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 $(U,F),(V,F)$ and $(W,F)$ - it is important they are over the same field - a bilinear map 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 $W$ it maps to the field that the vector spaces $U$ and $V$ are over (which in this case was $F$). An inner product is a special case of that. See the pages:

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

Kernel of a bilinear map

Here $f:U\times V\rightarrow W$ is a bilinear map

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

---

Let $u\in U$ and $v\in V$ be given such that either one or both is 0.

• If $u=0$ then (by definition we have) $\forall x\in U[0x=0]$ (note the first 0 is a scalar, the second the 0 vector)

  Let $x\in U$ be given

  Now $$f(0,v)=f(0x,v)$$

  Using $$\lambda f(a,b)=f(\lambda a,b)$$ (where $a=x$ and $\lambda=0$) we see

  $$f(0,v)=f(0x,v)=0f(x,v)$$

  But $0$ multiplied by any vector is the $0$ vector (in this case of $W$) so

  $$f(0,v)=0f(x,v)=0$$ (where this 0 is understood to be $\in W$)

  so $$f(0,v)=0$$

  • We now know for whatever value of $v$ (zero or not) that $f(0,v)=0$, so $\forall v\in V[(0,v)\in\text{Ker}(f)]$
• If $v=0$ then (by definition we have $\forall y\in V[0y=0]$ (note the first 0 is a scalar, the second the 0 vector)

  Let $y\in V$ be given

  Now $$f(u,0)=f(u,0y)$$

  Using $$\lambda f(a,b)=f(a,\lambda b)$$ (where $b=y$ and $\lambda=0$) we see

  $$f(u,0)=f(u,0y)=0f(u,y)$$

  But $0$ multiplied by any vector is the $0$ vector (in this case of $W$) so

  $$f(u,0)=0f(u,y)=0$$ (where this 0 is understood to be $\in W$)

  so $$f(u,0)=0$$

  • We now know for whatever value of $u$ (zero or not) that $f(u,0)=0$, so $\forall u\in U[(u,0)\in\text{Ker}(f)]$

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 $\tau:U\times V\rightarrow W$ be a bilinear map

Or something explicit.

Examples of bilinear maps

• The Tensor product
• The Vector dot product - although this is an example of an inner product

See next

• Bilinear form

See also

• Bilinear form
• Inner product
• Linear map
• Tensor product