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.

  • $\begingroup$ The best $C(\alpha)$ is $\alpha^2$. For a proof, compute the determinant. $\endgroup$ – Mikael de la Salle Jul 19 '13 at 16:55
  • 1
    $\begingroup$ This is false, I believe, see my example above. $\endgroup$ – Matthias Ludewig Jul 19 '13 at 21:12
  • $\begingroup$ You might benefit from Ky-Fan's theorem that for any matrix $A$, $\text{Re}\lambda(A) \prec \lambda(\text{Re}A)$, followed by the Fan-Hoffman theorem: $\lambda_j(Re(A)) \le \sigma_j(A)$ (evals and svals are sorted in decreasing order). $\endgroup$ – Suvrit Jul 19 '13 at 23:13
  • $\begingroup$ Kofi: in your question you are asking for a lowerbound of the biggest eigenvalue. 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
  • $\begingroup$ Yes sorry, I fixed that mistake. But what happens if I replace the 1 with a large X in the example? If you can answer the question, I would like to invite you to just post an answer! $\endgroup$ – Matthias Ludewig Jul 20 '13 at 7:52

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 $0$.


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