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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!

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  • $\begingroup$ If $D$ is not invertible then what about this identity $\det(\mathbf{S})=\det(AD-BD^{-1}CD)$ $\endgroup$
    – Sry
    Jun 20, 2015 at 6:54

1 Answer 1

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The answer is affirmative. See, e.g., theorem 1 of John Silvester, Determinants of Block Matrices.

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  • $\begingroup$ Great! Thanks for the reference $\endgroup$
    – Gordon
    May 24, 2015 at 9:33

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