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How can I show that there exists a Linear Transformation $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ s.t. $f(l_i)=m_i$ for two sets of three distinct lines ${l_1,l_2,l_3}$ and ${m_1,m_2,m_3}$ each of which passes through the origin.

I was trying to do this with a matrix equation where a $2 \times 2$ matrix represented the linear transformation and each line was represented by a vector. This gave rise to a system of six equations and six unknowns but I don't know how to guarantee a solution exists.

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It's almost always the case that the worst thing you can do in linear algebra is to start working with matrices. Now, can you show the statement is true if you just have two lines in each set instead of three? –  Amit Kumar Gupta Sep 16 '11 at 0:38

2 Answers 2

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Pick three nonzero vectors $u_1$, $u_2$, $u_3$ parallel to $l_1$, $l_2$, $l_3$ and three nonzero vectors $v_1$, $v_2$, $v_3$ parallel to $m_1$, $m_2$, $m_3$, respectively. Now, $u_3 = \alpha_1 u_1 + \alpha_2 u_2$ and $v_3 = \beta_1 v_1 + \beta_2 v_2$ for some nonzero $\alpha_1, \alpha_2, \beta_1, \beta_2 \in \mathrm R$. Then define $T: \mathrm R^2 \rightarrow \mathrm R^2$ by $Tu_1 = \dfrac{\beta_1}{\alpha_1} v_1$ and $Tu_2 = \dfrac{\beta_2}{\alpha_2} v_2$.

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Ok so I think the main points are that multiplying a line by a constant yields the same line and that any two distinct lines generate the plane. –  user9352 Sep 16 '11 at 14:52

Hint: can you do it if $l_1$ and $m_1$ are horizontal and $l_2$ and $m_2$ are vertical?

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Well this would just be adding some constant to x component and adding some other constant to the y component –  user9352 Sep 16 '11 at 0:56
    
@user9352 Yes. In fact you are at liberty to pick your constants, as long as they are both nonzero (why should they be nonzero?). So, how would you pick the constants? –  Srivatsan Sep 16 '11 at 1:13
    
Not adding, multiplying. –  Robert Israel Sep 16 '11 at 5:10

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