# 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

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.

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 $T^{\star}$.
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 $B=P^{T}CP$.