A linear transformation $T : \mathbb{R}^n \to \mathbb{R}^n$ sends straight lines to straight lines. What can you say about $T$? What can you say about the matrix?

The only information I know about this linear transformation is that it will send any line $L$ to a another line may be $L'$. But I dont know where it will send any plane or any particular point. One example of this type of linear transformation is rotation. It will send straight lines to straight lines. May be it is a very basic question, but I don't understand what will be the general form of the matrix.

  • 1
    $\begingroup$ Welcome to Math.SE. A bit of clarification would improve your Question. By "sends straight lines to themselves" you perhaps mean that any for any line $L$, the image of $L$ under $T$ is again $L$. Unless you restrict yourself to lines through the origin the only linear transformation that would satisfy this is the identity map (since by intersecting two lines one can show every point is fixed by $T$). This seems too trivial to be your intended Question. $\endgroup$ – hardmath Nov 18 '16 at 13:56
  • $\begingroup$ i edited my question. i also dont understand why it will be identity .may be it is too basic .but i dont understand it -@hardmath $\endgroup$ – mathstudent Nov 18 '16 at 15:15
  • $\begingroup$ I see the update, that the image of any line $L $ under $T $ is a possibly different line. Assuming that points are not considered lines, this tells us only that $T $ is 1-1 and therefore an invertible linear transformation. $\endgroup$ – hardmath Nov 18 '16 at 15:21
  • $\begingroup$ pardon me i dont understand why it would be 1-1.it may send a plane to a straight line or to a point.here i am considering linear map from $R^n$ to $R^n$ @hardmath $\endgroup$ – mathstudent Nov 18 '16 at 15:25
  • $\begingroup$ If not 1-1, then $T$ will send some line to a point (the origin). $\endgroup$ – hardmath Nov 18 '16 at 15:27

Any line $\ell\subset{\mathbb R}^n$ has parametric representations of the form $$\ell:\quad t\mapsto {\bf a}+t{\bf p}\qquad(-\infty<t<\infty)$$ with ${\bf p}\ne{\bf 0}$. If $\>T\colon\>{\mathbb R}^n\to{\mathbb R}^n$ is a linear map then the image of $\ell$ will have the representation $$T(\ell):\quad t\mapsto T({\bf a}+t{\bf p})= T{\bf a}+t\>T{\bf p}\qquad(-\infty<t<\infty)\ ,$$ which is again a line if $T{\bf p}\ne{\bf 0}$, and is the single point ${\bf b}:=T{\bf a}$ if $T{\bf p}={\bf 0}$.

If you want that all lines are mapped into lines again you should be sure that ${\bf p}\ne{\bf 0}$ guarantees $T{\bf p}\ne{\bf 0}$. The latter is the case iff ${\rm rank}(T)=n$. If the rank of $T$ is $<n$ then ${\rm ker}(T)\ne0$, hence there are vectors ${\bf p}\ne{\bf 0}$ with $T{\bf p}={\bf 0}$. Lines parallel to such ${\bf p}$ will then be mapped to points.

The answer to your question therefore is: $${\rm rank}(T)=n\ .$$


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