Linear transformation of skew-symmetric block matrix

Question

Suppose that we have the following special skew-symmetric block matrix $$ J = \begin{bmatrix} J_1 & -A & -B\\ A^T & J_2 & -C \\ B^T & C^T & J_3 \end{bmatrix} \in R^{(2n+m) \times (2n+m)}$$ where $J_1,J_2, A \in R^{n \times n}$ , $B, C \in \mathbb{R}^{n\times m}$, $J_3 \in \mathbb{R}^{m\times m}$ and $\text{rank }B=m\leq n$. Furthermore $J_1, J_2$ and $J_3$ are skew-symmetric. Is there any nonsingular matrix $T\in \mathbb{R}^{(2n+m) \times (2n+m)}$ such that $$TJ=\begin{bmatrix} \tilde{J_1} & -\tilde{A} & -\tilde{B}\\ \tilde{A}^T & 0 & 0 \\ \tilde{B}^T & 0 & 0 \end{bmatrix}$$ where $\tilde{J}_1 \in \mathbb{R}^{n \times n}$ is skew-symmetric, $\tilde{A}\in \mathbb{R}^{n\times n}$, $\tilde{B} \in \mathbb{R}^{n\times m}$ and $\text{rank }\tilde{B}=m\leq n$? I hope you can give me a hint or any ideas. Thanks in advance.

Answer 1

This is not always possible. As $TJ$ contains the submatrix $\pmatrix{\tilde J&-\tilde B\\ \tilde B^T&0}$ and $\operatorname{rank}\tilde{B}=m$, the rank of $TJ$ is at least $2m$. Since $T$ is nonsingular, the rank of $J$ must also be at least $2m$. However, this condition is not always satisfied, despite $\operatorname{rank}B=m$.

For example, consider $m=n=2,\,J_1=J_2=K:=\pmatrix{0&-1\\ 1&0},\,B=I,A=C=0$ and $$ J=\pmatrix{K&0&-I\\ 0&0&0\\ I&0&K}. $$ Clearly $J$ is skew-symmetric. Since $K^2=-I$, the third block row of $J$ is just $-K$ times the first block row. Hence the rank of $J$ is precisely the rank of its first block row, which is only $m$.