My question follows from https://math.stackexchange.com/questions/3857976/inverse-inequality-of-symmetric-matrix. Suppose we assume that $A$ and $B$ are two positive definite matrices with positive entries and $A\geq B $ entry wise.
Can we say that $A^{-1}\leq B^{-1}$ entry wise?
I tried with numeric examples in Matlab, but I am not getting any counter-example.
Any help would be really great.
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1 Answer
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3
$$A=\left( \begin{array}{cc} 1 & \frac{1}{10} \\ \frac{1}{10} & 1 \\ \end{array} \right),\;\;A^{-1}=\left( \begin{array}{cc} \frac{100}{99} & -\frac{10}{99} \\ -\frac{10}{99} & \frac{100}{99} \\ \end{array} \right),\;\;B=\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)=B^{-1}$$ is a counter example.
answered Oct 19, 2020 at 14:30
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$\begingroup$ Can I take the liberty to ask you here itself, will the inequality hold true if we put any more conditions on A? Even a hint would be enough. $\endgroup$ Commented Oct 19, 2020 at 15:54
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3$\begingroup$ sorry, no idea why this inequality should hold in some general sense. $\endgroup$ Commented Oct 19, 2020 at 16:24
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$\begingroup$ Slightly more generally, suppose $A$ is a $2 \times 2$ positive definite symmetric matrix. The partial derivative of $(A^{-1})_{1,1}$ with respect to the entry $a_{1,2}$ is $2 a_{12} a_{22}/(\det A)^2$. Thus if $a_{12} > 0$, increasing $a_{12}$ (and keeping $a_{21}=a_{12}$) increases $(A^{-1})_{11}$ rather than decreasing it. $\endgroup$ Commented Oct 19, 2020 at 23:38