Difference between revisions of "Bilinear map"

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A bilinear map combines elements from 2 [[Vector space|vector spaces]] to yield and element in a third (in contrast to a [[Linear map|linear map]] which takes a point in a vector space to a point in a different vector space)
 
A bilinear map combines elements from 2 [[Vector space|vector spaces]] to yield and element in a third (in contrast to a [[Linear map|linear map]] which takes a point in a vector space to a point in a different vector space)
  
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It is sometimes called a "Bilinear form"
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A ''[[Bilinear form|bilinear form]]'' is a special case of a bilinear map, and an ''[[Inner product|inner product]]'' is a special case of a bilinear form.
  
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==Definition==
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__TOC__
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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 is a [[Function|function]]:
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==[[Bilinear map/Definition|Definition]]==
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{{:Bilinear map/Definition}}
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==Relation to bilinear forms and inner products==
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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:
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* [[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"/>
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* [[Inner product]] - a bilinear form that is either ''symmetric'', ''skew-symmetric'' or ''alternate'' (see the [[Bilinear form]] for meanings)<ref name="Roman"/>
  
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<math>\tau:(U,F)\times(V,F)\rightarrow(W,F)</math><br />
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==Kernel of a bilinear map==
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or<br />
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Here {{M|f:U\times V\rightarrow W}} is a bilinear map
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<math>\tau:U\times V\rightarrow W</math> (in keeping with [[Mathematicians are lazy|mathematicians are lazy]])
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{{Begin Theorem}}
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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}})
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{{Begin Proof}}
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: Let {{M|u\in U}} and {{M|v\in V}} be given such that either one or both is 0.
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:* 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)
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:*: Let {{M|x\in U}} be given
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:*:: Now <math>f(0,v)=f(0x,v)</math>
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:*::: Using <math>\lambda f(a,b)=f(\lambda a,b)</math> (where {{M|1=a=x}} and {{M|1=\lambda=0}}) we see
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:*:: <math>f(0,v)=f(0x,v)=0f(x,v)</math>
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:*::: But {{M|0}} multiplied by any vector is the {{M|0}} vector (in this case of {{M|W}}) so
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:*:: <math>f(0,v)=0f(x,v)=0</math> (where this 0 is understood to be {{M|\in W}})
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:*: so <math>f(0,v)=0</math>
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:** 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)]}}
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:* 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)
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:*: Let {{M|y\in V}} be given
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:*:: Now <math>f(u,0)=f(u,0y)</math>
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:*::: Using <math>\lambda f(a,b)=f(a,\lambda b)</math> (where {{M|1=b=y}} and {{M|1=\lambda=0}}) we see
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:*:: <math>f(u,0)=f(u,0y)=0f(u,y)</math>
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:*::: But {{M|0}} multiplied by any vector is the {{M|0}} vector (in this case of {{M|W}}) so
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:*:: <math>f(u,0)=0f(u,y)=0</math> (where this 0 is understood to be {{M|\in W}})
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:*: so <math>f(u,0)=0</math>
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:** 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)]}}
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This completes the proof
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{{End Proof}}{{End Theorem}}
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==Common notations==
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* If an author uses <math>T</math> for [[Linear map|linear maps]] they will probably use <math>\tau</math> for bilinear maps.
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* If an author uses <math>L</math> for [[Linear map|linear maps]] they will probably use <math>B</math> for bilinear maps.
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As always I recommend writing:
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{| class="wikitable" border="1"
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|-
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| Let {{M|\tau:U\times V\rightarrow W}} be a bilinear map
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|}
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Or something explicit.
  
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Such that it is linear in both parts. Which is to say that the following "Axioms of a bilinear map" hold:
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==Examples of bilinear maps==
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===Axioms of a bilinear map===
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* The [[Tensor product]]
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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:
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* The [[Vector dot product]] - although this is an example of an ''[[Inner product|inner product]]''
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# <math>\tau(\lambda u+\mu v,a)=\lambda \tau(u,a)+\mu \tau(v,a)</math>
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# <math>\tau(u,\lambda a+\mu b)=\lambda \tau(u,a)+\mu \tau(u,b)</math>
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==Common notations==
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If an author uses <math>T</math> for [[Linear map|linear maps]] they will probably use <math>\tau</math> for bilinear maps.
+
  
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If an author uses <math>L</math> for [[Linear map|linear maps]] they will probably use <math>B</math> for bilinear maps.
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==See next==
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* [[Bilinear form]]
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==See also==
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* [[Bilinear form]]
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* [[Inner product]]
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* [[Linear map]]
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* [[Tensor product]]
  
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==References==
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<references/>
  
 
{{Definition|Linear Algebra}}
 
{{Definition|Linear Algebra}}

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 [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:

  • [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)

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

For a 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:

  1. [math]\tau(\lambda u+\mu v,a)=\lambda \tau(u,a)+\mu \tau(v,a)[/math]
  2. [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 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: [math]\{(u,v)\in U\times V|\ u=0\vee v=0\}\subseteq\text{Ker}(f)[/math], 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 [math]f(0,v)=f(0x,v)[/math]
    Using [math]\lambda f(a,b)=f(\lambda a,b)[/math] (where [ilmath]a=x[/ilmath] and [ilmath]\lambda=0[/ilmath]) we see
    [math]f(0,v)=f(0x,v)=0f(x,v)[/math]
    But [ilmath]0[/ilmath] multiplied by any vector is the [ilmath]0[/ilmath] vector (in this case of [ilmath]W[/ilmath]) so
    [math]f(0,v)=0f(x,v)=0[/math] (where this 0 is understood to be [ilmath]\in W[/ilmath])
    so [math]f(0,v)=0[/math]
    • 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 [math]f(u,0)=f(u,0y)[/math]
    Using [math]\lambda f(a,b)=f(a,\lambda b)[/math] (where [ilmath]b=y[/ilmath] and [ilmath]\lambda=0[/ilmath]) we see
    [math]f(u,0)=f(u,0y)=0f(u,y)[/math]
    But [ilmath]0[/ilmath] multiplied by any vector is the [ilmath]0[/ilmath] vector (in this case of [ilmath]W[/ilmath]) so
    [math]f(u,0)=0f(u,y)=0[/math] (where this 0 is understood to be [ilmath]\in W[/ilmath])
    so [math]f(u,0)=0[/math]
    • 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 [math]T[/math] for linear maps they will probably use [math]\tau[/math] for bilinear maps.
  • If an author uses [math]L[/math] for linear maps they will probably use [math]B[/math] for bilinear maps.

As always I recommend writing:

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

Or something explicit.

Examples of bilinear maps

See next

See also

References

  1. ↑ 1.0 1.1 1.2 1.3 Advanced Linear Algebra - Steven Roman - Third Edition - Springer Graduate texts in Mathematics
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