0
$\begingroup$

Let $H=H_n$ be a positive definite Hankel matrix of size $n$ with $\lambda_n$ is it's smallest eigenvalue. What bounds are known on $\lambda_n$ in terms of the entries on $H$.
I can see some results in terms of an underlying distribution and the entries of $H$
being moments over this distribution. What i'm looking for is a bound in terms of the entries of the matrix. If the bound is stated in terms of moments please give an explanation as to how to construct the distribution given the matrix.

$\endgroup$
2
  • 3
    $\begingroup$ For any positive definite $H$ with eigenvalues $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n > 0$ you can compute $\Delta := \det H = \prod_k \lambda_k$ and $t := \mathop{\rm tr} H = \sum_k \lambda_k$ and deduce a lower bound on $\lambda_n$ from the AM-GM inequality: $$ \lambda_n = \Delta \, \big/ \, \prod_{k=1}^{n-1} \lambda_k \geq \Delta \, \big/ \, \bigl(\frac1{n-1} \sum_{k=1}^{n-1} \lambda_k)\bigr)^{n-1} = \Delta \, \big/ \, \bigl((t - \lambda_n) / (n-1)\bigr)^{n-1}. $$ I doubt that the Hankel assumption will let you do improve on this by much if at all. $\endgroup$ Jun 15, 2015 at 18:57
  • $\begingroup$ Thanks, however there seems to be a lot of works dealing with Hankel matrix $\lambda_n$ specifically. The conclusions there are in terms of moments. I was hoping for an explanation building on these results. $\endgroup$
    – gil
    Jun 16, 2015 at 13:49

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.