Bilinear Forms

Bilinear forms provide a different and more general way of looking at the relationship between a vector space and its dual.

Definition. A map $\phi: V \times W \to \mathbb{F}$ is a bilinear form if it is linear in both arguments, i.e. if $\phi(v, -): W \to \mathbb{F} \in W^\ast$ for all $v \in V$ and $\phi(-, w): V \to \mathbb{F} \in V^\ast$ for all $w \in W$.

Definition. A bilinear form $\phi: V \times W \to \mathbb{F}$ induces linear maps $\phi_L: V \to W^\ast$ and $\phi_R: W \to V^\ast$ defined by

\[\phi_L(v)(w) = \phi(v, w) = \phi_R(w)(v)\]

for $v \in V$ and $w \in W$.

Base on the above, we can see that $\phi: V \times V^\ast \to \mathbb{F}, \phi(v, \theta) = \theta(v)$ is a bilinear form with $\phi_L: V \to V^{\ast\ast} = \text{ev}$ and $\phi_R: V^\ast \to V^\ast = \iota_{V^\ast}$.

Proposition. $\phi_R = \phi_L^\ast \circ \text{ev}$ and $\phi_L = \phi_R^\ast \circ \text{ev}$.

Proof.

$\phi_L^\ast: W^{\ast\ast} \to V^\ast$ with $\phi_L^\ast(\theta)(v) = \theta(\phi_L(v)) \in \mathbb{F}$ for $\theta \in W^{\ast\ast} = \mathcal{L}(W^\ast, \mathbb{F})$ and $\text{ev}: W \to W^{\ast\ast}$ so $\phi_L^\ast \circ \text{ev}: W \to V^\ast$ and the domain and codomain matches that of $\phi_R$. Hence,

\[(\phi_L^\ast \circ \text{ev}(w))(v) = \text{ev}(w)(\phi_L(v)) = \phi_L(v)(w) = \phi_R(w)(v)\]

Definition. $\ker \phi_L$ is the left kernel of $\phi$ and $\ker \phi_R$ is the right kernel of $\phi$, i.e.

\[\begin{align*} \ker \phi_L &= \Set{v \in V : \forall w \in W, \phi(v, w) = 0} \\ \ker \phi_R &= \Set{w \in W : \forall v \in V, \phi(v, w) = 0} \end{align*}\]

By considering the subsets $S \subset V$ and $T \subset W$, it is a more general idea of annihilator.

Definition. Suppose that $V$ and $W$ are vector spaces with subspaces $S$ and $T$ respectively and $\phi: V \times W \to \mathbb{F}$ is a bilinear form. Then we define

\[\begin{align*} S^\perp &= \Set{w \in W : \forall s \in S, \phi(s, w) = 0} \\ {}^\perp T &= \Set{v \in V : \forall t \in T, \phi(v, t) = 0} \end{align*}\]

Non-degeneracy

We can now explore how bilinear form relates to dual spaces.

Definition. A bilinear form $\phi: V \times W \to \mathbb{F}$ is non-degenerate if $\ker \phi_L = 0$ and $\ker \phi_R = 0$. Otherwise $\phi$ is degenerate.

Proposition. Suppose that $V$ and $W$ are finite dimensional vector spaces and $\phi: V \times W \to \mathbb{F}$ is a non-degenerate bilinear form. Then both $\phi_L: V \to W^\ast$ and $\phi_R: W \to V^\ast$ are isomorphisms and $\dim V = \dim W$.

Proof.

$\ker \phi_L = 0$ implies $\phi_L$ is injective and $\ker \phi_R = 0$ implies $\phi_R$ is injective. Therefore, $\dim W^\ast \ge \dim V$ and $\dim V^\ast \ge \dim W$ but $\dim V = \dim V^\ast$ so we have $\dim V = \dim W$. Hence, $\phi_L$ and $\phi_R$ are isomorphisms.

Base on that, the following definition of dual spaces is rather natural.

