Prove:
If L is a line in $R^2$
and if T: $R^2 \rightarrow R^2$ is a linear transformation,
then T(L) is also a line in $R^2$
Please, help

Question

Prove:
If L is a line in $R^2$ 
and if T: $R^2 \rightarrow R^2$ is a linear transformation, 
then T(L) is also a line in $R^2$
Please, help

loser66 · Answer

@perl

perl · Answer

we have to use the definition of linear transformation

perl · Answer

do you have the book of this online

loser66 · Answer

I have definition of linear transformation from the book I took the course last semester. hihihi. Let me write it out

loser66 · Answer

attachment

perl · Answer

it is blurry, can you do the pic again

perl · Answer

a line in R^2 has the form  ax +by = c

loser66 · Answer

I am sorry, my scanner is stupid, let me type it down

loser66 · Answer

A linear transformation (or operator) A on a vector space V is a correspondence that assigns to every vector  x in V a vector Ax in V, in such a way that 
$$A(\alpha x+\beta y)= \alpha Ax +\beta A y$$

loser66 · Answer

We can do something like:
Let $\vec x=\left [\begin {matrix}x_1\x_2\end{matrix}\right]$ and $\vec y=\left [\begin {matrix}y_1\y_2\end{matrix}\right]$ in $R^2$

loser66 · Answer

$\alpha$, $\beta$ are scalar 
Since T is linear transformation , we have $T(\alpha\vec x)=T\left [\begin {matrix}\alpha x_1\\alpha x_2\end{matrix}\right]$ 
and $T(\beta \vec y)=T\left [\begin {matrix}\beta y_1\\beta y_2\end{matrix}\right]$

loser66 · Answer

and then?? hihihi

loser66 · Answer

oh, and then the distance between x and y = distance between Tx and Ty. Is it valid??

perl · Answer

well you want a line

perl · Answer

how we do we describe a line in vector notation
r(t) = ro + t*V

loser66 · Answer

no idea!! :(

perl · Answer

https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/vectors/v/linear-algebra--parametric-representations-of-lines

xapproachesinfinity · Answer

@Loser66 are you a linear algebra fan hehe!
when i took this course, i was so lost haha

xapproachesinfinity · Answer

well i actually passed it, i would say it was B minus in american system of grading

xapproachesinfinity · Answer

oh i see
you are still better anyway, so why you say shame on you haha

xapproachesinfinity · Answer

bad scanner bro haha

kirbykirby · Answer

This is slightly distant in my memory.. but if your line $L$ is written as $\vec{x}=\alpha\vec{m}+\vec{b}, \alpha\in \mathbb{R}$, then $T(L)=T(\alpha\vec{m}+\vec{b})=\alpha T(\vec{m})+T(\vec{b})$ since $T$ is a linear mapping. And since $T: \mathbb{R}^2\rightarrow \mathbb{R}^2$ , then $T(\vec{m})\in \mathbb{R}^2 $ and $T(\vec{b})\in \mathbb{R}^2$ , and so you can think of the new line being like $\alpha \vec{v}+\vec{w}$ where $\vec{v}=T(\vec{m})$ and $\vec{w} =T(\vec{b})$

I'm not 100% sure though.

loser66 · Answer

I got all but the very first expression $$\vec{x}=\alpha\vec{m}+\vec{b}, \alpha\in \mathbb{R}$$
How can we write the line under vector form like this?

perl · Answer

that is the vector form

loser66 · Answer

Got it, hihihi

loser66 · Answer

Thank you for the help, friends. I appreciate. :)

perl · Answer

yes it looks like the proof works for any line in R^n

perl · Answer

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