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

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

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