948: Multilinear Map Is Not Necessarily Linear

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description/proof of that multilinear map is not necessarily linear

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Topics

About: module

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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 multilinear map.
• The reader knows a definition of linear map.

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Target Context

• The reader will have a description and a proof of the proposition that a multilinear map is not necessarily linear.

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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:
$R$: $\in \{\text{ the rings }\}$
$\{M_1,...,M_k\}$: $\subseteq \{\text{ the }R\text{ modules }\}$ where $2\le k$
$M_1\times ...\times M_k$: $=\text{ the product module }$
$M$: $\in \{\text{ the }R\text{ modules }\}$
$f$: $:M_1\times ...\times M_k\to M$, $\in \{\text{ the multilinear maps }\}$
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Statements:
not necessarily $f\in \{\text{ the linear maps }\}$
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2: Note

This proposition may be a matter-of-of-course, and in fact, it is, but this is a reminder for not making some cursory mistakes by being deluded by the "linear" part of "multilinear".

Of course, there are some multilinear maps that are linear: any $0$ map is so.

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3: Proof

Whole Strategy: Step 1: see a counterexample.

Step 1:

A counterexample suffices.

Let $R=\mathbb{R}$, $M_1\times ...\times M_k=\mathbb{R}\times \mathbb{R}$, $M=\mathbb{R}$, and $f:\mathbb{R}\times \mathbb{R}\to \mathbb{R},(v_1,v_2)\mapsto v_1.v_2$, which is the usual inner product.

In fact, in this case, $R$ is a field, $M_1\times ...\times M_k$ and $M$ are some vectors spaces, and $f$ is a map between vectors spaces, but anyway, any field is a ring, any vectors space is a module, and any map between vectors spaces is a map between modules.

Let us see that $f$ is indeed multilinear.

$f((r{v}_1+r^{\prime }{v}_1^{\prime },v_2))=(r{v}_1+r^{\prime }{v}_1^{\prime }).v_2=r{v}_1.v_2+r^{\prime }{v}_1^{\prime }.v_2=rf(v_1,v_2)+r^{\prime }f(v_1^{\prime },v_2)$; $f((v_1,r{v}_2+r^{\prime }{v}_2^{\prime }))=v_1.(r{v}_2+r^{\prime }{v}_2^{\prime })=r{v}_1.v_2+r^{\prime }{v}_1.v_2^{\prime }=rf(v_1,v_2)+r^{\prime }f(v_1,v_2^{\prime })$.

Let us see that $f$ is not linear.

$f(r(v_1,v_2))=f((r{v}_1,r{v}_2))=r^2{v}_1.v_2=r^{2}f(v_1,v_2)$, but when $r\ne 0$ and $r\ne 1$, $r^2\ne r$, and when $v_1,v_2\ne 0$ (which implies that $f((v_1,v_2))\ne 0$), $r^{2}f(v_1,v_2)\ne rf(v_1,v_2)$, because if $r^{2}f(v_1,v_2)=rf(v_1,v_2)$, $r^{2}f(v_1,v_2)-rf(v_1,v_2)=0$, but $=(r^2-r)f(v_1,v_2)$, which would imply that $f(v_1,v_2)=(r^2-r{)}^{-1}(r^2-r)f(v_1,v_2)=(r^2-r{)}^{-1}0=0$, a contradiction.

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References

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