$A^T A$ eigenvalue bounding Let $A$ be a square matrix with real entries.

Is there anything like any eigenvalue of $A^tA \leq \max({1,\lambda^2})$ where 
$\lambda$ is an eigenvalue of $A$ and max is taken over all eigenvalues?
Consider $A=\begin{bmatrix} 0 &amp; 2 \\ 0 &amp; 0 \end{bmatrix}$
We know that $A$ and $A^t$ have the same set of eigenvaluse, and we also know that the matrix products $AB$ and $BA$ have the same sets of eigenvalues. These two bits of information may be of help. Good luck!