1022: For Multilinear Map from Finite Product Vectors Space, There Is Unique Linear Map from Tensor Product of Finite Vectors Spaces s.t. Multilinear Map Is Linear Map after Canonical Map from Product Vectors Space into Tensor Product

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description/proof of that for multilinear map from finite product vectors space, there is unique linear map from tensor product of finite vectors spaces s.t. multilinear map is linear map after canonical map from product vectors space into tensor product

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of product vectors space.
• The reader knows a definition of multilinear map.
• The reader knows a definition of tensor product of $k$ vectors spaces.
• The reader knows a definition of linear map.
• The reader admits the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.

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Target Context

• The reader will have a description and a proof of the proposition that for any multilinear map from any finite product vectors space, there is the unique linear map from the tensor product of the finite number of vectors spaces such that the multilinear map is the linear map after the canonical map from the product vectors space into the tensor product.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\text{ the fields }\}$
$\{V_1,...,V_k\}$: $\subseteq \{\text{ the }F\text{ vectors spaces }\}$
$V$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$V_1\times ...\times V_k$: $=\text{ the product vectors space }$
$V_1\otimes ...\otimes V_k$: $=\text{ the tensor product }$
$f$: $:V_1\times ...\times V_k\to V$, $\in \{\text{ the multilinear maps }\}$
$g$: $:V_1\times ...\times V_k\to V_1\otimes ...\otimes V_k,(v_1,...,v_k)\mapsto [((v_1,...,v_k))]$
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Statements:
$!\mathrm{\exists }{f}^{\prime }:V_1\otimes ...\otimes V_k\to V\in \{\text{ the linear maps }\}(f=f^{\prime }\circ g)$
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2: Note

$V_j$ or $V$ does not need to be finite-dimensional.

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3: Proof

Whole Strategy: Step 1: see that there is only 1 option for $f^{\prime }$ to satisfy the requirements; Step 2: see that it is well-defined; Step 3: see that it satisfies the requirements.

Step 1:

Let $[v]\in V_1\otimes ...\otimes V_k$ be any.

Although $v$ is not uniquely determined, once $v$ is determined, $v\in F(V_1\times ...\times V_k,F)$ and $v$ is uniquely expressed as $v=r^1((v_{1,1},...,v_{1,k}))+...+r^l((v_{l,1},...,v_{l,k}))$, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique, because $\{((v_1,...,v_k))|v_j\in V_j\}$ is a basis for $F(V_1\times ...\times V_k,F)$.

$[v]=[r^1((v_{1,1},...,v_{1,k}))+...+r^l((v_{l,1},...,v_{l,k}))]$.

$=r^1[((v_{1,1},...,v_{1,k}))]+...+r^l[((v_{l,1},...,v_{l,k}))]$.

As $f^{\prime }$ needs to be linear, $f^{\prime }([v])=f^{\prime }(r^1[((v_{1,1},...,v_{1,k}))]+...+r^l[((v_{l,1},...,v_{l,k}))])=r^1{f}^{\prime }([((v_{1,1},...,v_{1,k}))])+...+r^l{f}^{\prime }([((v_{l,1},...,v_{l,k}))])$ is required.

$f((v_{j,1},...,v_{j,k}))=f^{\prime }\circ g((v_{j,1},...,v_{j,k}))$ is required.

$=f^{\prime }([((v_{j,1},...,v_{j,k}))])$, so, $f^{\prime }([v])$ is required to be $r^{1}f((v_{1,1},...,v_{1,k}))+...+r^{l}f((v_{l,1},...,v_{l,k}))$.

So, if any $f^{\prime }$ exists, $f^{\prime }$ is uniquely determined.

But of course, we need to confirm that $f^{\prime }([v])=r^{1}f((v_{1,1},...,v_{1,k}))+...+r^{l}f((v_{l,1},...,v_{l,k}))$ is well-defined.

Step 2:

Once $v$ is determined, $r^{1}f((v_{1,1},...,v_{1,k}))+...+r^{l}f((v_{l,1},...,v_{l,k}))$ is uniquely determined.

The only issue is that it does not depend on the choice of $v$.

