bilinear form

Definition.

Let $U,V,W$ be vector spaces over a field $K$. A bilinear mapis 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 $B:V\times V\to W$. Oneoften encounters bilinear forms with additional assumptions. Abilinear 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 impliesskew-symmetric. Further if $K$ is not of characteristic $2$, thenskew-symmetric implies alternating.

Left and Right Maps.

Let $B:U\times V\to W$ be a bilinear map. We may identify $B$ with thelinear map $B_{\otimes }:U\otimes V\to W$ (see tensor product). We mayalso 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 vectorspace of $V$. We can therefore say that $B$ is symmetric if and onlyif $B_L=B_R$ 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, infact, dual homomorphisms.

Rank.

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

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

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

We say that $B$ isnon-degenerate if both the left and right map arenon-degenerate. Note that in for $B$ to be non-degenerate it isnecessary that $dimU=dimV$. If this holds, then $B$ isnon-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 asubspace. The left and right orthogonal complements of $S$ aresubspaces ${}^{⟂}S,S^{⟂}\subset V$ 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 analogousstatement can be made for ${}^{⟂}S$.

Next, suppose that $B$ is non-degenerate.By the rank-nullitytheorem 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 orskew-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-degenerateand if $S$ is a non-degeneratesubspace, we have

$$V=S\oplus S^{⟂}=S\oplus {}^{⟂}S.$$

Finally, note that if$B$ is positive-definite, then $B$ is necessarily non-degenerate andthat every subspace is non-degenerate. In this way we arrive at thefollowing well-known result: if $V$ is positive-definite inner productspace, then

$$V=S\oplus S^{⟂}$$

for every subspace $S\subset V$.

Adjoints.

Let $B:V\times V\to K$ be a non-degenerate bilinearform, and let $T\in L(V,V)$ be a linear endomorphism. We define theright 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 orthogonalbasis (if one exists), by the matrix transpose.

2. If $B$ is a symmetric, non-degenerate bilinear formthen $T\in L(V,V)$ is then said to be a normal operator (withrespect 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 becongruent 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 standardEuclidean inner product on ${ℝ}^n$.