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.
Best Answer
No, there is no such $C(\alpha)$, even for fixed $n>1$.
Indeed, consider the matrix $A = \begin{pmatrix} \alpha & x \\ 0 & \alpha\end{pmatrix}$.
The product of the eigenvalues of $A^t A$ does not depend on $x$ (it is equal to $det(A)^2=\alpha^4$), whereas their sum goes to $\infty$ (it is $Tr(A^tA)=2\alpha^2+x^2$). Hence one the the eigenvalues goes to $\infty$, and the other to $0$.