View source for Characteristic property of the tensor product

{{Stub page|grade=A*|msg=Important linear algebra here. Totally}}
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==[[/Statement|Statement]]==
{{/Statement|full=true}}
==Proof of claims==
===Claim 1: {{M|p}} is [[multilinear map|multilinear]]===
* Let {{M|i\in\{1,\ldots,k\} }} be given
** {{M|p(v_1,\ldots,v_{i-1},v_i+\lambda u,v_{i+1},\ldots,v_k)\eq v_1\otimes\cdots\otimes v_{i-1}\otimes (v_i+\lambda u)\otimes v_{i+1}\otimes\cdots\otimes v_k}}
*: {{M|\eq (v_1\otimes\cdots\otimes v_k)+(v_1\otimes \cdots\otimes v_{i-1}\otimes \lambda u\otimes v_{i+1}\otimes\cdots\otimes v_k)}}
*: {{M|\eq (v_1\otimes\cdots\otimes v_k)+\lambda(v_1\otimes\cdots\otimes v_{i-1}\otimes u\otimes v_{i+1}\otimes\cdots\otimes v_k)}}
** Next notice: {{M|p(v_1,\ldots,v_k)+\lambda p(v_1,\ldots,v_{i-1},u,v_{i+1},\ldots,v_k) }} 
*: {{M|\eq (v_1\otimes\cdots\otimes v_k)+\lambda(v_1\otimes\cdots\otimes v_{i-1}\otimes u\otimes v_{i+1}\otimes \cdots \otimes v_k)}}
** We see: {{M|p(v_1,\ldots,v_{i-1},v_i+\lambda u,v_{i+1},\ldots,v_k)}}{{M|\eq p(v_1,\ldots,v_k)+\lambda p(v_1,\ldots,v_{i-1},u,v_{i+1},\ldots,v_k)}}
* Since {{M|i\in \{1,\ldots,k\} }} was arbitrary, we see it is [[linear map|linear]] in each {{M|i}} - the definition of a [[multilinear map]]
==Proof==
<div style="float:right;margin:0px;margin-left:0.2em;overflow:hidden;max-width:40em;margin-bottom:0.2em;">
{| class="wikitable" border="1" style="margin:0px;"
|-
| <center><span style="font-size:12pt;"><m>\xymatrix{ 
\mathcal{F}(V_1\times\cdots\times V_k) \ar@{-->}@/^1.8em/[rrr]^{\tilde{A} } \ar[dr]_\pi & V_1\times\cdots\times V_k \ar@{_{(}->}[l]_-i \ar@2{->}[d]_p \ar@2{->}[rr]^-A & & W \\
 & V_1\otimes\cdots\otimes V_k \ar@{.>}[urr]_-{\overline{A} }
}</m></span></center>
|-
! Diagram
|}
</div><!--
END OF DIAGRAM
-->The gist of the proof is as follows:
* First note that by the [[characteristic property of the free vector space]] that {{M|\overline{A} }} [[extension (function)|extends]] uniquely to a [[linear map]] {{M|\tilde{A}:\mathcal{F}(V_1\times\cdots\times V_2)\rightarrow W}}
* Now we want to [[factor (linear map)|factor]] {{M|\overline{A} }} through {{M|\pi}}, this will yield us another [[linear map]]:
*: {{MM|\overline{A}:\underbrace{\dfrac{\mathcal{F}(V_1\times\cdots\times V_K)}{\mathcal{R} } }_{\text{i.e. }V_1\otimes\cdots\otimes V_k} \rightarrow W}} - as required
** To do this we use the [[group factorisation theorem]]:
*** If {{M|\mathcal{R}\eq\text{Ker}(\pi)\subseteq\text{Ker}(\tilde{A})}} then {{M|\tilde{A} }} [[factor (function)|factors]] through {{M|\pi}} to yield (in this case) {{M|\overline{A} }}, and furthermore {{M|\overline{A} }} is a [[linear map]].
**** Let {{M|(v_1,\ldots,v_{i-1},av_i,v_{i+1},\ldots,v_k)\in\mathcal{R} }} be given
***** Then {{M|\tilde{A}(v_1,\ldots,v_{i-1},av_i,v_{i+1},\ldots,v_k)}} is:
****** {{M|\eq A(v_1,\ldots,v_{i-1},av_i,v_{i+1},\ldots,v_k)}} as it extends {{M|A}} and {{M|(v_1,\ldots,v_{i-1},av_i,v_{i+1},\ldots,v_k)\in V_1\times\cdots\times V_k}}
****** {{M|\eq aA(v_1,\ldots,v_k)}} as {{M|A}} is [[multilinear map|multilinear]]
****** {{M|\eq a\tilde{A}(v_1,\ldots,v_k)}} as {{M|\tilde{A} }} is an extension of {{M|A}} 
****** {{M|\eq \tilde{A}(a(v_1,\ldots,v_k))}} as {{M|\tilde{A} }} is [[linear map|linear]]

{{Requires proof|grade=A|msg=The gist is above. Tidy up a bit, basically done, not quite sure where to go, demote to {{C|E}} once the outline is complete}}
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==Notes==
<references group="Note"/>
==References==
<references/>
{{Theorem Of|Linear Algebra|Abstract Algebra}}