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The objective is to minimize the diagonal elements of a symmetric positive definite matrix. The expression of this matrix is a little bit nasty and its inverse is much easier to deal with.
Can I claim that minimizing the diagonal entries of this matrix is equivalent to maximizing those of its inverse ?

If it helps, the matrix is of the form:
$A = B - BC^{T}\Lambda^{-1}CB$.
Its inverse is of the following form:
$A^{-1}=B^{-1}+kC^{T}C$. For some known "$k$".
$B$ is a toeplitz symmetric positive definite matrix.
$\Lambda$ is a symmetric positive definite matrix.
$C$ is a $1\times2$ block matrix where the first part is diagonal and the second block is the all zero matrix. (To match the dimensions).

Thank you all for your time, your help is much appreciated.

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  • $\begingroup$ By "minimise the diagonal elements" do you mean minimise the sum of the diagonal elements? $\endgroup$
    – lemon
    Commented Jul 20, 2014 at 0:22
  • $\begingroup$ After some analysis, minimizing the diagonal entries individually is equivalent to minimizing the trace (in this special case). $\endgroup$ Commented Jul 20, 2014 at 0:24

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For symmetric positive definite $A$, $\exists$ $D$ with $D_{ii}\neq 0$ s.t. $A=PDP^{-1}$ and $A^{-1}=PD^{-1}P^{-1}$ where $D_{ii}\cdot(D^{-1})_{ii}=1$. So yes, minimising $A$ is equivalent to maximising $A^{-1}$.

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  • $\begingroup$ Oh I have no idea how I missed that although I was toying around this property a little bit :P. Thank you :) $\endgroup$ Commented Jul 20, 2014 at 1:01

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