Show that $$f(x) = \ln \det \left( \mbox{diag}(x)^{-1} + A \right)$$ is convex when $x \in \mathbb R^n_{++}$ and $A$ is given and positive definite.
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2026-03-26 12:08:26.1774526906
Convexity of $\ln \det(\mbox{diag}(x)^{-1} + A)$ with $A$ given and positive definite.
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One can use that the function $$ X \mapsto \log\det( I + X^{-1})$$ is convex on symmetric, positive definite $X$, see here.
Indeed, $$ \log\det( x^{-1} + A) = \log\det( A^{-1} \, x^{-1} + I) + \log\det(A) = \log\det( (x \, A)^{-1} + I) + \log\det(A), $$ where $x^{-1}$ short for $\operatorname{diag}(x)^{-1}$.