Let $f : H_{n\times n} (\mathbb{C}) \rightarrow \mathbb{R}$ be the function on Hermitian, positive semidefinite matrices $f(A) = \frac{M_i (A)}{\det(A)}$ where $M_i(A)$ is the determinant of the the submatrix formed by the $A_{kl}$ where $1\leq k, l \leq i$, i.e., it is the $i\times i$ principal minor of $A$. Is the function $f$ convex ?
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2 Answers
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From Theorem 9, of this article it follows that $A \mapsto \log\frac{M_i(A)}{\det(A)}$ is convex on the set of positive definite matrices. The alleged convexity in the OP is a simple consequence of this stronger log-convexity result.
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2$\begingroup$ More easily, this also follows from Jacobi's identity: $\det(A[S]) = \det(A)\det(A^{-1}[\bar{S}]$, where $A[S]$ is the principal submatrix index by $S$, which $\bar{S}$ denotes the complement of $S$. $\endgroup$– SuvritCommented Mar 9, 2015 at 16:22
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The answer is No.
If you consider only diagonal matrices then the question is equivalent to whether the function $B(x_{1},\ldots, x_{n})=\prod_{j=1}^{m}x_{j}^{-1}$ is convex function over $\mathbb{R}^{m}$ (for $1 \leq m \leq n-1$).
For example the function $B(x,y)=\frac{1}{xy}$ is concave function in the domain $x \geq 0 $ and $y \leq 0$.
However, the function $f$ might be convex for positive semidefinite matrices.
answered Mar 9, 2015 at 3:10
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$\begingroup$ Yes. I forgot to add the assumption of semidefiniteness. $\endgroup$– VamsiCommented Mar 9, 2015 at 16:20