characterization of uniform ellipticity Let $B$ be a $n\times n$ matrix over $\mathbb{R}$ and define $A:=BB^*$. I read in a paper that the following two statements are equivalent:
(1) the matrix $A$ is uniformly elliptic; i.e. for all vectors $y\in\mathbb{R}^n$ and some constant $N>0$, it holds that $$y^*Ay \geq \frac{1}{N^2} |y|^2.$$
(2) for the same constant $N$, the Frobenius norm of the inverse of $B$ is bounded above: $$\text{tr}(B^{-1}(B^{-1})^*) \leq N^2.$$
Can anyone see why this is true?
 A: I'll help you with one direction; the other implication is done likewise.
Let's take $\det B\ne 0$ (otherwise the $A$ is not elliptic and $B^{-1}$ is not defined). $\|\cdot\|$ stands for Euclidean norm and the induced matrix norm.
For direct implication we take $N^2=n\|B^{-1}\|^2$.
$$\frac{y^\ast Ay}{\|y\|^2}\ge \inf_{y\ne 0}\frac{\|B^\ast y\|^2}{\|y\|^2}=\inf_{z\ne 0}\frac{\|z\|^2}{\| (B^{-1})^\ast  z\|^2}=\frac{1}{\|(B^{-1})^\ast\|^2}=\frac{n}{N^2}\ge \frac{1}{N^2}.$$
On the other hand, $\|(B^{-1})^\ast\|^2$ is the largest eigenvalue of $ B^{-1} (B^{-1})^\ast$, and the Frobenius norm $tr (B^{-1} (B^{-1})^\ast)$ is the sum of all eigenvalues of the same matrix, therefore $$tr (B^{-1} (B^{-1})^\ast)\le n \|(B^{-1})^\ast\|^2 =N^2.$$
A: The Frobenius norm of $B^{-1}$ is the square root of the sum of the squares of the reciprocals of the singular values of $B$.  And the eigenvalues of $A$ will be the squares of the singular values of $B$.  The best constant in (1) is the smallest singular value of $B$.
So there will be a relationship between the two $N$'s, but they won't be identical.  Let's call the best $N$ in (1) $N_1$, and the best $N$ in (2) $N_2$.
$$ N_2 = \left(\sum_{i=1}^n \sigma_i^{-2}\right)^{1/2} $$
$$ N_1 = \max_{1\le i \le n} \sigma_i^{-1} .$$
Hence $N_1 \le N_2 \le \sqrt n N_1$.
