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A question from my Linear Algebra textbook:

Prove that there is an additive function $T : R \to R$ that is not linear. Hint: Let $V$ be the set of real numbers regarded as a vector space over the field of rational numbers. By the corollary to Theorem 1.13 (p. 61), $V$ has a basis $\beta$. Let $x$ and $y$ be two distinct vectors in $\beta$ and define $f: \beta \to V$ by $f(x) = y$, $f(y) = x$, and $f(z) = z$ otherwise. There exists a linear transformation $T: V \to V$ such that $T(u) =f(u)$ for all u in $\beta$. Then $T$ is additive, but for $c =y/x$, $T(cx) \neq cT(x)$.

I also know that if $T: V \to W$ is additive and $V$ and $W$ are vector spaces over the field of rational numbers then $T$ must be linear. So my question is how does this hint make sense? How can $T$ be linear and not be closed under scalar multiplication?

Nick
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    Linearity is written here with two meanings, an oversight. $T$ is $\mathbb Q-$ linear. But 'not linear' in the first sentence means not $\mathbb R-$ linear. – geetha290krm Nov 15 '23 at 07:42
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    The "hint" is actually not a hint - it's the solution. What only remains to be shown is that $x^2\neq y^2$. Note that $V = (\mathbb R,\mathbb Q)$ and $T : V\to V$ is linear. But $T : (\mathbb R,\mathbb R)\to (\mathbb R,\mathbb R)$ is NOT linear. Here, the pair $(W,K)$ means the vector space $W$ over the field $K$. – amsmath Nov 15 '23 at 07:49
  • Very much related to, but not a duplicate of, https://math.stackexchange.com/questions/423492/overview-of-basic-facts-about-cauchy-functional-equation ("Overview of basic facts about Cauchy functional equation"). As other commenters have noted, the property of being "linear" depends upon a choice of a field of scalars – leslie townes Nov 15 '23 at 08:39

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