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

2 Answers 2

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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 $\rho (A) = 0$.

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    I see, I was thinking non-negative definite...2012-08-29
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The second part of the question is easier to answer, with the counterexample $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}.$