Prove $\sup_x \frac{\left \| x \right \|_2}{\left \| Ax \right \|_2} = \frac{1}{\lambda_{\min}(A^TA)^{\frac{1}{2}}}$

Question

I've been trying to prove this following property: $\sup_x \frac{\left \| x \right \|_2}{\left \| Ax \right \|_2} = \frac{1}{\lambda_{\min}(A^TA)^{\frac{1}{2}}}$ for a rectangular (not necessarily square!) matrix with full rank. I still haven't found a way to prove it without needing the inverse. I thought there should be a way to prove this using the pseudoinverse but I'm still stuck.

Answer 1

Hints: The matrix $A^T A$ is symmetric and positive definite, therefore it is orthogonally diagonalisable. The 2-norm is invariant under orthogonal transformations, which are bijective. Also maybe $$\sup_{x \ne 0}\frac{\| x \|}{\| A x \|} = \sup_{\| x \| = 1} \frac{1}{\| A x \|} = \frac{1}{\inf_{\| x \| = 1} \| A x \|}.$$ is helpful.