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 $\rho\geq \frac{1}{N}$.
Is it true that $\rho=\frac{1}{N}$, i.e., for any $\varepsilon >0$ there exists $C=C_\varepsilon >0$ such that
$$ \lambda_1(\vez) \geq C\bigl( \det A(\vez)\bigr)^{\frac{1}{N}+\varepsilon} ? $$Can one indicate another explicit and notrivial lower bound for
$\lambda_1(\vez)$?
Best Answer
The answer on the first question is no. Consider what happens if two points (say $z_1$ and $z_2$), approach each other. Then only one eigenvalue approaches zero (just by trivial rank consideration), and $\lambda_1\leq C\det A$.