I showed as an exercise that for an additive function $T: V\to W$, if $T(u+v)= T(u)+T(v)$, then if the field is the rationals, we get that the linearity implies scalar multiplication, i.e. $T(qv)=qT(v)$ for all q in rationals and v in V and so T is actually a linear transformation. My question is, if the field was all of R, can we conclude the same thing? I think not, but is there a way to extend this result by maybe introducing limiting procedures into the vector space? Thanks!
Can we deduce that any additive map from V to W, over the field $R$ is actually a linear transformation?
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linear-algebra
vector-spaces
linear-transformations
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0If $T$ is continuous, then yes: it must be a linear transformation. See [Cauchy's functional equation](https://en.wikipedia.org/wiki/Cauchy's_functional_equation). Also, I'm pretty sure this has been asked on this site before. – 2017-02-07
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0See also [here](http://math.stackexchange.com/questions/957274/fab-fafb-but-f-is-not-linear?noredirect=1&lq=1). – 2017-02-07