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In Hubbard and Hubbard, the authors define a linear transformation with two properties:

(1) $T(v+w)=T(v)+T(w)$

(2) $T(cv)=cT(v)$

My question is this: isn't property 2 redundant? For instance, since $T(2v)=T(v+v)$, by property 1 it would obviously have to be $T(v)+T(v)=2T(v)$. Is there something about the definition of transformations that makes it not so obvious that $T(v)+T(v)$ equals $2T(v)$?

FD_bfa
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Notograptus
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    That argument only works when $c$ is an integer – Sam Ballas May 22 '24 at 16:18
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    what about $T(\frac{1}{2}v)=\frac{1}{2} T(v)$? – Sujit Bhattacharyya May 22 '24 at 16:20
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    @SujitBhattacharyya would have actually followed from property (1) alone since we would have $T(v)=T(\frac{1}{2}v + \frac{1}{2}v) = T(\frac{1}{2}v)+T(\frac{1}{2}v) = 2T(\frac{1}{2}v)$ and so dividing by $2$ we have $\frac{1}{2}T(v)=T(\frac{1}{2}v)$. We might have been safe for rational numbers in general if this were the field in question... – JMoravitz May 22 '24 at 16:22
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    @Notograptus If $T$ is defined over a real vector space, you can deduce that $T(cv) = cT(v)$ will hold for rational numbers $c$, but not that the statement holds for arbitrary real $c$. If $T$ is a continuous linear transformation (at any point within the vector space), then $T(cv) = cT(v)$ must hold for real $c$. – Ben Grossmann May 22 '24 at 16:40
  • @Notograptus By the way, the condition that $T(v + w) = T(v) + T(w)$ is sometimes called Cauchy's functional equation – Ben Grossmann May 22 '24 at 16:43

1 Answers1

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Yes, from $T(v+w)=T(v)+T(w)$ for each two vectors $v$ and $w$, you can deduce that $T(2v)$ is always equal to $2T(v)$. However, it does not follow that, say, $T(\pi v)=\pi T(v)$.

Another User
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