Answer
See below
Work Step by Step
a) Since $A$ is $3 \times 2$ matrix, $T:R^3 \rightarrow R^2$
Then $T(0,0,1)=T[\frac{2}{3}(1,0,0)+\frac{1}{3}(0,1,0)-\frac{1}{3}(2,1,-3)]\\
=\frac{2}{3}T(1,0,0)+\frac{1}{3}T(0,1,0)-\frac{1}{3}T(2,1,-3)\\
=\frac{2}{3}(4,5)+\frac{1}{3}(-1,1)-\frac{1}{3}(7,-1)\\
=(0,-4)$
The matrix is now $A=\begin{bmatrix}
4 & -1 & 0\\
5 & 1 & 4
\end{bmatrix}$
b) Obtain: $Ker(T)=\{x \in R^3: T(x)=0\}\\
=\{ x \in R^3: Ax=0\}\\
=\{x \in R: \begin{bmatrix}
4 & -1 & 0\\
5 & 1 &4
\end{bmatrix}\begin {bmatrix} x \\ y \\ z\end{bmatrix}=\begin {bmatrix} 0 \\ 0\end{bmatrix}\}\\
=\{x \in R: \begin{bmatrix}
4x-y\\
5x+y+4z
\end{bmatrix}=\begin{bmatrix}
0 \\ 0
\end{bmatrix}\}\\
=\{(x,4x,-\frac{9}{4}x): x\in R\}\\
=\{x(1,4,-\frac{9}{4}): x \in R\}\\
=span \{(1,4,-\frac{9}{4})\}$
Hence, $T$ is not one-to-one.
Apply Rank Nullity Theorem:
$\dim [Ker(T_2T_1)]+\dim [Rng(T)=\dim R\\
1+\dim [Rng(T)]=3\\
\dim [Rng(T)]=2$
$\dim R^2=2=\dim [Rng(T)]$
$\rightarrow T$ is onto.
