I am just curious if there is a transformation that does not satisfy $\;T(x+y) = T(x) + T(y),\;$ but satisfies $\;T(cx)=cT(x).\;$

I cannot think of any. Thanks for any help people.

  • $\begingroup$ I think the answer's no. Just put $c=2$ and you will get a contradiction. $\endgroup$ – Shuchang Sep 26 '13 at 14:38
  • $\begingroup$ @ShuchangZhang: That only shows that $T(x+x)=T(x)+T(x)$. It could be the case that more general additivity doesn't hold. I think that you can find such a transformation with the Axiom of Choice, but I don't know any details. $\endgroup$ – William Ballinger Sep 26 '13 at 14:40
  • $\begingroup$ $x\in \mathbb{R}$ ? $\mathbb{R}^n$? Do you know the notion of homogeneous function ? $\endgroup$ – user37238 Sep 26 '13 at 15:02
  • $\begingroup$ What do you mean by transformation? $\endgroup$ – Thomas Andrews Oct 15 '13 at 21:12

It depends what kind of map you are looking at. For $T:\mathbb R\to\mathbb R$ the answer is no, since we have $$T(x+y)=T((x+y)\cdot1)=(x+y)\cdot T(1)=x\cdot T(1)+y\cdot T(1) =T(x)+T(y)$$ in that case.

For maps $T:\mathbb R^n\to\mathbb R^n$ the answer is certainly yes. An easy example is to think of a map $T:\mathbb R^2\to\mathbb R^2$ that maps $x\mapsto \varphi(x) x$, where $\varphi:\mathbb R^2\to\mathbb R$ is any function that only depends on the direction of $x$, i.e. the ratio $(x_1:x_2)$. Then $$T(cx)=\varphi(cx)cx=\varphi(x)cx=cT(x),$$ but $$T(x+y)=\varphi(x+y)(x+y)$$ and $\varphi$ does not have to be linear.


How about $T\colon\mathbb{R}^2\to\mathbb{R}^2$ by:

$$T(v)=\begin{cases}v,&\text{$v=(\lambda,0)$ for $\lambda\in\mathbb{R}$}\\0,&\text{else}\end{cases}$$

Writing $V$ for the span of $(1,0)$, we have $v\in V$ iff $cv\in V$ for every $c\in\mathbb{R}$, so $T(cv)=cT(v)$ (both sides $0$ if $v\notin V$ and both sides $cv$ if $v\in V$. However, $$T((1,0)+(0,1))=T(1,1)=0\ne (1,0)=T(1,0)+T(0,1).$$

Edit: This is a special case of Christoph's construction, with $\varphi(v)=1$ if $v\in V$ and $\varphi(v)=0$ otherwise.

  • $\begingroup$ That's a good example! It's a special case of the class of maps I described in my answer, where $\varphi$ is $1$ along the $x$-axis and $0$ in all other directions. $\endgroup$ – Christoph Sep 26 '13 at 15:02
  • $\begingroup$ @ChristophPegel Haha, as well as getting the answer in faster, you got this comment in while I was editing! $\endgroup$ – mdp Sep 26 '13 at 15:05

Let's consider $T\colon \mathbb{R}^2 \to \mathbb{R}^2$. Let $\psi (t) = t + \frac14\sin (2t)$. Then you can define

$$T(r\cos \varphi, r\sin\varphi) = (r\cos \psi(\varphi), r\sin \psi(\varphi)),$$

where the choice of $(r,\varphi)$ such that $(x,y) = (r\cos \varphi, r\sin\varphi)$ does not matter. For $(x,y) \neq (0,0)$, $r = \pm \sqrt{x^2+y^2}$ is only determined up to sign, choosing a different sign translates the admissible choices of $\varphi$ by $\pi$, and $\psi(t+\pi) = t+\pi + \frac14\sin (2(t+\pi)) = t+\pi + \frac14\sin (2t) = \psi(t)+\pi$, so $T$ is well-defined.

$T$ rotates each line through the origin by an angle, but that angle depends on the line, hence $T$ is not linear. But $T(c\cdot x) = c\cdot T(x)$ for all $c\in\mathbb{R}$ and $x\in\mathbb{R}^2$.


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