For $n\times n$ matrix A, if $\|A\|_\infty=O(1/n)$, can I deduce $\|A^{-1}\|_\infty=O(n)$? Here, $\|A\|_{\infty}=\max _{1 \leq i \leq n} \sum_{j=1}^{n}\left|a_{i j}\right|$.
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2No. Imagine a $2\times 2$ diagonal matrix with entries $1/n$ and $e^{-n}$, say. – Berci Dec 07 '20 at 13:10
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Berci's exmaple shows that even with $\Vert A \Vert \in\Theta(1/n)$, it doesn't work (which is more interesting, cause your statement with $O(1/n)$ doesn't even hold for real functions). But you may be interested in this question that asks for an upper bound on the norm of an inverse matrix. – Milten Dec 07 '20 at 13:22
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@Milten I'm sorry but what does $|A|\in \Theta(1/n)$ mean? And thank you for your posted question to let me understand that I should start from the condition number. – Patrick Star Dec 07 '20 at 13:43
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@PatrickStar See https://en.wikipedia.org/wiki/Big_O_notation – Milten Dec 07 '20 at 18:33