Prove $\|A\|_2 \le |s|$ if and only if $s^2I-A^TA$ is positive semi-definite where $A \in R^{p \times q}$

Question

How to show $\|A\|_2 \le |s|$ if and only if $s^2I-A^TA$ is positive semi-definite where $A$ is a matrix of $p \times q$?

What I know is $\|A\|_2=\sigma_{max}(A)$, $\sigma_{max}$ is the maximum singular value of A.

Answer 1

Write $A^TA = \sum_{k=1}^q\sigma_k^2 u_ku_k^T$ for an eigendecomposition of $A^TA$. These $\sigma_k$ are the nonzero singular values of $A$. Since $s^2I = \sum_{k=1}^q s^2 u_ku_k^T$, the difference is $$s^2I-A^TA = \sum_{k=1}^q(s^2-\sigma_k^2)u_ku_k^T$$ which is PSD if and only if the coefficients $s^2-\sigma_k^2$ are all nonnegative, in particular $\| A\|_2\leq |s|$.