Finding the dimension of vector space

Question

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?

Answer 1

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 $V$.