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June 9

Why is the only multilinear map linear in all arguments the zero map?

I wonder if there's an elementary proof that a multilinear map with multiple arguments cannot be simultaneously linear in all of them unless it's the zero map, as stated in the article. I attempted the following proof by contradiction:

Let T by a bilinear map that is also linear in both its arguments. Then T(2a,0) = 2T(a, 0) = T(a, 2*0) = T(a,0). But since 2T(a,0) = T(a,0), T must necessarily be zero for all a if this is not to be a contradiction. However I am not sure this covers the general case of all arguments nonzero.--Jasper Deng (talk) 06:03, 9 June 2015 (UTC)[reply]

Actually you only need additive properties. If T is both linear and bilinear then T(a, b) = T((a, 0) + (0, b)) = T(a, 0) + T(0, b) by linearity, and T(a, b) = T(a + 0, b) = T(a, b) + T(0, b) by bilinearity. Cancelling, T(a, b) = T(a, 0). Similarly T(a, b) = T(0, b), so T( a, b) = T(0, 0) = 0. --RDBury (talk) 06:32, 9 June 2015 (UTC)[reply]

The statement is ill-defined. A "linear map" on several arguments clearly means as a map from the cartesian product of the input spaces. Yet linearity of such a map requires that addition be defined on the cartesian product (and the mention of linearity requires further that the scalar action be defined). The sum and scalar action are defined on say the direct sum (equivalently the direct product) of vector spaces (and even then only over a common base field for the scalar product). The article should be updated, since as it stands, the statement is both confusing and invalid due to its unstated necessary premise. —Quondum 12:30, 9 June 2015 (UTC)[reply]

I may be missing something, but the article does not speak of linearity of a map on the cartesian product, but rather on the restriction of such a map to subspaces spanned by individual factors in the cartesian product, where it is clear what "linear" means. I don't object against a full blown out definition of everything though. YohanN7 (talk) 13:03, 9 June 2015 (UTC)[reply]

FWIW, the article defines a multilinear map to be a map

f:V_{1}\times \cdots \times V_{n}\to W

. So "linearity" here obviously implicitly means linearity of f as a function on the Cartesian product. But the statement could certainly be clarified. A very short proof could even be included. Sławomir Biały (talk) 13:31, 9 June 2015 (UTC)[reply]

Referring to linearity of

f | SUBSPACE

is not the same as implicitly referring to "linearity of

f

". I still may be missing something and still don't mind definitions/proofs because misunderstanding is still possible. YohanN7 (talk) 13:45, 9 June 2015 (UTC)[reply]

I agree that the meaning of "linear map" is clear from context here. I agree with Quondum though in that the statement in the article is confusing; imo it should just be removed. It's the kind of thing that's put as an exercise to make sure the reader understands the difference between linear and bilinear, but other than that I doubt it has any application. The article is poorly referenced though, so it's difficult to tell what is from the source and what is OR. --RDBury (talk) 14:05, 9 June 2015 (UTC)[reply]

I'm with RDBury on this: its only purpose in the article is evidently pedagogical; it addresses a question that is not within the scope of the article. I'll trim it out. —Quondum 15:39, 9 June 2015 (UTC)[reply]