-dimensional space "V". Define the $n\; imes\; n$ - matrix "A" by $(A\_\{ij\})=B(e\_\{i\},e\_\{j\})$. Then if the $n\; imes\; 1$ matrix x represents a vector "v" with respect to this basis, and analogously, "y" represents "w", then:

:$B(v,w)\; =\; x^\{T\}\; A\; y,$

Suppose " C' " is another basis for "V", with : with "S" an invertible $n\; imes\; n$ - matrix.Now the new matrix representation for the symmetric bilinear form is given by :

$A\text{'}\; =S^\{T\}\; A\; S$

Maps to the dual space

Every bilinear form "B" on "V" defines a pair of linear maps from "V" to its dual space "V"*. Define $B\_1,B\_2colon\; V\; o\; V^*$ by:$B\_1(v)(w)\; =\; B(v,w)$:$B\_2(v)(w)\; =\; B(w,v)$This is often denoted as:$B\_1(v)\; =\; B(v,\{cdot\})$:$B\_2(v)\; =\; B(\{cdot\},v)$where the ($cdot$) indicates the slot into which the argument for the resulting linear functional is to be placed.

If either of "B"₁ or "B"₂ is an isomorphism, then both are, and the bilinear form "B" is said to be nondegenerate.

If "V" is finite-dimensional then one can identify "V" with its double dual "V"**. One can then show that "B"₂ is the transpose of the linear map "B"₁ (if "V" is infinite-dimensional then "B"₂ is the transpose of "B"¹ restricted to the image of "V" in "V"**). Given "B" one can define the "transpose" of "B" to be the bilinear form given by:$B^*(v,w)\; =\; B(w,v).$

If "V" is finite-dimensional then the rank of "B"₁ is equal to the rank of "B"₂. If this number is equal to the dimension of "V" then "B"₁ and "B"₂ are linear isomorphisms from "V" to "V"*. In this case "B" is nondegenerate. By the rank-nullity theorem, this is equivalent to the condition that the kernel of "B"₁ be trivial. In fact, for finite dimensional spaces, this is often taken as the "definition" of nondegeneracy. Thus "B" is nondegenerate if and only if:$B(v,w)=0mbox\{\; for\; all\; w\}Rightarrow\; v=0.$

Given any linear map "A" : "V" → "V"* one can obtain a bilinear form "B" on "V" via:$B(v,w)\; =\; A(v)(w)$This form will be nondegenerate if and only if "A" is an isomorphism.

If "V" is finite-dimensional then, relative to some basis for "V", 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 associated matrix is non-singular. These statements are independent of the chosen basis.

Reflexivity and orthogonality

A bilinear form

:"B" : "V" × "V" → "F"

is "reflexive" if

:$B(v,w)=0Longleftrightarrow\; B(w,v)=0.$

Reflexivity allows us to define orthogonality: two vectors "v" and "w" are "orthogonal" with respect to the reflexive bilinear form if and only if :

:$B(v,w)=0$ or $B(w,v)=0$

The radical of a bilinear form is the subset of all vectors orthogonal with every other vector. A vector "v", with matrix representation "x", is in the radical of a bilinear form with matrix representation "A", if and only if :$A\; x=\; 0\; Longleftrightarrow\; x^\{T\}\; A=0$ The radical is always a subspace of "V". It is trivial if and only if the matrix "A" is nonsingular, and thus if and only if the bilinear form is nondegenerate.

Suppose "W" is a subspace. Define :$W^\{perp\}=\{v|\; B(v,w)=0\; forall\; win\; W\}$

When the bilinear form is nondegenerate, the map $Wleftarrow\; W^\{perp\}$ is bijective, and the dimension of $W^\{perp\}$ is dim("V")-dim("W").

One can prove that "B" is reflexive if and only if it is "either":
*symmetric : $B(v,w)=B(w,v)$ for all $v,win\; V$; or
*alternating if $B(v,v)=0$ for all $vin\; V$

Every alternating form is skew-symmetric ($B(v,w)=-B(w,v)$). This may be seen by expanding "B"("v"+"w","v"+"w").

If the characteristic of "F" is not 2 then the converse is also true (every skew-symmetric form is alternating). If, however, char("F") = 2 then a skew-symmetric form is the same thing as a symmetric form and not all of these are alternating.

A bilinear form is symmetric (resp. skew-symmetric) if and only if its coordinate matrix (relative to any basis) is symmetric (resp. 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 char("F") ≠ 2).

A bilinear form is symmetric if and only if the maps $B\_1,B\_2colon\; V\; o\; V^*$ are equal, and skew-symmetric if and only if they are negatives of one another. If char("F") ≠ 2 then one can always decompose a bilinear form into a symmetric and a skew-symmetric part as follows:$B^\{pm\}\; =\; frac\{1\}\{2\}\; (B\; pm\; B^*)$where "B"* is the transpose of "B" (defined above).

Also if char("F") ≠ 2 then one can define a quadratic form in terms of its associated symmetric form. One can likewise define quadratic forms corresponding to skew-symmetric forms, Hermitian forms, and skew-Hermitian forms; the general concept is ε-quadratic form.

Different spaces

Much of the theory is available for a bilinear mapping

:"B": "V" × "W" → "F".

In this situation we still have linear mappings of "V" to the dual space of "W", and of "W" to the dual space of "V". It may happen that both of those 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), nondegenerate is a weaker notion: a pairing can be nondegenerate without being a perfect pairing, for instance $mathbf\{Z\}\; imes\; mathbf\{Z\}\; o\; mathbf\{Z\}$ via $(x,y)\; mapsto\; 2xy$ is non-degenerate, but induces multiplication by 2 on the map $mathbf\{Z\}\; o\; mathbf\{Z\}^*$

Relation to tensor products

By the universal property of the tensor product, bilinear forms on "V" are in 1-to-1 correspondence with linear maps "V" ⊗ "V" → "F". If "B" is a bilinear form on "V" the corresponding linear map is given by:$votimes\; wmapsto\; B(v,w).$The set of all linear maps "V" ⊗ "V" → "F" is the dual space of "V" ⊗ "V", so bilinear forms may be thought of as elements of:$(Votimes\; V)^\{*\}\; cong\; V^\{*\}otimes\; V^\{*\}.$Likewise, symmetric bilinear forms may be thought of as elements of "S"²"V"* (the second symmetric power of "V"*), and alternating bilinear forms as elements of Λ²"V"* (the second exterior power of "V"*).

On normed vector spaces

A bilinear form on a normed vector space is bounded, if there is a constant $C$ such that for all $u,\; vin\; V$:$B(u,v)\; le\; C\; |u|\; |v|.$

A bilinear form on a normed vector space is elliptic, or coercive, if there is a non-zero constant $c$ such that for all $uin\; V$:$B(u,u)\; ge\; c\; |u|^2.$

ee also

*bilinear operator
*multilinear form
*quadratic form
*sesquilinear form
*inner product space

References

*citation | last=Shilov | first=Georgi E. | title=Linear Algebra | editor-last=Silverman | editor-first=Richard A. | date=1977 | publisher=Dover | isbn=0-486-63518-X.

External links

*planetmath reference|id=1612|title=Bilinear form