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I want to prove (or to find a counterexample) for the next inequality

$$c_1 \leq c_2$$ where the relations among the scalar constants are $$x^T(A+A^T)x \leq c_2x^Tx$$ $$x^T(A^TA)x \geq c_1^2x^Tx$$ I know that $x^T(A+A^T)x>0 \implies x^TAx>0$


You were right. The second inequality was wrong written, otherwise it does not make sense. Also you were right about the factorization of A, so I have dropped that part.

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That $A$ is non-singular does not imply that it can be written as $P^TDP$ (or $P^{-1}DP$) with a diagonal $D$. Consider for example $$A=\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}.$$ – Mårten W Mar 4 '13 at 8:48
@Mårten: But that $A$ is singular? – joriki Mar 4 '13 at 8:50
@joriki: No, neither all singular, nor all non-singular matrices are automatically diagonalisable. – Mårten W Mar 4 '13 at 8:51
@Mårten: I don't understand. I didn't write anything about what sort of matrices are diagonalisable. I merely pointed out that the matrix you seemed to be offering as a counterexample for a claim about non-singular matrices was in fact not non-singular. – joriki Mar 4 '13 at 8:52
@joriki: Sorry, my mistake. It should have ones on the diagonal. – Mårten W Mar 4 '13 at 8:55

Since $x^T x$ is non-negative, taking it to the other side of your inequalities doesn't change their direction. Thus you have lower bounds on both $c_2$ and $|c_1|$. In order to derive an inequality between them, you'd need an upper bound on one and a lower bound on the other.

To construct a simple counterexample, just choose $A$ to be the identity and $x$ an arbitrary unit vector; then the given inequalities are $2\le c_2$ and $1\le c_1^2$, and they're satisfied by $c_1=3$ and $c_2=2$ with $c_1\not\le c_2$.

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You are right. The second inequality was wrong written. My misstake – noether Mar 4 '13 at 17:42

First, you probably mean the infimum of those $c_1$ and $c_2$ that fulfill your inequalities (i.e. the least upper bound, not just any upper bound). Otherwise the whole problem does not make any sense, as you can see in joriki's answer. So, what you probably mean is: \begin{align} c_2 &= \sup_{x\not=0}\frac{x^T(A+A^T)x}{x^Tx}\\ c_1 &= \left(\sup_{x\not=0}\frac{x^T(A^TA)x}{x^Tx}\right)^{1/2} \end{align} In the equation for $c_2$, adding the matrix transpose is pointless, since $x^TA^Tx=x^TAx$. So you can just as well write $$c_2 = 2\sup_{x\not=0}\frac{x^TAx}{x^Tx}.$$ Now, on the right you have (two times) the Rayleigh quotient, and taking the supremum of that gives you (two times) the maximum eigenvalue $\lambda_{\max}$ of $A$.

In the equation for $c_1$, you have the spectral norm on the right, which is the same as the largest singular value $\sigma_{\max}$ of $A$. So what you want to prove or disprove is effectively the following:

For all square matrices $A$ it holds that: $\sigma_{\max}(A) \leq 2 \lambda_{\max}(A).$

Obviously, this cannot hold for matrices with only negative eigenvalues. Just take $A=-I$, and you have a counterexample, since $\sigma_{\max}=1$ and $\lambda_{\max}=-1$.

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Thank you very much for your answer. I wrote down the second inequality wrong, now it is fixed. So sorry for the inconvenience :( – noether Mar 4 '13 at 17:53

First not that you do not need $A$ be singular to do the factorization $A=P^{-1}DP$. Second even if $A$ is nonsingular, there is no guarenty that $D$ becomes diagonal. Third, even if all these variables were scalers, no conclusion could be made for the inequalities.

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