Are Frobenius norm of $A$ and $A^TA$ comparable in general?

Question

Suppose $A \in M_n(\mathbb R)$. No particular structure is assumed. I am trying to see whether $\|A\|_F$ and $\|A^T A\|_F$ comparable. More specifically, I would like to get some constant $c$ (independent of $A$) such that $c \|A^TA\|_F \ge \|A\|_F$. My guess the answer is no. If we let $\{\sigma_1, \dots, \sigma_s\}$ denote the singular values of $A$, apparently $\|A\|_F = \sum_{j=1}^s \sigma_j$ and $\|A^TA\|_F = \sum_{j=1}^s \sigma_j^2$. Without more assumptions, it seems we can never find such constant $c$.

Answer 1

Your intuition is spot on. For $\varepsilon > 0$, consider the matrix $$A_\varepsilon = \operatorname{diag}(\varepsilon, 0, 0, \ldots, 0) \in M_n(\mathbb{R}).$$ Then $$A_\varepsilon^T A_\varepsilon = A_\varepsilon^2 = \operatorname{diag}(\varepsilon^2, 0, 0, \ldots, 0).$$ Note that $\|A_\varepsilon\| = \varepsilon$ and $\|A_\varepsilon^T A_\varepsilon\| = \varepsilon^2$, so if such a $c$ existed, we'd have $$c \varepsilon^2 \ge \varepsilon \implies c \ge \frac{1}{\varepsilon}.$$ But $\varepsilon$ can be made arbitrarily small, which means $c$ must be arbitrarily large.