Parity of a generalized characteristic polynomial Let
$$Q(\lambda)=\det\begin{pmatrix}-\lambda C_{11}&C_{12}\\C_{21}&-\lambda C_{22}\end{pmatrix}$$
where $C_{ij}$ are (not necessary square) matrices.
In this answer, it is claimed that $Q(\lambda)$ has a parity,i.e. $Q(\lambda)=\pm Q(-\lambda)$, but no explanation is given (it is probably obvious for some reason which currently eludes me).
Why does $Q(\lambda)$ have a parity?

Alternatively, suppose $C_{11}$ is $m \times k$, so that $C_{22}$ is $(n-m) \times (n-k)$ where the overall matrix is $n \times n$. Then
$$
\pmatrix{-\lambda C_{11}&C_{12}\\ C_{21}&-\lambda C_{22}}
=\pmatrix{-I_m\\ &I_{n-m}}
\pmatrix{\lambda C_{11}&C_{12}\\ C_{21}&\lambda C_{22}}
\pmatrix{I_k\\ &-I_{n-k}}.
$$
Taking determinants on both sides, you get $Q(\lambda)=\pm Q(-\lambda)$, where the sign of the "$\pm$" is given by $\det(-I_m)\det(-I_{n-k})=(-1)^{m+n-k}$.
