Lower bound of determinant of a Gram matrix given its main diagonal elements I encounter with this question where I want to find a positive lower bound of the determinant of a complex Gram matrix, say $\Omega$ with size $P \times P$. However, all I know is the main diagonal elements, $\Omega$ is a positive definite Hermitian matrix, i.e., the corrsponding vectors are indenpendent, and $\Omega$ is a diagonally dominant matrix. Also finding the eigenvalues of $\Omega$ is quite difficult. Is there something that may help me with this question? Thank you.
 A: Unless $\Omega$ is $1\times1$, there isn't any positive lower bound, because the infimum is zero. In fact, given any $\omega_{11},\omega_{22},\ldots,\omega_{nn}>0$, if we pick some $\epsilon\in(0,\sqrt{\omega_{11}})$ and define
$$
X=\pmatrix{\epsilon&\sqrt{\omega_{11}-\epsilon^2}\\ 0&\sqrt{\omega_{22}}}\oplus\operatorname{diag}(\sqrt{\omega_{33}},\ldots,\sqrt{\omega_{nn}}),
$$
then the Gram matrix $\Omega=XX^T$ is nonsingular and its diagonal is equal to $(\omega_{11},\omega_{22},\ldots,\omega_{nn})$, but $\det\Omega=(\det X)^2=\epsilon^2\prod_{j=2}^n\omega_{jj}$ can be arbitrarily close to zero.
A: Here is a different approach.
Let $a$ and $b$ be two parameters with 
$$0<a<b\tag{1}$$
Consider
$$X=\begin{pmatrix}a& 0& 0& 0& 0& b\\
 b& a& 0& 0& 0& 0\\
 0& b& a& 0& 0& 0\\
  0& 0& b& a& 0& 0\\
  0& 0& 0& b& a& 0\\
  0& 0& 0& 0& b& a \end{pmatrix}$$
with structure 
$$X=aI_6+bC_6\tag{2}$$ 
where $C_6$ is a circulant matrix.
One can easily prove that 
$$\det(X)=a^6-b^6$$
The Gram matrix associated with $X$ is 
$$G=X^TX=(aI+bC^T)(aI+bC)=a^{2}I+ab(C+C^T)+b^2\underbrace{CC^T}_{I}$$
i.e.,
$$G=\begin{pmatrix}
  a^2 + b^2&       ab&         0&         0&         0&       ab\\
        ab& a^2 + b^2&       ab&         0&         0&         0\\
          0&       ab& a^2 + b^2&       ab&         0&         0\\
          0&         0&       ab& a^2 + b^2&       ab&         0\\
          0&         0&         0&       ab& a^2 + b^2&       ab\\
        ab&         0&         0&         0&       ab& a^2 + b^2\end{pmatrix}$$
$G$ is strictly diagonally dominant (indeed $a^2+b^2 > 2ab$) with 
$$\det(G)=(a^6-b^6)^2 > 0$$
proving that $G$ is positive definite.
This determinant can be made arbitrarily small ; therefore, the smallest eigenvalue of $G$ can itself be made arbitrarily small. One can exhibit the smallest eigenvalue ; in fact, the spectrum of $G$ can be sorted in the general case in this way :
$$ \underbrace{(a - b)^2}_{\text{smallest eigenvalue}} \ < \ a^2 - ab + b^2 \ < \ a^2 + ab + b^2 \ < \ (a + b)^2 \tag{3} $$
(the two central eigenvalues being with multiplicity 2).
Remark : 
We have chosen here to work with a $6 \times 6$ matrix for didactic reasons, but the results just shown can be extended in a straightforward manner to a $(n+1) \times (n+1)$ matrix with the same structure (see (2)). In this general case, we still continue to deal with eigenvalues of a circulant matrix, which are in the general case :
$$\lambda_k=a^2+b^2+2ab \cos\left(\frac{k \pi}{n}\right) \ \ \ \ (k=0,1,\cdots n)$$ 
A: Of course, as indicated in the other answers, there is no general lower bound. But maybe the following is useful:
Call $\lambda$ an eigenvalue of $G$ with smallest absolute value. If $G$ is the Gram matrix of some tuple $(e_1, \ldots, e_n)$, then
$$|\lambda| = \min_{(c_i) \in \mathbb C^n - \{0\}} \frac{\left\Vert \sum_{i=1}^n c_i e_i \right \Vert}{\sqrt{\sum_{i=1}^n |c_i|^2}} \,.$$
So if somehow you are in a situation where you can bound the norms $\left\Vert \sum_{i=1}^n c_i e_i \right \Vert$ from below (think of this as bounding the distance from the origin to an ellipsoid) then you can bound $\det(G)$ from below.
