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. 1.

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

2. 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

 $\displaystyle B_{L}:U\to L(V,W),\qquad B_{L}(x)(y)=B(x,y),\;x\in U,\;y\in V;$ $\displaystyle B_{R}:V\to L(U,W),\qquad B_{R}(y)(x)=B(x,y),\;x\in U,\;y\in V.$

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 $\operatorname{rank}B_{L}=\operatorname{rank}B_{R}$. We call this integer $\operatorname{rank}B$, the of $B$. Applying the rank-nullity theorem to both the left and right maps gives the following results:

 $\displaystyle\dim U$ $\displaystyle=\dim\ker{B_{L}}+\operatorname{rank}B$ $\displaystyle\dim V$ $\displaystyle=\dim\ker{B_{R}}+\operatorname{rank}B$

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 $\dim U=\dim V$. If this holds, then $B$ is non-degenerate if and only if $\operatorname{rank}B$ is equal to $\dim U,\dim V$.

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 ${{}^{\perp}}S,S^{\perp}\subset V$ defined as follows:

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

We may also realize $S^{\perp}$ 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, $S^{\perp}=\ker B^{\prime}_{R}$. An analogous statement can be made for ${}^{\perp}S$.

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

 $\displaystyle\dim V$ $\displaystyle=\dim S+\dim S^{\perp}$ $\displaystyle=\dim S+\dim{}^{\perp}S.$

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

 $\dim S^{\perp}=\dim{}^{\perp}S.$

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

 ${}^{\perp}S=S^{\perp}.$

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 $S\cap S^{\perp}=\{0\}$, and also $S\cap{}^{\perp}S=\{0\}$. Hence, if $B$ is non-degenerate and if $S$ is a non-degenerate subspace, we have

 $V=S\oplus S^{\perp}=S\oplus{}^{\perp}S.$

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

 $V=S\oplus S^{\perp}$

for every subspace $S\subset V$.

Let $B:V\times V\rightarrow K$ 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

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

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

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

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.

1. 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. 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. 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. 4.

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

 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