$A^T A$ 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?

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How to show it @JonasMeyer? Yes yor assumptions should be in my guess! – Salih Ucan Jan 28 '13 at 2:56
Yobo: I added the assumption of real entries to the question. I deleted my comment in part because it implied an incorrect statement base on my backwards thinking. – Jonas Meyer Jan 28 '13 at 3:40

Consider $A=\begin{bmatrix} 0 & 2 \\ 0 & 0 \end{bmatrix}$

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This is a thought: Let svd of $A=U\Sigma V^T$. Thus $A^TA=V\Sigma^2 V^T$. Thus, eigenvalues of $A^TA$ are squares of singular values of those of $A$. Thus, what you actually require is that the squared singular values of a matrix should be less than the maximum among the absolute squared values of its eigenvalues, i.e \begin{align} \sigma_i^2\leq max(|\lambda_k|^2),~\forall i \end{align} where $\lambda_k$ are the eigenvalues of $A$. I don't think it is true in general, unless $A$ has some kind of properties.

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This should be correct always?? There is no suspicious here?? – Salih Ucan Jan 28 '13 at 3:57
Yes, it should be. – dineshdileep Jan 28 '13 at 4:16

Presumably $A$ is a real $n\times n$ matrix with all its eigenvalues being real. The inequality is false in general (see chaohuang's answer for a counterexample), but it is true in each of the following circumstances (exercises):

• $A$ is real symmetric (in this case, $A$ is guaranteed to have only real eigenvalues),
• $A$ is a doubly stochastic matrix (hint: Perron-Frobenius theorem),
• the norm of every column of $A$ does not exceed $\frac1{\sqrt{n}}$ (hint: for any unit vector $x$, we have $\|Ax\|\le\sum_i|x_j|\|a_{\ast j}\|\le\|x\|\ \left\|\left(\|a_{\ast 1}\|,\ldots,\|a_{\ast n}\|\right)\right\|$ by Cauchy-Schwarz inequalty),
• the norm of every row of $A$ does not exceed $\frac1{\sqrt{n}}$,
• $\|A\|_1\|A\|_\infty\le1$, where $\|A\|_1$ and $\|A\|_\infty$ are respectively the maximum absolute column sum norm and the maximum absolute row sum norm of $A$ (hint: $\rho(A^TA)\le\|A^TA\|_1\le\|A^T\|_1\|A\|_1$).
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Why is it false in general could you give a counterexample? Thank you.. – Salih Ucan Jan 28 '13 at 7:43
@Yobo See the answer of chaohuang for a counterexample. – user1551 Jan 28 '13 at 7:50

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!

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