I know that given a $2N \times 2N$ block matrix with $N \times N$ blocks like
$$\mathbf{S} = \begin{pmatrix}
A & B\\
C & D
\end{pmatrix}$$
we can calculate $$\det(\mathbf{S})=\det(AD-BD^{-1}CD)$$ and so clearly if $D$ and $C$ commute this reduces to $\det(AD-BC)$, which is a very nice property.
My question is, for a general $nN\times nN$ matrix with $N\times N$ blocks where all of the blocks commute with each other, can we find the determinant in a similar way? That is, by first finding the determinant treating the blocks like scalars and then taking the determinant of the resulting $N\times N$ matrix.
Thanks in advance!
Best Answer
The answer is affirmative. See, e.g., theorem 1 of John Silvester, Determinants of Block Matrices.