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Let $A,B$ be positive definite matrices, and assume that $$ a_{i,j}<{b_{i,j}} $$ for all $1\leq i,j\leq n$, where $a_{i,j}$ is the $(i,j)$ element of the matrix $A$ and $b_{i,j}$ is the $(i,j)$ element of the matrix $B$. Is it true that $$ \text{trace}(A^{-1})\geq\text{trace}(B^{-1}). $$ ?

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No, it isn't true.

Take $A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, B = \begin{bmatrix} 2 & \sqrt{\frac{10}{3}} \\ \sqrt{\frac{10}{3}} & \frac{7}{3} \end{bmatrix}$. Trace of $B$ is $\frac{13}{3}$ and determinant of B is $\frac{4}{3}$, therefore its eigenvalues are $4$ and $\frac{1}{3}$, and $$trace(B^{-1}) = \frac{1}{4} + 3 > 2 = trace(A^{-1})$$

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Thanks you !!!! –  user1 Mar 8 '13 at 6:32

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