description/proof of that multilinear map is not necessarily linear
Topics
About: module
The table of contents of this article
Starting Context
- The reader knows a definition of multilinear map.
- The reader knows a definition of linear map.
Target Context
- The reader will have a description and a proof of the proposition that a multilinear map is not necessarily linear.
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.
Main Body
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.
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.
\(f ((r v_1 + r' v'_1, v_2)) = (r v_1 + r' v'_1) . v_2 = r v_1 . v_2 + r' v'_1 . v_2 = r f (v_1, v_2) + r' f (v'_1, v_2)\); \(f ((v_1, r v_2 + r' v'_2)) = v_1 . (r v_2 + r' v'_2) = r v_1 . v_2 + r' v_1 . v'_2 = r f (v_1, v_2) + r' f (v_1, v'_2)\).
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 \(r \neq 0\) and \(r \neq 1\), \(r^2 \neq r\), and when \(v_1, v_2 \neq 0\) (which implies that \(f ((v_1, v_2)) \neq 0\)), \(r^2 f (v_1, v_2) \neq r f (v_1, v_2)\), because if \(r^2 f (v_1, v_2) = r f (v_1, v_2)\), \(r^2 f (v_1, v_2) - r f (v_1, v_2) = 0\), 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.
