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

Basis of R2 = {e1,e2}.where e1={1,0} and e2={0,1}.
T(x1,x2)=(x1−x2,5x1).
T(1,0)=(1,5).  T(0,1)=(-1,0).So the associated matrix is
A=[1−1;5 0] .The number of non-zero rows=2
              rank(A)=2
              rank(T)=2
Using the rank-nullity theorem,we have 
rank(T)+nullity(T)=2.
which implies nullity(T)=0.
