For an $m×n$ matrix A with $m \geq n$. We define
$ k(A) = \frac{\max_{\lVert x\rVert=1} \lVert Ax \rVert}{\min_{\lVert x\rVert=1} \lVert Ax \rVert} $
For the euclidean norm, that is, $\lVert A \rVert$ is the maximum eigenvalue of $A^tA$, I try to show that $k(A^tA) = (k(A))^2$.
I appreciate your help.
If $A=UDV^\top$ is the SVD of $A$ with singular values $\sigma_1 \ge \cdots \ge \sigma_n$ then $\kappa(A) = \sigma_1 / \sigma_n$.
Then $A^\top A = VD^\top D V^\top$ so the eigenvalues of $A^\top A$ are $\sigma_1^2 \ge \cdots \ge \sigma_n^2$ and $\kappa(A^\top A) = \sigma_1^2 / \sigma_n^2$.