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I wonder how to express the determinant of a block covariance matrix. For example, I have a covariance matrix

$\Sigma=\left[ \begin{array}{cc} \Sigma_1 & \Sigma_{12} \\ \Sigma_{21} & \Sigma_2 \\ \end{array} \right]$

where $\Sigma_1\in\mathbb{R}^{n_1}\times\mathbb{R}^{n_1}$, $\Sigma_2\in\mathbb{R}^{n_2}\times\mathbb{R}^{n_2}$ and hence $\Sigma_{12}=\Sigma_{21}^\intercal\in\mathbb{R}^{n_1}\times\mathbb{R}^{n_2}$.

Can I express the denterminant of $\Sigma$ in terms of the 3 blocks? That is:

$|\Sigma|=f(\Sigma_1,\Sigma_2,\Sigma_{12})$

I guess it is somehow related to Schur Complement, I am still studying..

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closed as off-topic by Suvrit, Chris Godsil, Stefan Waldmann, Stefan Kohl, Willie Wong Oct 28 '14 at 9:18

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A well-known block determinant formula: if $\Sigma_1$ is invertible, $$\det \Sigma = \det(\Sigma_1) \det(\Sigma_2 - \Sigma_{21} \Sigma_1^{-1} \Sigma_{12})$$ (and a similar formula if $\Sigma_2$ is invertible).

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  • $\begingroup$ This is called Schur complement formula. The Schur complement of $\Sigma_1$ in $\Sigma$ is precisely $\Sigma_2-\Sigma_{21}\Sigma_1^{-1}\Sigma_{12}$. $\endgroup$ – Denis Serre Jul 11 '17 at 8:11

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