Definition. A pair of finite dimensional vector spaces $V$ and $W$ are dual with respect to a bilinear form $\phi: V \times W \to \mathbb{F}$ if $\phi$ is non-degenerate, i.e. each vector space is isomorphic to the dual space of the other.

Matrix Representation

Definition. Let $(e_1, …, e_m)$ be a basis for $V$ and $(f_1, …, f_n)$ be a basis for $W$ and $\phi: V \times W \to \mathbb{F}$ a bilinear form. Then the $m \times n$ matrix $A$ representing $\phi$ with respect to $(e_1, …, e_m)$ and $(f_1, …, f_n)$ is given by $A_{ij} = \phi(e_i, f_j)$.

Consider vectors $v = \sum \lambda_i e_i$ and $w = \sum \mu_j f_j$ then

\[\phi(v, w) = \phi \left( \sum_i \lambda_i e_i, \sum_j \mu_j f_j \right) = \sum_i \sum_j \lambda_i \mu_j \phi(e_i, f_j)\]

Therefore, if $A$ is the matrix representing $\phi$ then

\[\phi(v, w) = \begin{pmatrix} \lambda_1 & \cdots & \lambda_m \end{pmatrix} A \begin{pmatrix} \mu_1 \\ \vdots \\ \mu_n \end{pmatrix}\]

Proposition. Let $(e_1, …, e_m)$ be a basis for $V$ with dual basis $(\epsilon_1, …, \epsilon_m)$ and $(f_1, …, f_n)$ be a basis for $W$ with dual basis $(\eta_1, …, \eta_n)$. If $A$ represents the bilinear form $\phi: V \times W \to \mathbb{F}$ with respect to $(e_i)$ and $(f_i)$ then $A$ represents $\phi_R$ with respect to $(f_i)$ and $(\epsilon_i)$ and $A^\intercal$ represents $\phi_L$ with respect to $(e_i)$ and $(\eta_i)$.

Proof.

Suppose that $B$ is matrix representing $\phi_R$ and $C$ be that representing $\phi_L$. Then

\[A_{ij} = \phi(e_i, f_j) = \phi_R(f_j)(e_i) = \left( \sum_k B_{kj} \epsilon_k \right) (e_i) = \sum_k B_{kj} \delta_{ik} = B_{ij}\]

and

\[A_{ij} = \phi(e_i, f_j) = \phi_L(e_i)(f_j) = \left( \sum_k C_{ki} \eta_k \right) (f_j) = \sum_k C_{ki} \delta_{kj} = C_{ji} = C_{ij}^\intercal\]

Proposition. A bilinear form $\phi$ represented by matrix $A$ is non-degenerate iff $A$ is invertible.

Proof.

$\phi$ is non-degenerate iff $\ker \phi_L = 0$ and $\ker \phi_R = 0$ iff $A$ is invertible.

Proposition. Suppose that $(e_1, …, e_m)$ and $(e_1’, …, e_m’)$ are two bases of $V$ such that $e_i’ = \sum P_{ki} e_k$ and $(f_1, …, f_n)$ and $(f_1’, …, f_n’)$ are two bases of $W$ such that $f_j’ = \sum Q_{lj} f_l$ and $\phi: V \times W \to \mathbb{F}$ is a bilinear form represented by $A$ with respect to $(e_i)$ and $(f_i)$ and by $B$ with respect to $(e_i’)$ and $(f_i’)$. Then

\[B = P^\intercal A Q\]

Proof.

\[\begin{align*} B_{ij} &= \phi(e_i', f_j') \\ &= \phi \left( \sum_k P_{ki} e_k, \sum_l Q_{lj} f_l \right) \\ &= \sum_{k,l} P_{ki} Q_{lj} \phi(e_k, f_l) \\ &= \sum_{k,l} P_{ki} Q_{lj} A_{kl} \\ &= (P^\intercal A Q)_{ij} \end{align*}\]

Definition. The rank of $\phi$, denoted by $r(\phi)$, is defined to be the rank of any matrix representing $\phi$.

We can see that the rank is independent of any choices and $r(\phi) = r(\phi_R) = r(\phi_L)$.

Reference