Let $A$ be a real $n\times n$ matrix and let $\mu_1, \dots, \mu_n$ the (generalized, complex) eigenvalues of $A$. Assume that $$ 0 < \alpha < \mathrm{Re}(\mu_1) < \dots < \mathrm{Re}(\mu_n).$$

I am interested in a lower bound on the eigenvalues of $A^t A$ in relation to $\alpha$.

If $A$ is symmetric, then the lowest eigenvalue of $A^tA$ is bigger than $\alpha^2$. But if $A$ is non-symmetric, then this is not true.

**Question:** Is there a positive number $C(\alpha)$ of $\alpha$ such that we have: If for a matrix $A$, the above inequality holds, then the smallest eigenvalue of $A^tA$ is bigger than $C(\alpha)$? And what is the best such $C(\alpha)$?

\Edit: For example, for the matrix $$A := \begin{pmatrix} \alpha & 1 \\ 0 & \alpha \end{pmatrix},$$ the matrix $A^t A$ has the smallest eigenvalue $\alpha^2 + \frac{1}{2}\bigl( 1 - \sqrt{4\alpha^2 + 1}\bigr)$, which is always smaller than $\alpha^2$.

\Edit: Fixed a typing error.

biggesteigenvalue. If the question is about the smallest, replace 1 by a big x in your example. $\endgroup$ – Mikael de la Salle Jul 20 '13 at 7:44