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It's easy to show that given a linear transformation $T:\mathbb{R}^n \rightarrow \mathbb{R}^m$ lines are mapped to lines and the origin stays fixed (as long as its rank $=n$).

Yet is the converse true?

More precisely, if $T:\mathbb{R}^n \rightarrow \mathbb{R}^m$ is a function that maps lines to lines in the sense that for any pair of vectors $a, b$ there exists vectors $c, d$ such that $T(a+tb)=c+td$ & $T(0)=0$ can we deduce that $T(x+y)=T(x)+T(y)$ for all vectors $x, y$?

Would appreciate any help.

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    $\begingroup$ I think it probably doesn't even have to be continuous to do this, but it's just a 2:37 am hunch. $\endgroup$ – Matt Samuel Sep 19 at 6:37
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    $\begingroup$ There are at least two things you might mean by "maps lines to lines": you might want $T$ to restrict to a linear map from one line to the other (which looks like what you wrote but ideally you'd write down some quantifiers) or you might be happy for $T$ to send one line to the other regarded just as sets. $\endgroup$ – Qiaochu Yuan Sep 19 at 7:32
  • $\begingroup$ If you replace "lines" with "subspaces of dimension n - 1", and add that equally separated "lines" remain equally separated, then the answer is yes. Otherwise, as noted in the answer, it's easy to come up with counterexamples for R -> R. $\endgroup$ – Denziloe Sep 29 at 0:32
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Edit: We have to clarify: Do you actually mean "Lines are mapped to lines", i.e. the image of a line under $T$ is again a line (what I assumed), or do you mean actually mean that $T(a + tb) = c + td$ for all $t\in [0,1]$?


Not if $m=1$!

Take for example

$T: \mathbb{R}^n \rightarrow \mathbb{R}$

$T(x) = 2x_1$ if $x_1>0$

$T(x) = x_1$ if $x_1\leq0$

It projects the line to one dimension and stretches the line on the right half plane, but not on the left half plane. Its non linear around 0, but will still always project lines to lines.

From "(as long as its rank $=n$)" I assume you'd add the condition that $T$ has to have full rank, and than we could assume $m=n > 2$ and there my counter example obviously does not work any more.

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  • $\begingroup$ Yes. I meant for $T$ to have full rank. $\endgroup$ – Leo Sep 19 at 9:05
  • $\begingroup$ This seems not to be a counterexample, because if you take $a=-1$ then $T(-1+t) =-1+td$ implies for small t that $d=1$, which is false for $t=2$. $\endgroup$ – Andrea Marino Sep 19 at 9:29
  • $\begingroup$ Call the line you are projecting every other line to $L$. As far as I can see every line perpendicular to $L$ gets mapped into a point. $\endgroup$ – Leo Sep 22 at 10:01
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I think it is true. Recall the two equations $T(0)=0$ and for any $a,b$ there exist $c,d$ such that for any t $T(a+tb)=c+td$.

Substituting $t=0$ yields $c=T(a)$. In particular for $a=0$ we get $c=0$, i e. $T(tb)=td$. This means that every component of $T$ has derivatives in every direction around 0, and that $\partial_bT(0)=d$: just divide by $t$ and take the limit $t \to 0$. In particular, for $t=1$ we get $T(b)=d=\partial_bT(0)$.

Now we have the surprise: from the same equation, we get

$$ T(x+y) =\partial_{(x+y)}T(0) =\partial_xT(0) +\partial_yT(0) =T(x)+T(y)$$

Voilà ! To conclude, note that $T(tb)=td$ for $t=1$ gives $d=T(b)$, so that $T(tb)=td=tT(b)$, the second condition of linearity.

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  • $\begingroup$ Sorry if the layout is awful, I am from my phone: feel free to edit :) $\endgroup$ – Andrea Marino Sep 19 at 9:48
  • $\begingroup$ I can understand everything up to $T(b)=d=\partial _b T(0)$, yet I don't understand how this expression implies $T(x+y) = \partial _{(x+y)}T(0)$, since in the former $b$ is fixed, and in the latter $(x+y)$ is a variable. In other words, how do we know that $b$ isn't the only number for which $T(b)= \partial _b T(0)$? $\endgroup$ – Leo Sep 21 at 15:20
  • $\begingroup$ Recall that we derived $T(b)=\partial_bT(0)$ for any $b$. This holds in particular for $b=x+y$. I am not sure I got your doubt: both $b,x,y$ have the quantifier "for any", and are fixed from the beginning of the equations $\endgroup$ – Andrea Marino Sep 22 at 12:19
  • $\begingroup$ Sorry for the late reply, but could you elaborate on why every component of $T$ has a derivative around 0? $\endgroup$ – Leo Sep 28 at 12:55
  • $\begingroup$ You should show that the limit $\frac{T(tb)-T(0)}{t}$ for $t\to 0$ exists. But this is constantly $d$! (The notation is tha same as above) $\endgroup$ – Andrea Marino Sep 29 at 16:21
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Notice first that $T(0+tv)=0+tw$, thus $T(tv) = tT(v)$ (thanks Andrea).

Let $v_1$ & $v_2$ be linearlly independent vectors, and consider the lines $v_1+tv_2$ & $v_2+tv_2$. These cross each other precisely when $t=1$ at $p=v_1+v_2$.

Say $$T(v_1+tv_2)=w_1+tw_2' \rightarrow T(v_1) = w_1$$,

$$T(v_2+tv_1)=w_2+tw_1' \rightarrow T(v_2) = w_2$$

Since $v_1+1v_2=v_2+1v_1$ we must have $w_1+1w_2'=w_2+1w_1'$. Because of linear independency we must have $w_i' = w_i = T(v_i)$.

Therefore $T(v_1+v_2) = T(v_1)+T(v_2)$.

Let $x_1, ..., x_n$ be a basis of $\mathbb{R}^n$. Then $T(a_1x_1+...+a_nx_n)= a_1T(x_1+[a_2'x_2+...+a_n'x_n])=a_1T(x_1)+T(a_2x_2+...a_nx_n)$ where $a_i'=a_i/a_1$. Applying the trick repeadetly yields $T(a_1x_1+...+a_nx_n)=a_1T(x_1)+...+a_nT(x_n)$.

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