For any other choice, $v^{\prime }\in F(V_1\times ...\times V_k,F)$, such that $[v]=[v^{\prime }]$, $v^{\prime }-v\in (S)$, where $(S)$ is the sub-'vectors space' generated by $S:=\{((v_1,...,r{v}_j,...,v_k))-r((v_1,...,v_k))\in F(V_1\times ...\times V_k)|r\in F,v_1\in V_1,...,v_k\in V_k\}\cup \{((v_1,...,v_j+v_j^{\prime },...,v_k))-((v_1,...,v_j,...,v_k))-((v_1,...,v_j^{\prime },...,v_k))\in F(V_1\times ...\times V_k)|v_1\in V_1,...,v_k\in V_k,v_j^{\prime }\in V_j\}$.

As each element of $(S)$ is a linear combination of $S$, $v^{\prime }-v=s^1(((v_{1,1},...,r^1{v}_{1,j},...,v_{1,k}))-r^1((v_{1,1},...,v_{1,k})))+...+s^l(((v_{l,1},...,r^l{v}_{l,j},...,v_{l,k}))-r^l((v_{l,1},...,v_{l,k})))+t^1(((w_{1,1},...,w_{1,j_1}+w_{1,j_1}^{\prime },...,w_{1,k}))-((w_{1,1},...,w_{1,j_1},...,w_{1,k}))-((w_{1,1},...,w_{1,j_1}^{\prime },...,w_{1,k})))+...+t^m(((w_{m,1},...,w_{m,j_m}+w_{m,j_m}^{\prime },...,w_{m,k}))-((w_{m,1},...,w_{m,j_m},...,w_{m,k}))-((w_{m,1},...,w_{m,j_m}^{\prime },...,w_{m,k})))$.

Note that while each $((\bullet ))$ expression is a basis element for $F(V_1\times ...\times V_k,F)$, the above expression of $v^{\prime }-v$ may contain some duplications of basis elements, but nevertheless, for any $a^1((v_{1,1},...,v_{1,k}))+...+a^l((v_{l,1},...,v_{l,k}))$ with some possible duplications among the basis elements, $f^{\prime }([a^1((v_{1,1},...,v_{1,k}))+...+a^l((v_{l,1},...,v_{l,k}))])=a^{1}f((v_{1,1},...,v_{1,k}))+...+a^{l}f((v_{l,1},...,v_{l,k}))$ holds, because supposing $((v_{1,1},...,v_{1,k}))=((v_{2,1},...,v_{2,k}))$, $f^{\prime }([a^1((v_{1,1},...,v_{1,k}))+...+a^l((v_{l,1},...,v_{l,k}))])=f^{\prime }([(a^1+a^2)((v_{1,1},...,v_{1,k}))+a^3((v_{3,1},...,v_{3,k}))+...+a^l((v_{l,1},...,v_{l,k}))])=(a^1+a^2)f((v_{1,1},...,v_{1,k}))+a^{3}f((v_{3,1},...,v_{3,k}))+...+a^{l}f((v_{l,1},...,v_{l,k}))=a^{1}f((v_{1,1},...,v_{1,k}))+a^{2}f((v_{1,1},...,v_{1,k}))+a^{3}f((v_{3,1},...,v_{3,k}))+...+a^{l}f((v_{l,1},...,v_{l,k}))=a^{1}f((v_{1,1},...,v_{1,k}))+a^{2}f((v_{2,1},...,v_{2,k}))+a^{3}f((v_{3,1},...,v_{3,k}))+...+a^{l}f((v_{l,1},...,v_{l,k}))$.

So, $f^{\prime }([v^{\prime }])=f^{\prime }([v+s^1(((v_{1,1},...,r^1{v}_{1,j},...,v_{1,k}))-r^1((v_{1,1},...,v_{1,k})))+...+s^l(((v_{l,1},...,r^l{v}_{l,j},...,v_{l,k}))-r^l((v_{l,1},...,v_{l,k})))+t^1(((w_{1,1},...,w_{1,j_1}+w_{1,j_1}^{\prime },...,w_{1,k}))-((w_{1,1},...,w_{1,j_1},...,w_{1,k}))-((w_{1,1},...,w_{1,j_1}^{\prime },...,w_{1,k})))+...+t^m(((w_{m,1},...,w_{m,j_m}+w_{m,j_m}^{\prime },...,w_{m,k}))-((w_{m,1},...,w_{m,j_m},...,w_{m,k}))-((w_{m,1},...,w_{m,j_m}^{\prime },...,w_{m,k})))])=f^{\prime }[v]+s^{1}f((v_{1,1},...,r^1{v}_{1,j},...,v_{1,k}))-s^1{r}^{1}f((v_{1,1},...,v_{1,k}))+...+s^{l}f((v_{l,1},...,r^l{v}_{l,j},...,v_{l,k}))-s^l{r}^{l}f((v_{l,1},...,v_{l,k}))+t^{1}f((w_{1,1},...,w_{1,j_1}+w_{1,j_1}^{\prime },...,w_{1,k}))-t^{1}f((w_{1,1},...,w_{1,j_1},...,w_{1,k}))-t^{1}f((w_{1,1},...,w_{1,j_1}^{\prime },...,w_{1,k}))+...+t^{m}f((w_{m,1},...,w_{m,j_m}+w_{m,j_m}^{\prime },...,w_{m,k}))-t^{m}f((w_{m,1},...,w_{m,j_m},...,w_{m,k}))-t^{m}f((w_{m,1},...,w_{m,j_m}^{\prime },...,w_{m,k}))$.

