Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 $A=\begin{bmatrix} 0 & 2 \\ 0 & 0 \end{bmatrix}$

share|cite|improve this answer

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.

share|cite|improve this answer
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: $\rho(A^TA)\le\|A^TA\|_1\le\|A^T\|_1\|A\|_1$).
share|cite|improve this answer
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!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.