bilinear form

Definition.

Let $U,V,W$ be vector spaces over a field $K$. A bilinear map is a function $B:U\times V\to W$ such that

1. the map $x\mapsto B(x,y)$ from $U$ to $W$ is linear for each $y\in V$

2. the map $y\mapsto B(x,y)$ from $V$ to $W$ is linear for each $x\in U$.

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

Bilinear forms.

A bilinear form is a bilinear map $B:V\times V\to K$. A $W$-valued bilinear form is a bilinear map $B:V\times V\to W$. One often encounters bilinear forms with additional assumptions. A bilinear form is called

• symmetric if $B(x,y)=B(y,x)$, $x,y\in V$;

• skew-symmetric if $B(x,y)=-B(y,x)$, $x,y\in V$;

• alternating if $B(x,x)=0$, $x\in V$.

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 $B:U\times V\to W$ be a bilinear map. We may identify $B$ with the linear map $B_{\otimes }:U\otimes V\to W$ (see tensor product). We may also identify $B$ with the linear maps

$B_L:U\to L(V,W),B_L(x)(y)=B(x,y),x\in U,y\in V;$

$B_R:V\to L(U,W),B_R(y)(x)=B(x,y),x\in U,y\in V.$

called the left and right map, respectively.

Next, suppose that $B:V\times V\to K$ is a bilinear form. Then both $B_L$ and $B_R$ 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 $B_L=B_R$ and that $B$ is anti-symmetric if and only if $B_L=-B_R$. If $V$ is finite-dimensional, we can identify $V$ and $V^{**}$, and assert that $B_L={(B_R)}^{*}$; the left and right maps are, in fact, dual homomorphisms.

Rank.

Let $B:U\times V\to K$ be a bilinear form, and suppose that $U,V$ are finite dimensional. One can show that $\mathrm{rank}{B}_L=\mathrm{rank}{B}_R$. We call this integer $\mathrm{rank}B$, the of $B$. Applying the rank-nullity theorem to both the left and right maps gives the following results:

$dimU$ $=dim\ker B_L+\mathrm{rank}B$

$dimV$ $=dim\ker B_R+\mathrm{rank}B$

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 $dimU=dimV$. If this holds, then $B$ is non-degenerate if and only if $\mathrm{rank}B$ is equal to $dimU,dimV$.

Orthogonal complements.

Let $B:V\times V\to K$ be a bilinear form, and let $S\subset V$ be a subspace. The left and right orthogonal complements of $S$ are subspaces ${}^{⟂}S,S^{⟂}\subset V$ defined as follows:

${}^{⟂}S=\{u\in V\mid B(u,v)=0\text{for all }v\in S\},$

$S^{⟂}=\{v\in V\mid B(u,v)=0\text{for all }u\in S\}.$

We may also realize $S^{⟂}$ by considering the linear map $B_R^{\prime }:V\to S^{*}$ obtained as the composition of $B_R:V\to V^{*}$ and the dual homomorphism $V^{*}\to S^{*}$. Indeed, $S^{⟂}=\ker B_R^{\prime }$. An analogous statement can be made for ${}^{⟂}S$.

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

$dimV$ $=dimS+dim{S}^{⟂}$

$=dimS+dim{}^{⟂}S.$

Therefore, if $B$ is non-degenerate, then

$$dim{S}^{⟂}=dim{}^{⟂}S.$$

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

$${}^{⟂}S=S^{⟂}.$$

We say that $S\subset V$ is a non-degenerate subspace relative to $B$ if the restriction of $B$ to $S\times S$ is non-degenerate. Thus, $S$ is a non-degenerate subspace if and only if $S\cap S^{⟂}=\{0\}$, and also $S\cap {}^{⟂}S=\{0\}$. Hence, if $B$ is non-degenerate and if $S$ is a non-degenerate subspace, we have

$$V=S\oplus S^{⟂}=S\oplus {}^{⟂}S.$$

Finally, note that if $B$ 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

$$V=S\oplus S^{⟂}$$

for every subspace $S\subset V$.

Adjoints.

Let $B:V\times V\to K$ be a non-degenerate bilinear form, and let $T\in L(V,V)$ be a linear endomorphism. We define the right adjoint $T^{\star }\in L(V,V)$ to be the unique linear map such that

$$B(Tu,v)=B(u,T^{\star }v),u,v\in V.$$

Letting $T^{\ast }:V^{\ast }\to V^{\ast }$ denote the dual homomorphism, we also have

$$T^{\star }=B_R{}^{-1}\circ T^{\ast }\circ B_R.$$

Similarly, we define the left adjoint ${}^{\star }T\in L(V,V)$ by

$${}^{\star }T=B_L{}^{-1}\circ T^{\ast }\circ B_L.$$

We then have

$$B(u,Tv)=B({}^{\star }Tu,v),u,v\in V.$$

If $B$ is either symmetric or skew-symmetric, then ${}^{\star }T=T^{\star }$, and we simply use $T^{\star }$ to refer to the adjoint homomorphism.

Additional remarks.

1. if $B$ is a symmetric, non-degenerate bilinear form, then the adjoint operation is represented, relative to an orthogonal basis (if one exists), by the matrix transpose.

2. If $B$ is a symmetric, non-degenerate bilinear form then $T\in L(V,V)$ is then said to be a normal operator (with respect to $B$) if $T$ commutes with its adjoint $T^{\star }$.

3. An $n\times m$ matrix may be regarded as a bilinear form over $K^n\times K^m$. Two such matrices, $B$ and $C$, are said to be congruent if there exists an invertible $P$ such that $B=P^{T}CP$.

4. The identity matrix, $I_n$ on ${ℝ}^n\times {ℝ}^n$ gives the standard Euclidean inner product on ${ℝ}^n$.

Title	bilinear form
Canonical name	BilinearForm
Date of creation	2013-03-22 12:14:02
Last modified on	2013-03-22 12:14:02
Owner	rmilson (146)
Last modified by	rmilson (146)
Numerical id	54
Author	rmilson (146)
Entry type	Definition
Classification	msc 47A07
Classification	msc 11E39
Classification	msc 15A63
Synonym	bilinear
Related topic	DualityWithRespectToANonDegenerateBilinearForm
Related topic	BilinearMap
Related topic	Multilinear
Related topic	SkewSymmetricBilinearForm
Related topic	SymmetricBilinearForm
Related topic	NonDegenerateBilinearForm
Defines	rank of bilinear form
Defines	left map
Defines	right map