For each $s^{n}f((v_{n,1},...,r^n{v}_{n,j_n},...,v_{n,k}))-s^n{r}^{n}f((v_{n,1},...,v_{n,k}))$, it is $0$, because $s^{n}f((v_{n,1},...,r^n{v}_{n,j_n},...,v_{n,k}))=s^n{r}^{n}f((v_{n,1},...,v_{n,k}))$, because $f$ is multilinear.

For each $t^{n}f((w_{n,1},...,w_{n,j_n}+w_{n,j_n}^{\prime },...,w_{n,k}))-t^{n}f((w_{n,1},...,w_{n,j_n},...,w_{n,k}))-t^{n}f((w_{n,1},...,w_{n,j_n}^{\prime },...,w_{n,k}))$, it is $0$, because $t^{n}f((w_{n,1},...,w_{n,j_n}+w_{n,j_n}^{\prime },...,w_{n,k}))=t^{n}f((w_{n,1},...,w_{n,j_n},...,w_{n,k}))+t^{n}f((w_{n,1},...,w_{n,j_n}^{\prime },...,w_{n,k}))$, because $f$ is multilinear.

So, $f^{\prime }[v^{\prime }]=f^{\prime }[v]$.

Step 3:

Let us see that $f^{\prime }$ is indeed linear.

Let $[v],[v^{\prime }]\in V_1\otimes ...\otimes V_k$ be any.

$[v]=[r^1((v_{1,1},...,v_{1,k}))+...+r^l((v_{l,1},...,v_{l,k}))]$ and $[v^{\prime }]=[r^{\prime 1}((v_{1,1}^{\prime },...,v_{1,k}^{\prime }))+...+r^{\prime m}((v_{m,1}^{\prime },...,v_{m,k}^{\prime }))]$.

$r[v]+r^{\prime }[v^{\prime }]=[r{r}^1((v_{1,1},...,v_{1,k}))+...+r{r}^l((v_{l,1},...,v_{l,k}))+r^{\prime }{r}^{\prime 1}((v_{1,1}^{\prime },...,v_{1,k}^{\prime }))+...+r^{\prime }{r}^{\prime m}((v_{m,1}^{\prime },...,v_{m,k}^{\prime }))]$.

$f^{\prime }(r[v]+r^{\prime }[v^{\prime }])=r{r}^{1}f((v_{1,1},...,v_{1,k}))+...+r{r}^{l}f((v_{l,1},...,v_{l,k}))+r^{\prime }{r}^{\prime 1}f((v_{1,1}^{\prime },...,v_{1,k}^{\prime }))+...+r^{\prime }{r}^{\prime m}f((v_{m,1}^{\prime },...,v_{m,k}^{\prime }))$: as before any duplication of basis elements does not matter, $=r(r^{1}f((v_{1,1},...,v_{1,k}))+...+r^{l}f((v_{l,1},...,v_{l,k})))+r^{\prime }(r^{\prime 1}f((v_{1,1}^{\prime },...,v_{1,k}^{\prime }))+...+r^{\prime m}f((v_{m,1}^{\prime },...,v_{m,k}^{\prime })))=r{f}^{\prime }([v])+r^{\prime }{f}^{\prime }([v^{\prime }])$.

Let us reconfirm that $f=f^{\prime }\circ g$.

For each $(v_1,...,v_k)\in V_1\times ...\times V_k$, $f^{\prime }\circ g((v_1,...,v_k))=f^{\prime }([((v_1,...,v_k))])=f((v_1,...,v_k))$.

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References

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