I found that \begin{align*} \max_{|z_{mn}|=1} \sigma_2\left( \begin{bmatrix} z_{11} & z_{12} & z_{13}\\ z_{21} & z_{22} & z_{23}\\ z_{31} & z_{32} & z_{33} \end{bmatrix}\right)=2 \end{align*} by numerical simulation and found that the equality holds when \begin{align*} \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & -1\\ 1 & -1 & 1 \end{bmatrix}, \end{align*} where $\sigma_2(\cdot)$ is the second largest singular value of the matrix. I simulated with \begin{align*} \max_{|z_{mn}|=1} \sigma_2\left( \begin{bmatrix} 1 & 1 & 1\\ 1 & z_{22} & z_{23}\\ 1 & z_{32} & z_{33} \end{bmatrix}\right) \end{align*} in brute-force manner: divide the unit circle uniformly into 200 points (it is suffice to let entries to one since singular value is unitarily invariant). I want to prove it. Is there any keywords or idea related to this? Thank you.
2026-03-25 12:48:41.1774442921
Singular value of matrices whose entries are on the unit circle
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As you wrote, it is sufficient to consider the matrices $A_{a,b,c,d}=\begin{pmatrix}1&1&1\\1&e^{ia}&e^{ib}\\1&e^{ic}&e^{id}\end{pmatrix}$ where $a,b,c,d\in[0,2\pi]$.
Let $spectrum(AA^*)=\{\sigma_1^2\geq \sigma_2^2\geq \sigma_3^2\}$. Note that $\sum_j\sigma_j^2=9$, and, in particular, that $\sigma_1\leq 3$ and $\sigma_3^2\leq 3$.
Let $p(x)$ be the characteristic polynomial of $AA^*$. If $p(4)=0$, then $\sigma_2^2=4$ or-and $\sigma_1^2=4$.
Many numerical computations seem to "show" that the OP's conjecture is true, but I have no rigorous proof. I propose the following attack plan -in $2$ steps-.
$\textbf{Conjecture 1}$. If $\sigma_2=2$, then $\sigma_1=\sigma_2$. That is equivalent to show that
if $p(4)=0,p'(4)\leq 0$, then $p'(4)=0$.
$\textbf{Ideas:}$ we can locally verify the result, using the Maple's software NLPSolve in order to study the optimization
Let $U=\{a,b,c,d;p(4)=0,p'(4)\le 0\}$; show that $\min_Up'(4)=0$.
In particular, the matrices $A_{a,b,c,d}$ satisfying $\sigma_2=2$ are s.t $(a,b,c,d)=(a,a+\epsilon\pi,a+\epsilon\pi,a)$, where $\epsilon=\pm 1$ (they depend on one parameter).
$\textbf{Conjecture 2}$. $\sigma_2\leq 2$. Since the inflection point of the curve $y=p(x)$ is obtained for $x=3$, that is equivalent to show that
if $p(4)>0$, then $p'(4)>0$.
$\textbf{Idea:}$ we can locally verify the result, using the Maple's software NLPSolve in order to study the optimization
Let $U=\{a,b,c,d;p(4)> 0\}$; show that $\min_Up'(4)\geq 0$.