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Consider the following function:

$f(K)= y^{T} K y-log(det(K)) $

Where $y \in R^n $ is a constant vector. $K$ is a positive definite matrix that has the properties:

(1). The main diagonal entries of $K$ are 1, i.e $K_{i,i}=1 ; \forall i=1,2,..,n$

(2). Every non-diagonal entries of $K$ are positive but less than 1. i.e $0<K_{ij}<1 ; \forall i \neq j$.

How can I find the maximum and the minimum of $f$ in this case ? I guess we need to figure out the derivative We have: $$\frac{\partial f}{K}=yy^T -K^{-1} $$

So, if we want the derivative to be zero $K^{-1}=yy^T$ which is impossible, since $yy^T$ has rank 1. You can skip the constraint over the entries of $K$ for convenience.

Any thought is welcome.

Phong Le
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1 Answers1

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I'm not sure how to get an analytic answer, but you can formulate this question as a convex optimization problem, since $f(K)$ is convex as are your constraints.

$$ \mathrm{minimize}\,\,f(K)\\ \mathrm{subject\,\,to\,\,} K_{ii}=1, K_{ij} \in ]0,1[, K = K^T > 0 $$

John
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