It makes sense that the only linear transformations on $\mathbb{R}$ would be maps of the form $x\mapsto ax$ for $a\in\mathbb{R}$. But how do we know these maps are all the possible linear transformations on the reals? Is there a way to prove that there isn't some weird function out there that just happens to be linear but isn't just a multiplication function?
Why is $L(\mathbb{R})$ just the set of multiplication maps?
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linear-algebra
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0Try taking the derivative of a linear transformation from $\mathbb R$ to $\mathbb R$. – 2011-07-06
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6As soon as you know where one non-zero element e.g. $1$ is going you know where the rest must go by linearity. – 2011-07-06
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0@Jacob Schlather: Overkill. One does not know *a priori* that a linear operator on $\mathbb{R}$ is differentiable (although this is, of course, true). In order to prove this, it suffices to verify differentiability at $0$ by linearity but once you start working along these lines you are already making the problem much too complex. – 2011-07-06
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0@Amitest Good point. I tend to take it for granted. It is of course much simpler to just look at at T(1). – 2011-07-06
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0@Jacob: I removed an earlier comment similar to Amitesh's because your proposal doesn't help at all. After differentiating you're left with a linear map which you want to show not to be of erratic nature... – 2011-07-06
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4I suppose that we are implicitly assuming "$\Bbb R$-linear". Else, things become pretty intricate. – 2011-07-06
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0@Andrea In fact, I have noted some variations of this question in the exercises presented in my answer below. Of course, one can use the axiom of choice to construct non-trivial (i.e., non-$\mathbb{R}$-linear) $\mathbb{Q}$-linear maps of $\mathbb{R}$ (as a vector space over $\mathbb{Q}$). – 2011-07-06