Congruence of a matrix

113 Views Asked by Bumbble Comm At 27 Mar 2026 - 9:35

Let $X=\begin{bmatrix}a&b&0&0\\ c&d&0&0\\0&0&-a&-b\\0&0&-c&-d\end{bmatrix}$ where $a,b,c,d\in \mathbb{Z}$.

For a such given $X$, is there a $4\times 4$ matrix $P$ over $\mathbb{Z}$-coefficient satisfying the following?

$\operatorname{det}(P)=\pm 1$
$PXP^T=\begin{bmatrix} O&A\\B&O\end{bmatrix}$ where $A, B,O$ are $2\times 2 $ matrices and $O$ is the matrix with zero entries.

There are 1 best solutions below

Bumbble Comm On 27 Dec 2013 - 11:14

Try $$ P=\frac{1}{\sqrt{2}}\left(\begin{matrix} I&I\\ I&-I \end{matrix}\right), $$ where $I$ is the $2\times 2$ unit matrix.

It should be $P^{-1}=P$ and $PXP=\left(\begin{matrix} 0&A\\ A&0 \end{matrix}\right)$, with $A=\left(\begin{matrix} a&b\\ c&d \end{matrix}\right)$.