0
$\begingroup$

I want to answer this:

Given $T(x,y)=(ax+by,cx+dy)$ prove that:

  • T transforms straight lines into straight lines

  • T transforms parallel lines into parallel lines

I have no clue in how to show this. Any hint would be much appreciated. Thanks!

$\endgroup$
  • 2
    $\begingroup$ Could you elucidate me as to your definition of straight lines and parallel lines? $\endgroup$ – Benjamin Gadoua May 29 '16 at 4:46
  • $\begingroup$ What I am calling a straight line is something of the form y = mx + b . I'm not sure how would you name it, I just did what I thought was a good translation from the Spanish "Recta". since I'm not very good with math terminology in English. Parallel lines would be those with the same slope, denoted in the past equation by m. $\endgroup$ – user2021 May 29 '16 at 14:54
  • $\begingroup$ @Benjamin Gadoua Correction, as written in the answers a vectorial notation of the line would be more appropriate. $\endgroup$ – user2021 May 29 '16 at 15:24
2
$\begingroup$

The most general way to define a line is to specify a point on the line and a direction that the line runs parallel to. This way, the line can be written: $$L(t) = \vec{p} + \vec{d}t$$ where $\vec{p}$ is any point on the line, $\vec{d}$ is the direction it runs in, and $t$ is a real number parameter. Allowing $t$ to run from negative infinity to positive infinity gives all points on the line.

To show a linear transformation turns straight lines into straight lines, apply it to $L(t)$ and see if the result can be put into a similar form.

Lines are parallel if they have the same direction vector. If a transform preserves parallelness (parallelity? parallelocity? parallelitude? paralexity? I'll stop now.) then the resultant lines should have the same direction vector.

$\endgroup$
  • $\begingroup$ I have some moral problems with this answer without knowing more about how OP is trying to work. You're considering a setup where we're allowing affine lines, and I'm not sure if a linear transform will necessarily preserve parallel lines (I haven't spent any time to think about it and it's getting late). I think a more moral answer from a linear algebra point of view. I'm not sure what a concept of straight and parallel lines are here. $\endgroup$ – Benjamin Gadoua May 29 '16 at 5:56
  • $\begingroup$ Parallel lines will be parallel after a linear transformation since both resultant lines will have a direction vector $T(\vec{d})$. They may be degenerate if $T(\vec{p_1}) = T(\vec{p_2})$, but they will be parallel. $\endgroup$ – Mark H May 30 '16 at 1:36
1
$\begingroup$

One thing that is important to note, is because we're dealing with a linear transformation, we're not dealing with an affine space and we only have to consider "lines" that intersect the origin. In this case, that means that if you have two parallel lines, they actually have the same "slope-intersect form", which you may remember from high school math. Because they're the same and $T$ is obviously well-defined, it's going to send them to the same new "line", which is about as parallel as the previous line was with itself.

Consider the form of the linear equation,

$$ y = lx + m $$

Well, we know $m = 0$, so we have:

$$ y = lx $$

Furthermore,

$$ T(x,y) = T(x,lx) $$

From here, we already know that both terms of our image will be expressible with one variable, $x$, so we know that the second term will be expressible as $\hat{l}x $ for some constant $\hat{l}$, and we'll have a linear equation and it is a "straight line". We can go ahead and compute what $\hat{l} $ is anyways.

$$ T(x,lx) = (ax + blx, cx + dly) = (x(a+bl),x(c+dl)) $$

Now we can just take out tuple and re-express it as an equation:

$$ f(x(a+bl)) = x(c+dl) \Leftrightarrow f(x_1) = x_1\frac{c+dl}{a+bl} $$

$$ \Rightarrow \hat{l} = \frac{c+dl}{a+bl} $$

And we have our new equation,

$$ y_1 = \hat{l}x_1 $$

Which parameterizes a fantastically straight line.

What your teacher was probably trying to do, was be a little clever and ask you to prove the mapping $T$ is a linear transformation in the first place.

$\endgroup$
  • $\begingroup$ Vector spaces are required to contain the origin, but the straight lines in the question have no such requirement. Linear transformations preserve all straight lines and parallel lines are parallel after transformation, even if the lines do not go through the origin. $\endgroup$ – Mark H May 30 '16 at 1:41
  • $\begingroup$ Linear transformations are vector space homomorphisms. $\endgroup$ – John Cramerus May 30 '16 at 2:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.