Consider the $9\times 9$ matrix


$$M = \begin{pmatrix} i e\_3 \times{} & i & 0 \\ 
-i & 0 & -a \times{} \\
0 & a \times{} & 0 \end{pmatrix}$$


for some vector $a \in \mathbb R^3$, where $\times$ is the cross product.


It is claimed in [Fu and Qin - Topological phases and bulk-edge correspondence of magnetized cold plasmas](https://doi.org/10.1038/s41467-021-24189-3) that this can be reduced to the determinant of a $3\times3$ matrix, meaning


$$ \det(M-\omega ) = \det(N\_1-N\_2+N\_3)=0.$$


Here


$$N\_1 = \frac{aa^t}{\omega^2}, N\_2 =\frac{a^ta}{\omega^2}\operatorname{id},\text{ and }N\_3 = \begin{pmatrix} 1-\frac{1}{\omega^2-1} & i\frac{1}{\omega(\omega^2-1)} & 0 \\ -i\frac{1}{\omega(\omega^2-1)} &1-\frac{1}{\omega^2-1} & 0 \\ 0& 0 & 1-\frac{1}{\omega^2} \end{pmatrix}.$$


This determinant is stated as equation (1) versus the original $9\times9$ matrix is equation (11).


How can we derive the determinant of the $3\times 3$ matrix from the determinant of the $9\times 9 $ matrix without first expanding the determinant?



Reducing $9\times9$ determinant to $3\times3$ determinant
I think the simplest way to reduce $$A=M-\omega I$$ to a $3\times 3$ matrix is to use the Schur complement with respect to the $(\bar 2,\bar 2)$-elements of $A$,
\begin{align}
C = A/A\_{\bar 2,\bar 2} = A\_{2,2} - A\_{2,\bar 2} A\_{\bar 2,\bar 2}^{-1} A\_{\bar 2,2}.
\end{align}
Here, $\bar 2$ denotes the index complement of $2$, i.e., $\bar 2\equiv\{1,3\}$. We get
\begin{align}
C = -\omega(N\_1 - N\_2 + N\_3).
\end{align}
With
\begin{align}
\det A\_{\bar 2,\bar 2}=\omega^4(\omega^2-1)
\end{align}
and $\det A=\det A\_{\bar 2,\bar 2}\det C$ the result follows.


@Carlo: Thanks for posting the MMA notebook! Note that there is a small mistake in the definition of $N\_3$, which needs to be transposed.


