bilinear form (original) (raw)

Definition.

Let U,V,W be vector spaces over a field K. A _bilinear map_is a function B:U×V→W such that

1. 1. the map x↦B⁢(x,y) from U to W is linear for each y∈V
2. 1. the map y↦B⁢(x,y) from V to W is linear for each x∈U.

That is, B is bilinear if it is linear in each parameter, taken separately.

Bilinear forms.

A bilinear form is a bilinear map B:V×V→K. AW-valued bilinear form is a bilinear map B:V×V→W. One often encounters bilinear forms with additional assumptions. A bilinear form is called

• •
• •
• •

By expanding B⁢(x+y,x+y)=0, we can show alternating implies skew-symmetric. Further if K is not of characteristic 2, then skew-symmetric implies alternating.

Left and Right Maps.

Let B:U×V→W be a bilinear map. We may identify B with thelinear map B⊗:U⊗V→W (see tensor product). We may also identify B with the linear maps

called the left and right map, respectively.

Next, suppose that B:V×V→K is a bilinear form. Then bothBL and BR are linear maps from V to V*, the dual vector space of V. We can therefore say that B is symmetric if and only if BL=BR and that B is anti-symmetric if and only ifBL=-BR. If V is finite-dimensional, we can identify V andV**, and assert that BL=(BR)*; the left and right maps are, in fact, dual homomorphisms.

Rank.

Let B:U×V→K be a bilinear form, and suppose that U,V are finite dimensional. One can show that rank⁡BL=rank⁡BR. We call this integer rank⁡B, the of B. Applying the rank-nullity theorem to both the left and right maps gives the following results:

We say that B is_non-degenerate_ if both the left and right map are non-degenerate. Note that in for B to be non-degenerate it is necessary that dim⁡U=dim⁡V. If this holds, then B is non-degenerate if and only if rank⁡B is equal to dim⁡U,dim⁡V.

Orthogonal complements.

Next, suppose that B is non-degenerate. By the rank-nullity theorem we have that

Therefore, if B is non-degenerate, then

Indeed, more can be said if B is either symmetric or skew-symmetric. In this case, we actually have

We say that S⊂V is a non-degenerate subspace relative to Bif the restriction of B to S×S is non-degenerate. Thus, Sis a non-degenerate subspace if and only if S∩S⟂={0}, and also S∩S⟂={0}. Hence, if B is non-degenerate and if S is a non-degenerate subspace, we have

Finally, note that ifB is positive-definite, then B is necessarily non-degenerate and that every subspace is non-degenerate. In this way we arrive at the following well-known result: if V is positive-definite inner product space, then

for every subspace S⊂V.

Adjoints.

Let B:V×V→K be a non-degenerate bilinear form, and let T∈L⁢(V,V) be a linear endomorphism. We define theright adjoint T⋆∈L⁢(V,V) to be the unique linear map such that

Letting T∗:V∗→V∗ denote the dual homomorphism, we also have

Similarly, we define the left adjoint T⋆∈L⁢(V,V) by

We then have

If B is either symmetric or skew-symmetric, then T⋆=T⋆, and we simply use T⋆ to refer to the adjoint homomorphism.

Additional remarks.