Assume $\lambda_\min(A_kA_k^{\rm T})>\varepsilon$ and show that $A_k^{\rm T}(A_kA_k^{\rm T})^{-1}$ is bounded

Question

For all $k\in\mathbb{N}$, let $A_k\in\mathbb{R}^{n\times m}$, where $n\leq m$, and assume that there exists $\varepsilon>0$ such that for all $k\in\mathbb{N}$, $\lambda_\min(A_kA_k^{\rm T})>\varepsilon$, where $\lambda_\min$ denotes the minimum eigenvalue of a symmetric positive semidefinite matrix. Can we show that $A_k^{\rm T}(A_kA_k^{\rm T})^{-1}$ is bounded?

For the case where $n=m$, I can see that $A_k^{\rm T}(A_kA_k^{\rm T})^{-1}$ is bounded. My question is about the case where $n<m$. Any hint is appreciated.

Answer 1

Let $A= U \Sigma V^T$ where $\Sigma$ has the same form as $A$. Then $A^T (A A^T)^{-1} = V \Sigma^T (\Sigma \Sigma^T)^{-1}U^T$.

We have $\Sigma^T (\Sigma \Sigma^T)^{-1} = \begin{bmatrix} \operatorname{diag}({1 \over \sigma_1},\cdots, {1 \over \sigma_n}) \\ 0 \end{bmatrix}$. Hence $\|A^T (A A^T)^{-1}\| = {1 \over \sigma_n} < { 1 \over \sqrt{\epsilon}} $.

Note that by assumption $\lambda_\min (A A^T) = \sigma_n^2 > \epsilon$.