Difference between revisions of "Bilinear map"

Latest revision as of 15:44, 16 June 2015

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^[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:

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

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 $F$ )^[1]. 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 $F$ ^[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 $f:U\times V\rightarrow W$ is a bilinear map

[Expand]
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) ∀x∈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 (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

(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 ∀y∈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 (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

(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

References

↑ ^{Jump up to: 1.0} ^1.1 ^1.2 ^1.3 Advanced Linear Algebra - Steven Roman - Third Edition - Springer Graduate texts in Mathematics

[Roman-1] {Jump up to: 1.0} ^1.1 ^1.2 ^1.3 Advanced Linear Algebra - Steven Roman - Third Edition - Springer Graduate texts in Mathematics

[1]

@@ Line 4: / Line 4: @@
 __TOC__
-==Definition==
+==[[Bilinear map/Definition|Definition]]==
-Given the [[Vector space|vector spaces]] {{M|(U,F),(V,F)}} and {{M|(W,F)}} - it is important they are over the same field - a ''bilinear map''<ref name="Roman">Advanced Linear Algebra - Steven Roman - Third Edition - Springer Graduate texts in Mathematics</ref> is a [[Function|function]]:
+{{:Bilinear map/Definition}}
-*<math>\tau:(U,F)\times(V,F)\rightarrow(W,F)</math> or
-*<math>\tau:U\times V\rightarrow W</math> (in keeping with [[Mathematicians are lazy|mathematicians are lazy]])
-Such that it is [[Linear map|linear]] in both variables. Which is to say that the following "Axioms of a bilinear map" hold:
-For a [[Function|function]] <math>\tau:U\times V\rightarrow W</math> and <math>u,v\in U</math>, <math>a,b\in V</math> and <math>\lambda,\mu\in F</math> we have:
-# <math>\tau(\lambda u+\mu v,a)=\lambda \tau(u,a)+\mu \tau(v,a)</math>
-# <math>\tau(u,\lambda a+\mu b)=\lambda \tau(u,a)+\mu \tau(u,b)</math>
 ==Relation to bilinear forms and inner products==
 A ''[[Bilinear form|bilinear form]]'' is a special case of a bilinear map where rather than mapping to a vector space {{M|W}} it maps to the field that the vector spaces {{M|U}} and {{M|V}} are over (which in this case was {{M|F}})<ref name="Roman"/>. An ''[[Inner product|inner product]]'' is a special case of that. See the pages:
-* [[Bilinear form]]
+* [[Bilinear form]] - a map of the form {{M|\langle\cdot,\cdot\rangle:V\times V\rightarrow F}} where {{M|V}} is a vector space over {{M|F}}<ref name="Roman"/>
-* [[Inner product]]
+* [[Inner product]] - a bilinear form that is either ''symmetric'', ''skew-symmetric'' or ''alternate'' (see the [[Bilinear form]] for meanings)<ref name="Roman"/>
-For more information
+==Kernel of a bilinear map==
+Here {{M|f:U\times V\rightarrow W}} is a bilinear map
+{{Begin Theorem}}
+Claim: <math>\{(u,v)\in U\times V|\ u=0\vee v=0\}\subseteq\text{Ker}(f)</math>, that is if {{M|u}} or {{M|v}} (or both of course) are the zero of their vector space then {{M|1=f(u,v)=0}} (the zero of {{M|W}})
+{{Begin Proof}}
+: Let {{M|u\in U}} and {{M|v\in V}} be given such that either one or both is 0.
+:* If {{M|1=u=0}} then (by definition we have) {{M|1=\forall x\in U[0x=0]}} (note the first 0 is a scalar, the second the 0 vector)
+:*: Let {{M|x\in U}} be given
+:*:: Now <math>f(0,v)=f(0x,v)</math>
+:*::: Using <math>\lambda f(a,b)=f(\lambda a,b)</math> (where {{M|1=a=x}} and {{M|1=\lambda=0}}) we see
+:*:: <math>f(0,v)=f(0x,v)=0f(x,v)</math>
+:*::: But {{M|0}} multiplied by any vector is the {{M|0}} vector (in this case of {{M|W}}) so
+:*:: <math>f(0,v)=0f(x,v)=0</math> (where this 0 is understood to be {{M|\in W}})
+:*: so <math>f(0,v)=0</math>
+:** We now know for whatever value of {{M|v}} (zero or not) that {{M|1=f(0,v)=0}}, so {{M|\forall v\in V[(0,v)\in\text{Ker}(f)]}}
+:* If {{M|1=v=0}} then (by definition we have {{M|1=\forall y\in V[0y=0]}} (note the first 0 is a scalar, the second the 0 vector)
+:*: Let {{M|y\in V}} be given
+:*:: Now <math>f(u,0)=f(u,0y)</math>
+:*::: Using <math>\lambda f(a,b)=f(a,\lambda b)</math> (where {{M|1=b=y}} and {{M|1=\lambda=0}}) we see
+:*:: <math>f(u,0)=f(u,0y)=0f(u,y)</math>
+:*::: But {{M|0}} multiplied by any vector is the {{M|0}} vector (in this case of {{M|W}}) so
+:*:: <math>f(u,0)=0f(u,y)=0</math> (where this 0 is understood to be {{M|\in W}})
+:*: so <math>f(u,0)=0</math>
+:** We now know for whatever value of {{M|u}} (zero or not) that {{M|1=f(u,0)=0}}, so {{M|\forall u\in U[(u,0)\in\text{Ker}(f)]}}
+This completes the proof
+{{End Proof}}{{End Theorem}}
 ==Common notations==
 * If an author uses <math>T</math> for [[Linear map|linear maps]] they will probably use <math>\tau</math> for bilinear maps.
@@ Line 32: / Line 49: @@
 ==Examples of bilinear maps==
 * The [[Tensor product]]
-* The [[Dot product]] - although this is an example of an ''[[Inner product|inner product]]''
+* The [[Vector dot product]] - although this is an example of an ''[[Inner product|inner product]]''
 ==See next==
 * [[Bilinear form]]

Difference between revisions of "Bilinear map"

Latest revision as of 15:44, 16 June 2015

Contents

Definition

Relation to bilinear forms and inner products

Kernel of a bilinear map

Common notations

Examples of bilinear maps

See next

See also

References

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