Answer
See below
Work Step by Step
Define $T:R^3 \rightarrow V$ by $T(a,b,c)=\begin{bmatrix}
0 & a & b\\
-a & 0 & c\\
-b & -c & 0
\end{bmatrix}$
Assume $x_1=(a,b,c),(d,e,f) \in R^3\\
\alpha \in R$
Obtain: $T((a,b,c)+(d,e,f)=T((a+d,b+e,c+f))\\
=\begin{bmatrix}
0 & a+d & b+e\\
-(a+d) & 0 & c+f\\
-(b+e) & -(c+f) & 0
\end{bmatrix}\\
=\begin{bmatrix}
0 & a & b\\
-a & 0 & c\\
-b & -c & 0
\end{bmatrix}+\begin{bmatrix}
0 & d & e\\
-d & 0 & f\\
-e & -f & 0
\end{bmatrix}\\
=T(a,b,c)+T(d,e,f)$
then $T(\alpha (a,b,c))=T(\alpha a,\alpha b, \alpha c)\\
=\begin{bmatrix}
0 & \alpha a & \alpha b\\
-\alpha a & 0 & \alpha c\\
-\alpha b & -\alpha c & 0
\end{bmatrix}\\
=\alpha \begin{bmatrix}
0 & a & b\\
-a & 0 & c\\
-b & -c & 0
\end{bmatrix}\\
=\alpha T(a,b,c)$
Hence, $T$ is a linear transformation.
Obtain: $Ker(T)=\{x \in R^3:T (x)=0\}\\
=\{(a,b,c) \in R^3: T(a,b,c)=0\}\\
=\{(a,b,c) \in R^3:\begin{bmatrix}
0 & a & b\\
-a & 0 & c\\
-b & -c & 0
\end{bmatrix}=\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}\\
\rightarrow a=b=c=0$
then $Ker(T)=\{(0,0,0)\}=\{0\}$
$T$ 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.
