Are three distinct lines passing through origin isomorphic to every other set of three distinct lines passing through origin?

Question

Consider two pairs of DISTINCT numbers ($m_1$, $m_2$, $m_3$) and ($n_1$, $n_2$, $n_3$) [the tuples are actually representations of 3 distinct lines through origin in the same plane (say $\mathbb{R^{2}}$)]. Can I find an invertible matrix $M$ given

i) the tuples $(m_1, m_2, m_3)$ and ($n_1$, $n_2$, $n_3$)
ii) $m_1 \neq m_2 \neq m_3 $ and $n_1 \neq n_2 \neq n_3 $

$$M(m_1, m_2, m_3)^T = (n_1, n_2, n_3)^T$$ where $m_{i}, n_{i} \in \mathbb{R}~~$ $ \forall~ i\in\{1,2,3\}$

Answer 1

Answering my own question. The original question is related to Hyperplane Arrangements and was included in an assignment. The idea was to prove that three lines in a plane are isomorphic to other three lines in a plane. Observe the freedom that scaling operation provides.

Consider two sets of three lines in a plane $a_{1}, a_{2}~ \&~ a_{3}$ and $a_{1}^{'}, a_{2}^{'} ~ \&~ a_{3}^{'}$.
Let $a_{3} = c_{1}*a_{1} + c_{2}*a_{2}$ where $c_{1}, c_{2} \in \mathbb{R} $, $c_{1} \neq 0$ and $c_{2} \neq 0$
Let $a_{3}^{'} = c_{1}^{'}*a_{1}^{'} + c_{2}^{'}*a_{2}^{'}$ where $c_{1}^{'}, c_{2}^{'} \in \mathbb{R} $, $c_{1}^{'} \neq 0$ and $c_{2}^{'} \neq 0$

Let the transformation fulfilling the above requirement be $T$.
Operating on the basis vectors $a_{1}$ and $a_{2}$, let- $Ta_{1} = c_{1}^{'}/c_{1}*a_{1}^{'}$ and $Ta_{2} = c_{2}^{'}/c_{2}*a_{2}^{'}$

Then, $Ta_{3} =T (c_{1}*a_{1} + c_{2}*a_{2}) = c_{1}^{'}*a_{1}^{'} + c_{2}^{'}*a_{2}^{'} = a_{3}^{'}$

Hence, every set of three lines is isomorphic to every other set of three lines that lie in the same plane.

Answer 2

You can not. Assume $(n_1,n_2,n_3)^T$ has rank $2$, i.e. you have three distinct lines lying in a plane and $(m_1,m_2,m_3)^T$ has rank $3$. Then $M$ can not be invertible.

For example, take $n_1=m_1=e_1$, $n_2=m_2=e_2$, $m_3=e_3$ and $n_3=e_1+e_2$, where $(e_i)_{i\in \{1,2,3\}}$ is a basis of $\Bbb R^3$.

Edit: If you want all lines in a plane, for a counter example, just take $n_1=m_1=e_1$, $n_2=m_2=e_2$, so on the plane $\Bbb R^2$ the matrix $M$ is fixed. So for any choice of $m_3 \in \Bbb R^2$ the line determined by $n_3 \in \Bbb R^2$ is already fixed.