Linear Algebra Question concerning the trace of a symmetric positive definite matrix.

310 Views Asked by At

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.

1

There are 1 best solutions below

1
On BEST ANSWER

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 $A^{-1}$.