Rate at which $\kappa(n)$ approaches $1$

Description of constant

Given a real matrix $A$, let its condition number be \(\kappa(A):=\frac{\sigma_{\max}(A)}{\sigma_{\min}(A)},\) where $\sigma_{\min}(A)$ and $\sigma_{\max}(A)$ denote the smallest and largest singular values of $A$, respectively (with $\kappa(A)=\infty$ if $\sigma_{\min}(A)=0$). [AJM2025-def-kappaA]

For each positive integer $n$, let \(\kappa(n):=\min_{A\in\{\pm 1\}^{n\times n}}\kappa(A).\) [AJM2025-def-kappan]

Equivalently, $\kappa(n)=1$ precisely when there exists a Hadamard matrix of order $n$. [AJM2025-kappan-hadamard]

Problem 11 of [AJM2025] asks for the optimal decay exponent of $\kappa(n)-1$:

Problem 11. What is the largest $\alpha$ for which \(\kappa(n)=1+\frac{f(n)}{n^\alpha}\) for some subpolynomial $f$? [AJM2025-prob11]

Define $C_{23b}$ to be the largest $\alpha$ for which such a representation holds. [AJM2025-prob11]

Known upper bounds

Bound on $\alpha$ Reference Comments
$1$ [AJM2025] Stated in the discussion of Problem 11. [AJM2025-prob11] (A supporting mechanism is the lower bound $\kappa(n)\ge 1+c\frac{\log n}{n}$ for $n\not\equiv 0 \pmod 4$.) [AJM2025-thm6]

Known lower bounds

Bound on $\alpha$ Reference Comments
$0$ Trivial Take $\alpha=0$ and $f(n)=\kappa(n)-1$. Since $\kappa(n)\to 1$ as $n\to\infty$, $f$ is bounded (hence subpolynomial). [AJM2025-abstract-kappa-to-1]
$17/92 \approx 0.18478$ [AJM2025] The authors state that (unconditionally) their method permits $\kappa(n)\le 1+\frac{1}{n^\alpha}$ for all sufficiently large $n$ with $\alpha=17/92-\delta$ for any small $\delta>0$. [AJM2025-thm1-alpha]

References

Contribution notes

Prepared with assistance from ChatGPT 5.2 Pro.