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Let $S$ be the set of real positive matrices, $\lambda>0$ and $f:S\rightarrow\mathbb{R}$ defined by $$f(X)=\langle X,X\rangle-\lambda\log\det(X) $$

where $\langle X,X\rangle=\operatorname{trace}(X^\top X)$. How can one show that $f$ is coercive?

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Let $\mu = \max \{\det X : \langle X,X\rangle=1, X\ge 0\}$. The homogeneity of determinant implies that $\log \det X\le \log \mu+\frac{n}{2}\log \langle X,X\rangle$ for all $X\ge 0$. Therefore, $$f(X)\ge \langle X,X\rangle -\lambda \log \mu - \frac{\lambda n}{2}\log \langle X,X\rangle $$ which is $\ge \frac12 \langle X,X\rangle$ when $\langle X,X\rangle$ is large enough.

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