Rank of a linear transformation $T(x_1,x_2)=(x_1-x_2,5x_1)$ 
Find the associated matrix and compute the rank and nullity of the linear transformation $T:\mathbb{R}^2\to\mathbb{R}^2$ given by $T(x_1,x_2)=(x_1-x_2,5x_1)$.

The associated matrix $A$ is
$$
\left[\begin{matrix} T(\mathbf{e_1}) & T(\mathbf{e_2}) \end{matrix}\right]=\left[\begin{matrix} 1 & -1 \\ 5 & 0 \end{matrix}\right]
$$
To compute the nullity we must find vectors $\mathbf{x}$ such that $A\mathbf{x}=\mathbf{0}$. So by applying elementary row operations to the augmented matrix
$$
\left[\begin{matrix} 1 & -1 & 0 \\ 5 & 0 & 0 \end{matrix}\right] \to\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}\right]
$$
Thus we have $\mathbf{x}=\left[\begin{matrix} 0 \\ 0 \end{matrix}\right]$. And so the nullity of $T$ is $0$.
For the rank, is it enough to say that:


*

*$T(\mathbf{e_1})$ and $T(\mathbf{e_2})$ are linearly independent and so form a basis for the image of $T$. Hence the rank of $T$ is $2$?


EDIT: Just realised that we must have rank $T + $ nullity $T = 2$ so I must have a mistake somewhere.
EDIT: Corrected

To find the the rank, you can use the rank-nullity theorem: 
$$rank (T)+nullity(T)=2.$$
Since $nullity(T)=0$ as you have found, we have $rank (T)=2$. 
Or you can argue by saying that the columan space of $T$, $C(T)$, is spanned by $T(e_1)=\left[\begin{matrix} 1  \\  5 \end{matrix}\right]$ and $T(e_2)=\left[\begin{matrix} -1  \\  0 \end{matrix}\right]$, which implies that $C(T)=\mathbb{R}^2$. Therefore, 
$rank(T)=\dim C(T)=2$. 
Or you can look at the associated matrix $A=\left[\begin{matrix} 1 & -1 \\ 5 & 0 \end{matrix}\right]$, which has rref form given by $\left[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right]$ as you have done. So the rank of $A$ is the number of nonzero rows (or the number of pivots) in its rref form, which is $2$.
