To find matrix from given linear transformation Let $\vec{x}$ & $\vec{y}$ be linearly independent vectors in $\mathbb{R}^2$. Suppose T: $\mathbb{R}^2$ $\rightarrow$ $\mathbb{R}^2$ is a linear transformation such that $T \vec{y}$=$\alpha \vec{x}$ and $T \vec{x}=0$ with respect to some basis in $\mathbb{R}^2$. What is the form of T?
1.$$\left[
\begin {array}{c c}
a&0\\
0&a
\end {array}\right]$$a>0
2.$$\left[
\begin {array}{c c}
a&0\\
0&b
\end {array}\right]$$a,b>0,a not equal to  b
3.$$\left[
\begin {array}{c c}
0&1\\
0&0
\end {array}\right]$$
4.$$\left[
\begin {array}{c c}
0&0\\
0&0
\end {array}\right]$$

$\newcommand{\bx}{\mathbf{x}}\newcommand{\bb}{\mathbf{b}}\newcommand{\by}{\mathbf{y}}$ Suppose the basis is $\{ \bb_1,\bb_2\}$. Then the two columns of $T$ as a matrix are given by $T\bb_1$ and $T\bb_2$.
Now $\bx$ and $\by$ are independent, so they must form a basis for $\mathbb{R}^2$. Hence $\bb_1$ and $\bb_2$ are linear combinations of $\bx$ and $\by$. So write $\bb_1 = a_1\bx + c_1\by$ and $\bb_2 = a_2\bx + c_2\by$.
Then:
$$
T\bb_1 = T(a_1 \bx + c_1 \by)\\
= a_1 T\bx + c_1T\by\\
= a_1 \mathbf{0} + c_1 \alpha \bx\\
= c_1\alpha\bx
$$
And similarly
$$
T\bb_2 = c_2 \alpha \bx
$$
So the two columns of $T$ are both multiples of $\bx$. 
I'm not exactly sure if this is what the question is looking for, but it is an answer.
