Let $\widehat\Omega^{(k)}$ solve the $k$th diagonal-block problem and set
$$
\widetilde\Omega=
\begin{bmatrix}\widehat\Omega^{(1)}&0\\0&\widehat\Omega^{(2)}\end{bmatrix}.
$$
Its inverse is block diagonal. On each diagonal block, the <Graphical-Lasso Karush-Kuhn-Tucker conditions> hold by the definition of $\widehat\Omega^{(k)}$. On the off-diagonal blocks choose
$$
Z^{(12)}=-S^{(12)}/\lambda,
\qquad Z^{(21)}=-S^{(21)}/\lambda.
$$
The assumed inequalities $|S_{ij}|\leq\lambda$ ensure that every entry lies in $[-1,1]$, exactly the allowed <subgradient> at a zero entry of $\widetilde\Omega$.

Thus $-\widetilde\Omega^{-1}+S+\lambda Z=0$ on every block. The KKT conditions and the fact that the objective is <strictly convex> prove that $\widetilde\Omega=\widehat\Omega$, giving the claimed block decomposition.

Solution

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

{c}

For objective $-\log\det\Omega+\operatorname{Tr}(S\Omega)+\lambda\lVert\Omega\rVert_{1,\mathrm{entry}}$, the KKT conditions are $-\Omega^{-1}+S+\lambda Z=0$ with $Z_{ij}\in\partial|\Omega_{ij}|$.


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