A positive definite matrix fluctuation problem

Question

I have a positive definite matrix $A=(a_{ij})$ of dimension $n$ with all entries positive. Then the matrix flutuated to another positive definite matrix $A^*=(a^*_{ij})$ with the change in each entry bounded by $\left|a_{ij}-a^*_{ij}\right|\leq \epsilon a_{ij}$, where $\epsilon$ is a small constant. I want to derive the upper bound of the 2-norm of matrix $A^{-1}A^*$ by some expression of $\epsilon$. Is that possible?

Answer 1

Assuming "2 norm" means operator 2 norm, and defining the second matrix as $A'$ since $A^*$ universally refers to the adjoint in this setting,

$A' = A +B$
where you have an epsilon bound on $\vert b_{i,j}\vert$

Let $A$ have singular values $\sigma_1\geq \sigma_2\geq...\geq \sigma_n$
and let $R_A$ be the maximum L1 norm amongst the rows of $A$ and $C_A$ be the maximum L1 norm amongst the columns of $A$
$$ \begin{align} &\Big\Vert A^{-1}A'\Big\Vert_2\\ &=\Big\Vert A^{-1}\big(A+B\big)\Big\Vert_2 \\ &= \Big\Vert I+A^{-1}B\Big\Vert_2\\ &\leq \Big\Vert I\Big\Vert_2+\Big\Vert A^{-1}B\Big\Vert_2\\ &= 1+\Big\Vert A^{-1}B\Big\Vert_2\\ &\leq 1+\sigma_n^{-1}\cdot\Big\Vert B\Big\Vert_2\\ &\leq 1+\sigma_n^{-1}\cdot \sqrt{R_B \cdot C_B} \\ &\leq 1+\sigma_n^{-1}\cdot \sqrt{R_A\epsilon \cdot C_A\epsilon}\\ &= 1+\sigma_n^{-1}\cdot\epsilon \cdot \sqrt{R_A\cdot C_A} \\ \end{align} $$ note that as $\epsilon \to 0$ this gives an upper bound of 1 which is tight as $\Big\Vert A^{-1}A\Big\Vert_2 =1$

The key inequality used is that $\Big\Vert X \Big \Vert_2 \leq \sqrt{R_X\cdot C_X}$
which is the Schur Test and is provable by Cauchy-Schwarz