Denote by $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bT}{\mathbb{T}}$ $\bT^N$ the  real torus

$$\mathbb{T}^N :=\bigl\lbrace\vec{z}\in\bC^N;\;\;|z_1|=\cdots =|z_N|=1\bigr\rbrace$$ 

To each $\newcommand{\vez}{\vec{z}}$ $\vez\in\bT^N$  we associate the $N\times N$ Vandermonde matrix  $V(\vec{z})$

$$V_{ij}(\vec{z})= z_j^{i-1}. $$

Now form the hermitian and   positive semi-definite matrix

$$ A(\vez)= V(\vez)^\ast \cdot V(\vez). $$

As is well known $A(\vez)$ is invertible if and only if $\vez$ is  *nondegenerate*, i.e., the components $z_j$ are  pairwise distinct.     In general

$$\dim \ker A(\vez) = N-\nu(\vez), $$

where $\nu(\vez)$ denotes  the  number of distinct elements in the list $(z_1,\dotsc,z_N)$. Denote by $\lambda_1(\vez)$ the smallest eigenvalue of $A(\vez)$. The  map

$$\bT^N\ni \vez\mapsto \lambda_1(\vez) \in [0,\infty) $$

is continuous and  semi-algebraic and vanishes  exactly when $\det A(\vez)$ vanishes, where we recall that

$$ \det A(\vez)=\prod_{j > k} |z_j-z_k|^2. $$

We deduce from the Lojasewicz's  inequality that there exists  a positive  rational number $r$  and a constant $C=C_r>0$ such that

$$ \lambda_1(\vez)\geq  C(\det A(\vez) )^r,\;\;\forall\vec{z}\in\bT^N. \tag{1}  $$

Observe that if $(1)$ holds for some $r$ and $C$,  it also holds for any  given $r'>r$ (with a different constant $C$).   Denote by $R$ the set of $r$'s for which $(1)$ holds,  and set $\rho:=\inf R$. Note   that

$$\lambda_1(\vez) \leq \bigl( \det A(\vez)\bigr)^{\frac{1}{N}}, $$

which shows that $\rho\geq \frac{1}{N}$.


> 1. Is it true that   $\rho=\frac{1}{N}$, i.e., for any $\varepsilon >0$ there exists $C=C_\varepsilon >0$ such that
> $$ \lambda_1(\vez) \geq C\bigl( \det A(\vez)\bigr)^{\frac{1}{N}+\varepsilon} ? $$
> 
> 2. Can one indicate  another explicit and notrivial lower  bound for
> $\lambda_1(\vez)$?