Invertibility of $(\textbf{A}^T\textbf{A}+\epsilon \textbf{I})$? I'm given a problem:
$\sigma_1 \geq \sigma_2 \geq ... \geq \sigma_r$ are the nonzero singular values of $\textbf{A}\in\mathbb{R}^{M\times N}$. If $\epsilon \neq 0$ is a real scalar, s.t. $|\epsilon| < \sigma^{2}_r$, show that $(\textbf{A}^T\textbf{A}+\epsilon \textbf{I})$ is invertible.
I found the resources Why is $A^TA$ invertible if $A$ has independent columns? and Matrix inverse of $A + \epsilon I$, where $A$ is invertible
But I'm not sure how useful they are. The first is in the case where A has independent columns, which is not necessarily true here, and the second presumes A is invertible. I believe that $\textbf{A}^T\textbf{A}$ is invertible by definition, but I'm not sure if I can just plug $\textbf{A}^T\textbf{A}$ in everywhere that post uses A and follow through. That also wouldn't help me understand the problem, just blindly substitute into a solution.
Can anyone help me understand WHY $(\textbf{A}^T\textbf{A}+\epsilon \textbf{I})$ is invertible? And/or point me in the right direction to construct a proof of it?

You can show that $A^\top A$ is positive semi-definite (specifically, that it has nonnegative eigenvalues $\sigma_1^2, \ldots, \sigma_r^2, 0, \ldots, 0$).
[It is not always invertible. Specifically, if some of its eigenvalues are zero, then it is not invertible.]
Knowing this fact about $A^\top A$, can you explicitly write down the eigenvalues of $A^\top A + \epsilon I$? What values of $\epsilon$ make this matrix invertible or non-invertible?
