Preliminaries: The Frobenius of a matrix $A \in \mathbb{R}^{n \times n}$ is defined by
$$\|A\|_F := \sqrt{\mathrm{tr}(A^TA)}$$
i.e. the usual Euclidean norm of $A$ if we consider $A$ as a vector of length $n \times n$. The spectral radius of $A$ is defined by
$$\rho (A) := \max \{ |\lambda| \mid \lambda\text{ is an eigenvalue of }A\}. $$
Well-known statement: $\rho (A) \le \|A\|_F$.
My question: What is the smallest $c > 0$ such that $\|A \|_F \le c. \rho(A)$ for all matrices $A$?
I tried to look up several sources and they only mention the lower bound (as I stated in the well-known statement), not the upper bound of $\|A\|_F$. Any help would be greatly appreciated.
Edit: Thank you for your comment and answer. I changed the notation for the spectral radius.
Best Answer
There is no such nontrivial upper bound. For example if $A$ is nilpotent then its spectral radius (I would not use the notation $\sigma(A)$ for this but something like $\rho(A)$) is $0$ but its Frobenius norm can be nonzero.
On arbitrary matrices the spectral radius is not a norm (since, as we've seen, it can vanish on a nonzero matrix) so we don't expect it to be able to bound other norms. On the other hand we can say the following. The Frobenius norm can be written in terms of the singular values $\sigma_k(A)$ as
$$\| A \|_F = \sqrt{\sum \sigma_k(A)^2}.$$
which gives
$$\| A \|_F \le \sqrt{n} \sigma_1(A)$$
where $\sigma_1(A)$ is the largest singular value, which agrees with the operator norm of $A$ (with respect to the $\ell^2$ norm on vectors), and taking $A = \lambda I$ to be a scalar multiple of the identity shows that this bound is tight.
We also get the stronger lower bound $\sigma_1(A) \le \| A \|_F$ which implies the bound you give on the spectral radius since $\rho(A) \le \sigma_1(A)$. This last inequality is an equality if $A$ is symmetric (in which case more generally $\sigma_k(A)$ is the absolute value of the $k^{th}$ largest eigenvalue of $A$ in absolute value) but not in general.