Answer
See below
Work Step by Step
Let $x=(x,y)\in R^2$
We have: $T(x)=ax\rightarrow T(x,y)=\begin{bmatrix}
-\frac{5}{3} & -\frac{2}{3}\\
\frac{1}{3} & \frac{1}{3} \\
\frac{5}{3} & -\frac{5}{3}\\
-1 & 1
\end{bmatrix}\begin{bmatrix}
x\\
y \end{bmatrix}=(-\frac{5}{3}x-\frac{2}{3}y,\frac{1}{3}x+\frac{1}{3}y,\frac{5}{3}-\frac{1}{3}y,-x+y)$
Obtain: $Ker (T)=\{x \in R^3: T(x)=0\}\\
=\{(x,y,z)\in R^3: T(x,y,z)=(0,0,0)\\
=\{(x,y,z)\in R^3:(x+3y+2z,2x+6y+5z)=(0,0)$
We have the system: $-\frac{5}{3}x-\frac{2}{3}y=0\\
\frac{1}{3}x+\frac{1}{3}y=0\\
\frac{5}{3}x-\frac{5}{3}y=0\\
-x+y=0\\\rightarrow x=y=0$
Thus, $Ker (T)=\{(0,0):z \in R\}
\rightarrow \dim [Ker (T)]=1$
Then, $Rng(T)=\{ T(x):x \in R^3 \}\\
=span \{(-\frac{5}{3},\frac{1}{3},\frac{5}{3},-1);(-\frac{2}{3},\frac{1}{3},-\frac{1}{3},1)\}\\
\rightarrow Rng(T)=span (\{ -\frac{5}{3},\frac{1}{3},\frac{5}{3},-1);(-\frac{2}{3},\frac{1}{3},-\frac{1}{3},1)\}$
