Upper bounding the Frobenius norm of the inverse of a positive-definite symmetric matrix

Question

Let $\Sigma$ be a symmetric positive-definite $n \times n$ matrix. I want an upper bound on the Frobenius norm of $\Sigma^{-1}$ that does not involve calculating the determinant of $\Sigma$. The Frobenius norm for a generic $n\times n$ matrix $A$ with typical element $a_{ij}$ is

$\|A\|_F = \sqrt{\sum_{i=1}^n\sum_{j=1}^n a_{ij}^2}\,.$

I am looking for an upper bound on $\|\Sigma^{-1}\|_F$ that is a function of the elements of $\Sigma$.

Here's what I have so far. Let $\lambda_i$ be the eigenvalues of $\Sigma$ and $\sigma_{ij}$ be a typical element of $\Sigma$.

$\|\Sigma^{-1}\|_{F} = \sqrt{\text{tr}\left\{ \left(\Sigma^{-1}\right)^{\top}\Sigma^{-1}\right\}} = \sqrt{\text{tr}\left\{ \Sigma^{-1} \Sigma^{-1}\right\}} = \sqrt{\sum_{i=1}^{n} \frac{1}{\lambda_{i}^{2}}} \le \sum_{i=1}^{n} \frac{1}{\lambda_{i}} = \text{tr}\left\{\Sigma^{-1}\right\}$.

A few things that would work:

A lower bound on the minimal eigenvalue or an upper bound on the trace of the inverse of the matrix.

In this case I have found the following bound on the minimal eigenvalue:

$\lambda_{\text{min}} \ge \left(\frac{n-1}{\text{tr}\left\{\Sigma\right\}}\right)^{n-1} \times \det \Sigma$.

But it uses the determinant which is going to be tough to handle in my problem. It'd be great to get an upper bound on $\|\Sigma^{-1}\|_{F}$ that uses the trace of $\Sigma$, and/or $\|\Sigma\|_{F}$.

I know a lower bound because it's submultiplicative:

$\|I\|_{F} \le \left\|\Sigma^{-1}\right\|_{F} \| \Sigma \|_{F}$.

Answer 1

Perhaps you can use $\left\|M\right\|_F \leq \sqrt{r}\left\|M\right\|_2$ and $\left\|M^{-1}\right\|_2 = \frac{1}{\sigma_{min}(M)}$, where $M$ denotes any positive definite symmetric matrix, $r$ the rank of $M$ and $\sigma_{min}(M)$ the minimum singular value of $M$. Then one upper bound can be obtained by $\left\|M^{-1}\right\|_F \leq \sqrt{r}\left\|M^{-1}\right\|_2 = \frac{\sqrt{r}}{\sigma_{min}(M)}$.