 # Consider the 4-by-4 matrix <mi mathvariant="bold-italic">M = <mi mathvariant="bold-i Audrina Jackson 2022-07-08 Answered
Consider the 4-by-4 matrix $\mathbit{M}={\mathbit{M}}_{0}+{\mathbit{M}}_{1}$, where
${\mathbit{M}}_{0}=\alpha \left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& -1& 0\\ 0& 0& 0& -1\end{array}\right)$ and ${\mathbit{M}}_{1}=\beta \left(\begin{array}{cccc}0& \gamma & 0& -{\gamma }^{\ast }\\ {\gamma }^{\ast }& 0& -{\gamma }^{\ast }& 0\\ 0& \gamma & 0& -{\gamma }^{\ast }\\ \gamma & 0& -\gamma & 0\end{array}\right)$
where $\alpha$ and $\beta$ are constants and $\gamma ={\gamma }_{x}+i{\gamma }_{y}$ is complex.
Is it possible to unitary transform $\mathbit{M}$ into block off-diagonal form ${\mathbit{M}}_{B}$?
Namely, I want to find a unitary transform $\mathbit{U}$ so that I can write down ${\mathbit{M}}_{B}=\mathbit{U}\mathbit{M}{\mathbit{U}}^{\ast }$ (here ${\mathbit{U}}^{\ast }$ is the conjugate transpose).
Explicitly, the required block off-diagonal matrix is (in general form)
${\mathbit{M}}_{B}=\left(\begin{array}{cc}0& \mathbit{Q}\\ {\mathbit{Q}}^{\ast }& 0\end{array}\right)$ where $\mathbit{Q}=\left(\begin{array}{cc}{Q}_{z}& {Q}_{x}-i{Q}_{y}\\ {Q}_{x}+i{Q}_{y}& -{Q}_{z}\end{array}\right)$
Is there a general recipe to find such a unitary transformation matrix $\mathbit{U}$ which leads to the block off-diagonal form, $\mathbit{M}\to {\mathbit{M}}_{B}$?
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I think this $4×4$ matrix is the answer.
$\frac{1}{\sqrt{2}}\left(\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ -I& 0& I& 0\\ 0& -I& 0& I\end{array}\right)$
###### Not exactly what you’re looking for? daktielti
What do you mean by "transform"? Are you looking for a similar matrix? If not, what are you trying to do? Are you trying to compute a determinant?