The differential of the <log-determinant> is $d\log\det\Omega=\operatorname{Tr}(\Omega^{-1}d\Omega)$. The <subdifferential> of the entrywise $\ell^1$ norm consists of symmetric matrices $Z$ with
$$
Z_{ij}=\operatorname{sgn}(\Omega_{ij})\quad\hbox{if }\Omega_{ij}\ne0,
\qquad Z_{ij}\in[-1,1]\quad\hbox{if }\Omega_{ij}=0.
$$
The <Karush-Kuhn-Tucker conditions> for the <Graphical Lasso> are therefore
$$
-\widehat\Omega^{-1}+S+\lambda Z=0,
\qquad Z\in\partial\lVert\widehat\Omega\rVert_{1,\mathrm{entry}}.
$$
Because $-\log\det\Omega$ is <strictly convex> on the <positive-definite matrices>, these conditions characterize the unique minimizer whenever it exists.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 205 / 2 / b / Solution

{title2=$\log\det A$}

On <positive-definite matrices>, the log-determinant has differential $d\log\det A=\operatorname{Tr}(A^{-1}dA)$ and is a <strictly concave function>.

= Determinants
{synonym}

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 Incoming links: Log-determinant