# Bilinear map

A bilinear map combines elements from 2 vector spaces to yield and element in a third (in contrast to a linear map which takes a point in a vector space to a point in a different vector space)

A bilinear form is a special case of a bilinear map, and an inner product is a special case of a bilinear form.

## Definition

Given the vector spaces [ilmath](U,F),(V,F)[/ilmath] and [ilmath](W,F)[/ilmath] - it is important they are over the same field - a bilinear map[1] is a function:

• $\tau:(U,F)\times(V,F)\rightarrow(W,F)$ or
• $\tau:U\times V\rightarrow W$ (in keeping with mathematicians are lazy)

Such that it is linear in both variables. Which is to say that the following "Axioms of a bilinear map" hold:

For a function $\tau:U\times V\rightarrow W$ and $u,v\in U$, $a,b\in V$ and $\lambda,\mu\in F$ we have:

1. $\tau(\lambda u+\mu v,a)=\lambda \tau(u,a)+\mu \tau(v,a)$
2. $\tau(u,\lambda a+\mu b)=\lambda \tau(u,a)+\mu \tau(u,b)$

## Relation to bilinear forms and inner products

A bilinear form is a special case of a bilinear map where rather than mapping to a vector space [ilmath]W[/ilmath] it maps to the field that the vector spaces [ilmath]U[/ilmath] and [ilmath]V[/ilmath] are over (which in this case was [ilmath]F[/ilmath])[1]. An inner product is a special case of that. See the pages:

• Bilinear form - a map of the form [ilmath]\langle\cdot,\cdot\rangle:V\times V\rightarrow F[/ilmath] where [ilmath]V[/ilmath] is a vector space over [ilmath]F[/ilmath][1]
• Inner product - a bilinear form that is either symmetric, skew-symmetric or alternate (see the Bilinear form for meanings)[1]

## Kernel of a bilinear map

Here [ilmath]f:U\times V\rightarrow W[/ilmath] is a bilinear map

Claim: $\{(u,v)\in U\times V|\ u=0\vee v=0\}\subseteq\text{Ker}(f)$, that is if [ilmath]u[/ilmath] or [ilmath]v[/ilmath] (or both of course) are the zero of their vector space then [ilmath]f(u,v)=0[/ilmath] (the zero of [ilmath]W[/ilmath])

Let [ilmath]u\in U[/ilmath] and [ilmath]v\in V[/ilmath] be given such that either one or both is 0.
• If [ilmath]u=0[/ilmath] then (by definition we have) [ilmath]\forall x\in U[0x=0][/ilmath] (note the first 0 is a scalar, the second the 0 vector)
Let [ilmath]x\in U[/ilmath] be given
Now $f(0,v)=f(0x,v)$
Using $\lambda f(a,b)=f(\lambda a,b)$ (where [ilmath]a=x[/ilmath] and [ilmath]\lambda=0[/ilmath]) we see
$f(0,v)=f(0x,v)=0f(x,v)$
But [ilmath]0[/ilmath] multiplied by any vector is the [ilmath]0[/ilmath] vector (in this case of [ilmath]W[/ilmath]) so
$f(0,v)=0f(x,v)=0$ (where this 0 is understood to be [ilmath]\in W[/ilmath])
so $f(0,v)=0$
• We now know for whatever value of [ilmath]v[/ilmath] (zero or not) that [ilmath]f(0,v)=0[/ilmath], so [ilmath]\forall v\in V[(0,v)\in\text{Ker}(f)][/ilmath]
• If [ilmath]v=0[/ilmath] then (by definition we have [ilmath]\forall y\in V[0y=0][/ilmath] (note the first 0 is a scalar, the second the 0 vector)
Let [ilmath]y\in V[/ilmath] be given
Now $f(u,0)=f(u,0y)$
Using $\lambda f(a,b)=f(a,\lambda b)$ (where [ilmath]b=y[/ilmath] and [ilmath]\lambda=0[/ilmath]) we see
$f(u,0)=f(u,0y)=0f(u,y)$
But [ilmath]0[/ilmath] multiplied by any vector is the [ilmath]0[/ilmath] vector (in this case of [ilmath]W[/ilmath]) so
$f(u,0)=0f(u,y)=0$ (where this 0 is understood to be [ilmath]\in W[/ilmath])
so $f(u,0)=0$
• We now know for whatever value of [ilmath]u[/ilmath] (zero or not) that [ilmath]f(u,0)=0[/ilmath], so [ilmath]\forall u\in U[(u,0)\in\text{Ker}(f)][/ilmath]

This completes the proof

## Common notations

• If an author uses $T$ for linear maps they will probably use $\tau$ for bilinear maps.
• If an author uses $L$ for linear maps they will probably use $B$ for bilinear maps.

As always I recommend writing:

 Let [ilmath]\tau:U\times V\rightarrow W[/ilmath] be a bilinear map

Or something explicit.