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
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 $\rank B_L = \rank B_R$ . We call this integer $\rank B$ , the rank 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 order 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.

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:
We may also realize $S^\perp$ by considering the linear map $B_{R}': V\to S^*$ 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

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 $S\subset V$ .

Adjoints.

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

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 $\Rset^n\times \Rset^n$ gives the standard Euclidean inner product on $\Rset^n$ .