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