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Is $F(c\vec{v})$ = $cF(\vec{v})$ condition of linear transformation the same thing as $F(\vec{a})+F(\vec{b})=F(\vec{a}+\vec{b})$?

I am a physics undergraduate student and studying linear algebra.

The 2 conditions of linear transformation seem quite same to me. But, I know if these 2 were the same then it wouldn't be mentioned in explicitly in the literature.

Why do I think the 2 conditions are the same thing:

For any real number (rational or irrational) $c$, the vector $c\vec{v}$ can be expressed as a sum of $n$ equal vectors, $\frac{c}{n}\vec{v}$.

So, $$c\vec{v}=\sum_{i=1}^{n}\frac{c}{n}\vec{v}$$ Then, applying the first condition on 2 constituents we get $F(\frac{c}{n}\vec{v})+F(\frac{c}{n}\vec{v})=F(\frac{2c}{n}\vec{v})$ $\implies2F(\frac{c}{n})=F(\frac{2c}{n})$. Then applying induction, we can get the second condition $F(c\vec{v})$ = $cF(\vec{v})$

So, why is the second condition necessary?

One reasoning: If $c$ is an irrational number, then this repeated addition never ends; because $\frac{c}{n}$ is also an irrational number. While it may not be a problem to a physicist, it can be a real problem to a mathematician. But, I don't know that deep about number theory. So, I can't figure out the problem.

So, mathematicians, help me figure out the reason of explicitly stating the 2nd condition. You are welcome to use advanced stuff to make it clear, but first give a simple explanation, then rigorously explain it.

Partha Sarker
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    The first doesn't come close to implying the second. In $\mathbb R^2$ consider the function $f(\vec v)$ which is $\vec 0$ unless $\vec v=(a,0)$ for some $a\in \mathbb R$, and $f((a,0))=(a,0)$ That satisfies the first condition, but not the second. – lulu Feb 27 '18 at 13:32
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    People often imagine that the second condition implies the first which is true if you add some sort of continuity assumption, but which is false in general. See Cauchy's functional equation – lulu Feb 27 '18 at 13:34
  • Would you elaborate what you mean by 'continuity assumption'? @lulu – Partha Sarker Feb 27 '18 at 14:01
  • @Mockingbird360 e.g., if $f: \mathbb R^n \to \mathbb R^m$ is continuous and satisfies $f(a+b)=f(a)+f(b)$, then $f$ is linear over $\mathbb R$. – Dustan Levenstein Feb 27 '18 at 14:07
  • Sure. Working over $\mathbb R$, your argument shows that $T(\vec v+\vec w)=T(\vec v)+T(\vec w)\implies T(c\vec v)=cT(\vec v)$ IF $c\in \mathbb Q$. Alas, that is not enough to show it is true for $c\in \mathbb R$. If, however, you also assume that $T$ is continuous, it is enough (exercise). – lulu Feb 27 '18 at 14:08
  • @lulu If $T$ is continuous then as every $x\in\mathbb{R-Q}$ is arbitrarily close to another $y\in\mathbb{Q}$, then with assumption of continuity my result can be extended to irrationals also, am I right? – Partha Sarker Feb 27 '18 at 14:12
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    Well, that's a bit informal, but basically right. If ${c_i}$ is a sequence of rationals converging to $c\in \mathbb R$ then continuity tells us that $cT(\vec v)= \lim c_iT(\vec v)=\lim T(c_i \vec v)=T(\lim c_i \vec v)=T(c\vec v)$. – lulu Feb 27 '18 at 14:15
  • @lulu Related: https://math.stackexchange.com/questions/424809/why-do-we-define-a-linear-transformation-to-have-the-property-that-fcw-c-fw – Partha Sarker Feb 27 '18 at 14:21
  • @lulu Not related to the question, but asking out of curiosity; are there some nice classifications of non-linear transformation that follow some conditions? I know about affine transformation. – Partha Sarker Feb 27 '18 at 14:37
  • That's too broad...there are lots of non-linear maps. Physics is filled with them. – lulu Feb 27 '18 at 14:40
  • @Mockingbird360 Please note that if your OP is now resolved you can evaluate if accept an answer among the given as described here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Mar 02 '18 at 23:33

2 Answers2

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You have proved that if $F(a) + F(b) = F(a+b)$ then given any $n \in \mathbb{N}$ and any real $c$ and any vector $v$ $F(cv) = nF(\frac{cv}{n})$. Following the same argument you can prove that $F(nv) = nF(v)$ with $n \in \mathbb{N}$. Using this two proofs you managed to do it with rational numbers. As you say, you miss the irrational numbers, that is important, indeed they are almost all $\mathbb{R}$. And if you consider a $\mathbb{C}$-Vector Space you are missing much more. Now if you consider an arbitrary space vector, say $(V,\mathbb{K},+,.)$ with $\mathbb{K}$ an arbitrary field?

HFKy
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The two conditions are not equivalent since we need both to guarantee that a trasformation is linear.

The reason can be found in the properties which define a subsbace and notably

$0) \vec{0} \in W\\ 1) \vec{cv}\to c \cdot \vec{v}\in W\\ 2) \vec{v}+\vec{w} \in W$

Indeed let consider for example the following subset $$S=\{\vec v,\vec w:\vec v=t(1,0), \vec w=s(0,1)\}\subseteq\mathbb{R}$$

then "1" is satified but "2" is not.

We can also give of course examples of transformation for which

  • $F(c\vec{v})$ = $cF(\vec{v})$ is satisfied, but
  • $F(\vec{a})+F(\vec{b})=F(\vec{a}+\vec{b})$ is not

Let consider for example the transformation

$$\vec v(a,b) \to \begin{cases}\vec v \quad |a|\ge |b|\\-\vec v \quad |a|< |b| \end{cases} $$

user
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