The condition number of a nonsingular matrix $A$ is defined as $$ \kappa(A) = \mbox{cond}(A) = \Vert A \Vert \Vert A^{-1} \Vert \tag{1} $$ and its value is dependent on the choice of the norm in (1).
However, as all $p$-norms are equivalent in a finite-dimensional space, if a matrix $A$ is ill-conditioned in any $p$-norm, it will continue to be ill-conditioned in other $p$-norms also, where $p = 1, 2, \ldots..$ In fact, I also saw some results giving equivalence of $\Vert \cdot \Vert_1$ and $\Vert \cdot \Vert_\infty$ with some bounds.
My query is on numerical linear algebra, where I saw some results giving an estimate for $$ {1 \over \kappa(A)} $$ rather than $\kappa(A)$.
As we use ill-conditioned matrices (like Hilbert's matrix) for test matrices, $\kappa(A)$ will be quite large. Is my reasoning correct that numerical estimates are given for ${1 \over \kappa(A)}$ rather than $\kappa(A)$ as ${1 \over \kappa(A)}$ (reciprocals) will be closer to zero for an ill-conditioned matrix $A$, while $\kappa(A)$ can get quite large. This helps the accuracy of the numerical estimate.
I checked the available commands in MATLAB and I find that MATLAB uses "rcond(A)" for the reciprocal of condition number of a matrix $A$ using $1$-norm. As MATLAB does not give the paper used for finding this estimate, I like to know why specially $1$-norm is used in this estimate, as the default value of "cond(A)" in MATLAB uses the spectral norm or the $2$-norm. [In MATLAB, if the value of "rcond(A)" is closer to zero, it indicates that $A$ is very close to a singular matrix, while if the value of "rcond(A)" is closer to one, it indicates that $A$ is well-conditioned.]
Your comments are welcome!