Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$A$ is an $n\times n$ matrix (not symmetric). If $\rho(A)$, spectral radius of $A$, is less than or equal to 1, can we say that $x^TAx\leq x^Tx$?

In another word,

if $\rho(A)\leq 1$, then $\frac{1}{2}\rho(A+A^T)\leq 1$?

share|improve this question
add comment

2 Answers 2

No. Choose $A$ to be a matrix of all zeroes except $[A]_{1n} = n+1$. Let $x = (1,...,1)^T$. Then $x^T A x = n+1$, but $x^T x = n$. The spectral radius is $\rho (A) = 0$.

share|improve this answer
    
but I guess the the other way around is true that $\rho(A)>1$ then $\frac{1}{2}\rho(A+A^T)>1$? –  sara Aug 29 '12 at 18:37
    
No, Take $A = \begin{bmatrix}0 & 2 \\ -2 & 0 \end{bmatrix}$, then $\rho(A) = 2$, but $A+A^T = 0$, hence $\frac{1}{2} \rho(A+A^T) = 0$. –  copper.hat Aug 29 '12 at 18:44
    
Alright I checked Matrix Mathematics by D.S. Bernstein) Let $A\in\mathbb{R}^{n\times n}$, assume that $A$ is nonnegative, and let $\alpha\in[0,1]$. Then, $$\rho(A)\leq \rho(\alpha A+(1-\alpha)A^T)$$ Therefore, what I said for $\rho(A)>1$ must be true for nonnegative matrices. –  sara Aug 29 '12 at 18:49
    
I'm not sure what you mean by nonnegative. The matrix $A$ in my comment above satisfies $x^T A x \geq 0$ for all $x$, $\rho(A) = 2$ and, as above, $\rho(\frac{1}{2}(A+A^T)) = 0$. Perhaps the author is dealing with symmetric matrices? –  copper.hat Aug 29 '12 at 19:02
    
nonnegative means all the elements of $A$ are nonnegative. So the matrix in your example is not nonnegative. –  sara Aug 29 '12 at 19:09
show 1 more comment

The second part of the question is easier to answer, with the counterexample $$A=\begin{pmatrix}0&1\\0&0\end{pmatrix}.$$

share|improve this answer
add comment

Your Answer

 
discard

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.