Answer
See below
Work Step by Step
Define $T:R^3 \rightarrow V$ by $T(a,b,c)=\begin{bmatrix}
a & b\\ 0 &c \end{bmatrix}$
Assume $(a,b,c);(d,e,f) \in R^3\\
\alpha \in R$
Obtain: $T((a,b,c),(d,e,f))=T(a+d,b+e,c+f)\\
=\begin{bmatrix}
a+d & b+e\\0 &c +f \end{bmatrix}\\
=\begin{bmatrix}
a & b\\ 0 &c \end{bmatrix}+\begin{bmatrix}
d & e\\ 0 & f \end{bmatrix}\\
=T(a,b)+T(c,d)$
then $T(\alpha (a,b,c))=T(\alpha a, \alpha b,\alpha c)\\
=\begin{bmatrix}
\alpha a & \alpha b\\ 0 &\alpha c \end{bmatrix}\\
=\alpha\begin{bmatrix}
a & b\\ 0 &c \end{bmatrix}\\
=\alpha T(a,b)$
Hence, $T$ is a linear transformation.
Obtain: $Ker(T)=\{x \in R^3:T (x)=0\}\\
=\{(a,b,c) \in R^2: T(a,b,c)=0\}\\
=\{ (a,b,c)\in R^2: \begin{bmatrix}
a & b\\ 0 &c \end{bmatrix}=\begin{bmatrix}
0 & 0\\ 0 &0 \end{bmatrix}\}\\
\rightarrow a=b=c=0$
then $Ker(T)=\{\begin{bmatrix}
0 & 0\\ 0 &0 \end{bmatrix}\}=\{0\}$
$T$ is one-to-one (1)
Apply Rank Nullity Theorem:
$\dim [Ker(T)]+\dim [Rng(T)]=\dim R^3\\
0+\dim [Rng(T)]=3\\
\dim [Rng(T)]=3$
$Rng(T) \subset V$ and $\dim V=3=\dim [Rng(T)]$
$T$ is onto (2)
From $(1),(2)$, $T$ is an isomorphism.
