Answer
True
Work Step by Step
Define a linear transformation $T:V\rightarrow M_{32}(R)$
obtain: $T(\begin{bmatrix}
a & b & c \\ 0 & d & e \\ 0 & 0 & f
\end{bmatrix})=\begin{bmatrix}
a & b\\ c & d \\ e & f
\end{bmatrix}$
then $Ker(T)=\{A \in V: T(A)=0\}\\
=\{\begin{bmatrix}
a & b & c \\ 0 & d & e \\ 0 & 0 & f
\end{bmatrix}:T(\begin{bmatrix}
a & b & c \\ 0 & d & e \\ 0 & 0 & f
\end{bmatrix})=\begin{bmatrix}
0 & 0\\ 0 & 0 \\ 0 & 0
\end{bmatrix}: a,b,c,d,e,f \in R\}\\
=\{\begin{bmatrix}
a & b & c \\ 0 & d & e \\ 0 & 0 & f
\end{bmatrix}):\begin{bmatrix}
a & b\\ c & d \\ e & f
\end{bmatrix}=\begin{bmatrix}
0 & 0\\ 0 & 0 \\ 0 & 0
\end{bmatrix}: a,b,c,d,e,f \in R\}\\
\rightarrow a=b=c=d=e=f=0\\
\rightarrow Ker(T)=\{\begin{bmatrix}
0 & 0 & 0\\0 & 0 & 0 \\ 0 & 0 & 0
\end{bmatrix}\}=\{0\}$
Hence, there is one-to-one linear transformation $T$
Find $Rng(T)=\{T(A): A \in V\}\\
=\{\begin{bmatrix}
a & b\\ c & d \\e & f
\end{bmatrix}: a,b,c,d,e,f \in R\}\\
\rightarrow Rng(T)=M_{32}(R)$
We can say $T$ is onto.
Hence, $V \cong M_{32}(R)$
