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

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:

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

• Perhaps you can use $|\det(\Sigma)| \leq \|\Sigma\|_F^n$ Jan 18, 2016 at 2:11
• Thanks! From that, I got: $||\Sigma||_{F}^{-n} \le \text{det}\left(\Sigma\right)^{-1} \le ||\Sigma^{-1}||_{F}^{n}$. Which doesn't give me an upper bound to $||\Sigma^{-1}||_{F}$ yet, but I think it's closer...
– John
Jan 18, 2016 at 5:22
• @John Did you managed to solve this problem? Please share your solution if you have it. I'm looking for an upper bound too. Aug 2, 2017 at 16:09

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 $$\left\|M^{-1}\right\|_F \leq \sqrt{r}\left\|M^{-1}\right\|_2 = \frac{\sqrt{r}}{\sigma_{min}(M)}$$.