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) | ||
− | + | 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. | |
− | ==Definition== | + | __TOC__ |
− | + | ==[[Bilinear map/Definition|Definition]]== | |
+ | {{:Bilinear map/Definition}} | ||
+ | ==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]] - 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]] - a bilinear form that is either ''symmetric'', ''skew-symmetric'' or ''alternate'' (see the [[Bilinear form]] for meanings)<ref name="Roman"/> | ||
− | <math>\ | + | ==Kernel of a bilinear map== |
− | + | Here {{M|f:U\times V\rightarrow W}} is a bilinear map | |
− | <math>\ | + | {{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. | ||
+ | * If an author uses <math>L</math> for [[Linear map|linear maps]] they will probably use <math>B</math> for bilinear maps. | ||
+ | As always I recommend writing: | ||
+ | {| class="wikitable" border="1" | ||
+ | |- | ||
+ | | Let {{M|\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|inner product]]'' | |
− | + | ||
− | + | ||
− | + | ||
− | + | ||
− | + | ||
− | + | ==See next== | |
+ | * [[Bilinear form]] | ||
+ | ==See also== | ||
+ | * [[Bilinear form]] | ||
+ | * [[Inner product]] | ||
+ | * [[Linear map]] | ||
+ | * [[Tensor product]] | ||
+ | ==References== | ||
+ | <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.
Contents
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:
- [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 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])
- Now [math]f(0,v)=f(0x,v)[/math]
- 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]
- Let [ilmath]x\in U[/ilmath] be given
- 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])
- Now [math]f(u,0)=f(u,0y)[/math]
- 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]
- Let [ilmath]y\in V[/ilmath] be given
- 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)
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
- The Tensor product
- The Vector dot product - although this is an example of an inner product