I want a lower bound on the smallest eigenvalue of $AA^T$ in terms of the Frobenius norm of $A$

Question

Let $\lambda_n$ be smallest eigenvalue of the square matrix $AA^T$. How could I lower bound this in terms of Frobenius norm of $A$, i.e., getting a constant $c$ such that

$${\lambda _n}\left( {A{A^T}} \right) \ge c\left\| A \right\|_F^2$$

the square on the right hand side is a mere prediction. The matrix $A$ is not assumed to be symmetric. Assume also A is non-singular

Answer 1

The only such constant is $c = 0$. Consider $$ A = \pmatrix{1&0\\0&\epsilon}, $$ which is non-singular for all $\epsilon \neq 0$. We have $$ \lambda_n(AA^T) = \epsilon^2, \quad \|A\|_F^2 = 1 + \epsilon^2. $$ Rearranging your inequality yields $$ \frac{\lambda_n(AA^T)}{\|A\|_F^2} \geq c \implies\frac{\epsilon^2}{1 + \epsilon^2} \geq c. $$ However, as $\epsilon \to 0^+$, the left-hand side approaches $0$.