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.$$\left[
\begin {array}{c c}
0&0\\
0&0
\end {array}\right]$$
 A: $\newcommand{\bx}{\mathbf{x}}\newcommand{\bb}{\mathbf{b}}\newcommand{\by}{\mathbf{y}}$ 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.
A: $T \vec x = 0; \; \; T\vec y = \alpha \vec x. \tag{0}$
Note that
$T^2\vec y = \alpha T\vec x = 0, \tag{1}$
and since
$T\vec x = 0 \tag{2}$
we have
$T^2\vec x = 0 \tag{3}$
as well; thus
$T^2 = 0. \tag{4}$
From (2), we see that $\vec x$ is an eigenvector with eigenvalue $0$; from (4), we see that $0$ is the only possible eigenvalue, since for
$T\vec w = \mu \vec w \tag{5}$
we have
$0 = T^2 \vec w = \mu^2 \vec w, \tag{6}$
forcing $\mu = 0$ since the eigenvector $\vec w \ne 0$ by definition.  If $\vec w = a\vec x + b\vec y$ were any other eigenvector, linearly independent from $\vec x$, 
$b\alpha \vec x = bT\vec y = aT\vec x + bT\vec y = T\vec w = 0, \tag{7}$
forcing $b\alpha = 0$.  If $\alpha = 0$, then $T\vec y = 0$ so $T = 0$, since by ($0$) it vanishes on the basis $\{\vec x, \vec y \}$.  If $\alpha \ne 0$, then $b = 0$, and $\vec w = a\vec x$ is not linearly independent from $\vec x$, a contradiction.  Thus if $T \ne 0$, $\alpha \ne 0$ and $\vec x$ is the only possible eigenvector up to scalar multiples; $\langle \vec x \rangle$ is the only possible eigenspace.  Since
$T\vec y = \alpha \vec x, \tag{8}$
$T(\alpha^{-1}\vec y) - 0(\alpha^{-1}\vec y) = T(\alpha^{-1}\vec y) = \vec x, \tag{9}$
i.e., $\alpha^{-1}\vec y$ is a generalized eigenvector corresponding to the eigenvalue $0$.  If we define $E$ to be the matrix whose two columns are $\vec x$ and $\alpha^{-1}\vec y$,
$E = [x, \alpha^{-1}\vec y], \tag{10}$
then $E$ is nonsingular since $\vec x$ and $\alpha^{-1}\vec y$ are linearly independent, so the inverse matrix $E^{-1}$ exists and satisfies
$I = E^{-1}E = E^{-1}[\vec x, \alpha^{-1}\vec y] = [E^{-1}\vec x, E^{-1}(\alpha^{-1}\vec y)], \tag{11}$
where $I$ is the $2 \times 2$ identity matrix.  Furthermore,
$TE = T[\vec x, \alpha^{-1}\vec y] = [T\vec x, \alpha^{-1}T\vec y] = [0, \vec x], \tag{12}$
and by (11),
$E^{-1}TE = E^{-1}[0, \vec x] = [0, E^{-1}\vec x] = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}; \tag{13}$
this shows that if $T \ne 0$, it is similar to the nilpotent matrix
$\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}; \tag{14}$
writing
$T = E\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}E^{-1}, \tag{15}$
we deduce that the general form of $T$ is either
$T = 0 \Leftrightarrow \alpha = 0 \tag{16}$
or (15) if $\alpha \ne 0$, where the columns of $E$ may be any two linearly independent vectors $\vec x$, $\alpha^{-1}\vec y$.
Note:  This answer was prepared before our OP user61681 edited in the four possible choices for the form of $T$; based upon what we have seen here, (3.) must be the correct choice; indeed, (13) shows that $T$ takes the form (14) in the $\vec x$, $\alpha^{-1} \vec y$ basis.  (Provided of course $\alpha \ne 0$.)  End of Note.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
