True/False exam Question Given a pair of lines $L_{1}$ and $L_{2}$ in the plane, meeting only at the origin, there is a unique $2×2$ matrix $A$ so that $L_{1}$ is the $(\lambda = 2)$ eigenspace of $A$ while $L_{2}$ is the $(\lambda = 7)$ eigenspace of $A$
The question was in an exam paper I have and there are about 10 T/F question and you don't need to justify your answer he gives +1 for correct or -1 for incorrect. 
I said the answer was true because I was guessing but I felt it had something to do with linear transformation of the lines but I didn't know exactly why it was true. Can anyone explain what is being asked?
 A: You were right. The idea is the following, the two Lines span $\def\R{\mathbf R}\R^2$ and hence vectors from it form a basis of $\R^2$. Say $b_1 \in L_1 - \{0\}$, and $b_2 \in L_2 - \{0\}$. As $(b_1, b_2)$ is a basis of $\R^2$, any map linear map on $\R^2$ is uniquely defined by its values for $b_1$ and $b_2$. Hence, define $A \colon \R^2\to \R^2$ by $Ab_1 := 2b_1$ and $Ab_2 := 7b_2$. Then $b_1$ is a $2$-eigenvector (and hence $L_1$ the $2$-eigenspace) and $b_2$ the $7$-eigenvector. 
$A$ is uniquely defined by its values on $b_1$, $b_2$, hence its the unique linear map with this properties.
A: There are $u,v \in \mathbb R^2$ linear independent such that
$L_1=\{tu: t \in \mathbb R\}$ and $L_2=\{tv: t \in \mathbb R\}$.
You have to show that there is a unique $2×2$ matrix $A $ with:
$(*)$ $Au=2u$ and $Av=7v$.
Let $x \in \mathbb R^2$, then there are unique $p,q \in \mathbb R$ such that $x=pu+qv$.
Now its your turn to construct $A$ such that $(*)$ holds and to show that $A$ is uniquely determined by $(*)$
A: An eigenspace $E $ of a matrix A is a linear space such that $v \in E \rightarrow Av \in E $. Furthermore, all eigenspaces have associated eigenvalues. That is, if the eigenspace is a linear space of dimension 1, then it is spanned by one vector. 
Therefore, if $dim\ E = 1$, $v \in E$ and $Av \in E $, then $Av$ must be a multiple of $v $. Imagine $Av = \lambda v $. Then we say $\lambda $ is an eigenvalue of $A $ and $E $ is its associated space and $v $ is an eigenvector associated with that eigenvalue.
We call it an eigenspace because vectors in that space, after transformed, keep their direction. They only get scaled by the transformation.
Now imagine the two lines they talked about. If $L_1$ and $L_2$ only meet at the origin then they are not coincident nor parallel, but they intersect eachother. Take a vector $v_1$ that spans $L_1$ and a vector $v_2$ that spans $L_2$. Then $v_1$ and $v_2$ are linearly independent and form a basis for $\Bbb R^2$. Right?
Now imagine making a change of basis, making the new lines the coordinate system. You can either pick $L_1$ to be your new $x $-axis or $y $-axis. $L_2$ will be the other. Now you just define a new linear transformation such that $T(v_1) = 2v_1$ and $T(v_2) = 7v_2$. Since any other vector in $\Bbb R^2$ can be written as $\alpha v_1 + \beta v_2$, the linear transformation is defined for all vectors already.
Now all it takes is to check the uniqueness of the matrix.
Your matrix will be 2x2 and will have 0s on the positions 1.2 and 2.1 with a 2 and a 7 in the diagonal. Now if $L_1$ is the new $x$-axis, you get
$$\begin{bmatrix}
2 &0 \\ 
0 & 7
\end{bmatrix}$$
on the other hand, if $L_1$ was your new $y $-axis you get
$$\begin{bmatrix}
7 &0 \\ 
0 & 2
\end{bmatrix}$$
And thus there are 2 matrices that work fine.
