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?
 A: Suppose $C_{11}$ is $m \times k$, so that $C_{22}$ is $(n-m) \times (n-k)$ where the overall matrix is $n \times n$.
Consider the Leibniz formula.  Each term has $n$ factors that are matrix elements, one factor in each row and one factor in each column.  If a term has $a$ factors in the top left block, it must have $m-a$ in the top right, $k-a$ in the bottom left, and $n-m-k+a$in the bottom right.  That's a total of $n-m-k+2a$ factors with a $\lambda$, which is congruent to $n-m-k$ mod $2$.  Thus if $n-m-k$ is odd, each term has $\lambda$ to an odd power, which makes $Q(\lambda)$ an odd function, while if $n-m-k$ is even, each term has $\lambda$ to an even power, and $Q(\lambda)$ is an even function.
A: 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 $\det(-I_m)\det(-I_{n-k})=(-1)^{m+n-k}$.
