Consider a non-singular $A \in \mathbb{R}^{n\times n}$ with $\|\mathrm{col}_i(A)\| = 1$ ($i \in \{1,\ldots,n\}$), where $\mathrm{col}_i(A)$ is the $i$th column of $A$ and $\|\cdot\|$ is the Euclidean norm. Then, can we find an upper bound on $\|A^{-1}\|_{\infty}$?
Thanks!
PS: I find an upper bound $\|A^{-1}\|_{\infty} \leq \sqrt{n}$, but I am not sure if this bound is tight enough. The proof is as follows.
Let $B = A^{-1}$, and we have
$$\mathrm{row}_i(B) \mathrm{col}_j(A) = \begin{cases}1, & i = j,\\0, & i\neq j.\end{cases}$$
This means $\|\mathrm{row}_i(B)\| = 1$, which further implies $\|\mathrm{row}_i(B)\|_1 \leq \sqrt{n}$. Thus, $\|A^{-1}\|_{\infty} = \|B\|_{\infty} = \max_i \|\mathrm{row}_i(B)\|_1 \leq \sqrt{n}$.
Since $A$ can be arbitrarily close to a singular matrix, the norm of its inverse is not bounded above. E.g. $$ \left\|\pmatrix{1&\cos t\\ 0&\sin t}^{-1}\right\|_\infty =\left\|\pmatrix{1&-\frac{\cos t}{\sin t}\\ 0&\frac{1}{\sin t}}\right\|_\infty\ge\frac{1}{|\sin t|}\to\infty $$ as $t\to0$.