0

Let $T:\, \mathbb R^n\to\mathbb R^n\ \, (n\geq 2)$ be a bijective map that takes straight lines to straight lines, and $T(0)=0$. I want to show that $T$ is linear.

So far, I have proved that $T$ maps the 2-planes to the 2-planes, and the parallelograms to the parallelograms. I also showed that $T(u+v)=T(u)+T(v)$ for all linearly independent $u,v$.

But I got some impedent in showing the additive of $T$ for the linearly dependent vectors. Let $u,v\in\mathbb R^n$ be linearly dependent, then $u=\lambda v$ for some $\lambda$. Since both $T\big(\langle v\rangle\big)$ and $\big<T(v)\big>$ contains $0$ and $T(v)$, this implies $T\big(\langle v\rangle\big)=\big<T(v)\big>$. Hence, for all $\lambda\in\mathbb R$, there exists $\mu\in\mathbb R$ such that $\lambda T(v)=T(\mu v)$. And here is where I was struggling.

My attempt :

Take 3 parallel lines $d, d_1, d_2$ in a plane $(P)$ such that $d$ is equidistant from $d_1,d_2$. Let $ABCD$ be a parallelogram with center $I$ such that $A,B\in d_1$ and $C,D\in d_2$. Since $T$ is bjective and it takes parallelograms to parallelograms, $T(ABCD)$ is also parellogram with center $T(I)\in T(d)$. This implies $T$ maps equidistant parallel lines to equidistant parallel lines. Applying this, I can show that $T(2v) = 2T(v),\ \forall u$.

So what should I do next ? Could someone enlight me with some hints, some idea ? Thank you.

PermQi
  • 717
  • This is not true for $n=1$. See this link. – Bcpicao Dec 18 '23 at 12:38
  • @Bcpicao which part of my proof that you disagree ? – PermQi Dec 18 '23 at 12:42
  • How would they know that when you haven't provided a proof? But for a simple demonstration that they are right, $T(x) = \tanh x$ is a bijective map $\Bbb R \to \Bbb R$ which carries straight lines to straight lines (simply because $\Bbb R$ has only one straight line - itself - and $T$ is bijective), and $T(0) = 0$, but it is obviously not linear. – Paul Sinclair Dec 19 '23 at 12:43
  • This also points in the direction you need to examine to prove the case when $n > 1$. You need to make use of the fact in this case that there are $w$ which are independent of $v$. I suggest first that you revisit your proof of $T(u+v) = T(u) + T(v)$ for independent $u,v$ to see if you cannot expand it to be $T(au + bv) = aT(u) + bT(v)$, at least in the case $a + b = 1$. – Paul Sinclair Dec 19 '23 at 12:51
  • @PaulSinclair I think your idea would be hard to make. My proof of the additivity for independent vectors is using geometry. So for $u,v$ independent, $0,u,u+v,v$ form a parallelogram, hence its image through $T$ is a parallelogram formed by $0, T(u), T(u+v), T(v)$, which directly show that $T(u+v)=T(u)+T(v)$. But frome here, I think it is difficult to show $T(au+bv)=aTu+bTv$ – PermQi Dec 19 '23 at 14:28
  • @PaulSinclair may you see my link ? – PermQi Dec 28 '23 at 18:46

0 Answers0