Finding the dimension of vector space Let $J=
        \begin{pmatrix}
        0 & 1 & -9 \\
        -1 & 0 & 4 \\
        9 & -4 & 0 \\
        \end{pmatrix}
$, $V=\{A \in M_{3 \times 3}(\mathbb R)|AJ=JA^T \}$.
Find the dimension of V. Justify your answer.
I already know the answer of this question: the dimension of $V$ is 6.(This is not sure, since I solved it a long time ago.) I solved this question by letting $A=
        \begin{pmatrix}
        a_1 & a_2 & a_3 \\
        b_1 & b_2 & b_3 \\
        c_1 & c_2 & c_3 \\
        \end{pmatrix}
$ and just substituting into $AJ=JA^T$, but this solution is somewhat messy, since it includes system of linear equations with 9 variables.
Is there more elegant solution for this problem without having to deal with such complicated equations?

Notice that $J^T = -J$. For every $A \in V$ it holds $AJ = JA^T = -J^TA^T = - (AJ)^T$.
Therefore $AJ$ is antisymmetric and we have
\begin{equation}AJ=
\begin{pmatrix}
b_1 - 9 c_1 & b_2 - 9 c_2 & b_3 - 9 c_3\\
4 c_1 - a_1 & 4 c_2 - a_2 & 4 c_3 - a_3\\
9 a_1 - 4 b_1 & 9 a_2 - 4 b_2 & 9 a_3 - 4 b_3\\
\end{pmatrix}
\end{equation}
For $AJ$ to be antisymmetric the following system should be satisfied
\begin{align}
9a_2 -a_3 +4c_3 -4b_2 &= 0\\
9a_1 -4b_1+b_3-9c_3 &= 0\\
-a_1 +b_2 +4c_1-9c_2 &=0
\end{align}
EDIT: The diagonal elements should equal zero 
\begin{align}
b_1 - 9c_1 &=0\\
-a_2 +4c_2 &=0\\
9a_3 -4b_3 &=0 
\end{align}
Solving this system will give you the dimension of $V$.
