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I have been helping undergrads in an introduction to linear algebra course. When solving some exercise consisting in showing that a map is linear some get lazy after proving that it is closed under addition and do not prove the closure under scalar multiplication. I wanted to confront them with an example of a map closed under addition but not under the multiplication but could not come up with an example. Do you have any?

brandonscript
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4 Answers4

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If $T:\mathbb{C}\to\mathbb{C}$ is defined by $T(z)=\bar{z}$ then $T(z_1+z_2)=T(z_1)+T(z_2)$, but $T(cz)\neq cT(z)$

Pietro Paparella
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R.N
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Over the real numbers, this is tough. Any additive map is linear over $\mathbb{Q}$, so will be linear over $\mathbb{R}$ as soon as it's continuous. However, there are non-continuous additive maps, even $\mathbb{R}\to\mathbb{R}$, for instance, $\sqrt{2}\mapsto \pi, \pi\mapsto\sqrt{2}$, extend by $\mathbb{Q}$-linearity and fix everything else. $\pi$ and $\sqrt{2}$ are linearly independent over $\mathbb{Q}$, so this is well defined.

Kevin Carlson
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    Does "extend by $\mathbb{Q}$-linearity" use axiom of choice? – lisyarus Nov 06 '15 at 16:54
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    Given a real $x$, why should checking whether $x=r\pi$ for some rational $r$ depend on choice? – Zach Stone Nov 06 '15 at 17:27
  • Maybe it's just my low level of math, but what does "extend by $\mathbb{Q}$-linearity" mean? – Matt Gutting Nov 06 '15 at 17:36
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    Yes, it uses choice. When extending the map, you actually use that $\mathbb R$ has a basis as $\mathbb Q$-vectorspace (hence it is free). And this uses choice. – MooS Nov 06 '15 at 18:13
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    Original source: https://zbmath.org/?q=an%3A36.0446.04 :) – Carsten S Nov 06 '15 at 18:17
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    Just to be totally clear; there are models of ZF (that is, set theory without choice) in which every additive map from R to R is actually linear. So you will have to look elsewhere (eg. C) for nice answers. – Richard Rast Nov 06 '15 at 21:41
  • "extend by $\Bbb Q$-linearity and fix everything else" is meaningless in this context. One can only extend a map that is given by its values on a basis, and you did not mention any ($\Bbb Q$-)basis. That is where AC comes in. Aside from that, it should be "fix everything else" (i.e., fix remaining basis vectors) before extending by linearity. – Marc van Leeuwen Nov 07 '15 at 07:23
  • @MarcvanLeeuwen Most of your comment seems pretty pedantic. I disagree that "One can only extend a map that is given by its values on a basis-"for instance, one can extend a map defined on one term of a direct sum composition, which is how I'd think of this. Off course, you need AC to construct such a decomposition, so it's not much different from what you say. I have no idea why you want to fix everything else before extending, or how that even means anything different-I'm defining a function, not executing a time-dependent process. – Kevin Carlson Nov 07 '15 at 20:25
  • @KevinCarlson: I did not mean to be pedantic, just to point out how this works mathematically (which you seem to understand, but is not clear from your answer). Extending by linearity assumes you have defined a map on some generating set, and then let linearity take care of the rest. To do that you need (at least to mention) a generating set (or sum decomposition); in your construction the extension depends very heavily on which set is used. And "fix everything else" means be identity only on the other generators (summands); to me this comes logically before one can extend by linearity. – Marc van Leeuwen Nov 08 '15 at 07:52
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    @MarcvanLeeuwen OK, I think I see how we were both thinking now. I was meaning to extend by linearity on the span of $\sqrt 2$ and $\pi$, then to choose a complement and fix its generators. Saying "fix everything else" was too quick. – Kevin Carlson Nov 08 '15 at 16:22
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If you take any field $F$, and a homomorphism of additive groups $(F,+)\to (F,+)$ which does not preserve multiplication, then this will be just such a map.

This is will never exist when $F$ is the field of rational numbers, or more generally, when it is generated by the unity (such as any ${\bf F}_p$ for a prime $p$), because in those, we can define multiplication in terms of addition.

A similar thing happens if you look at continuous additive maps in a topological field generated topologically by the unity (that is, the smallest subfield is dense), like the reals or $p$-adics -- every continuous additive map is linear in this case.

On the other hand, if you don't care about continuity, it is pretty easy to define such a map when $F=K[a]$ is a finite extension of another field $K$. Then you can just take $f\colon F\to F$ as the map such that $f(a^n)=0$ for $0<n<\deg a$ and $f(k)=k$ for $k\in K$. For example, if $K={\bf R}$ and $a=i$, $f$ takes real part of a complex number. In this case, the map is even continuous.

For fields which are not finite extensions of other fields (such as $F={\bf R}$), the existence of such maps may require a nontrivial application of axiom of choice, or more precisely, basis theorem, and then we can proceed as in the preceding paragraph: if $K\subseteq F$ is a field extension, then the map $F\to F$ which is identity on $K$ and takes a basis complementary to $1$ to zero is $K$-linear, and therefore additive.

tomasz
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Consider any field homomorphism $L \to L$, and consider $L$ as a vector-space over a subfield, which is not fixed by the map.

From that point of view, the "easiest" example is Frobenius on $\mathbb F_4$.

Note that there are 8 additive maps on the field with 4 elements, which of 4 are also linear.

MooS
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