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Let $A$ be a non-negative irreducible matrix. By the Perron-Frobenius theorem, the eigenvalue of max. absolute value $\lambda$ is positive and has an eigenvector of all positive entries.

Is it true that

$$\max_{\mathbf{x}}\frac{\mathbf{x}^{T}\mathbf{A}\mathbf{x}}{\mathbf{x}^{T}\mathbf{x}}$$

where, perhaps, the max. is taken only over non-negative vectors $\mathbf x $?

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The answer is no. As a simple example, consider the matrix $$ A = \pmatrix{10 & 99\\1 & 10} $$ Its maximal eigenvalue is $10 + 3 \sqrt{11} \approx 19.95$. However, $$ \max_x \frac{x^TAx}{x^Tx} = \frac{(1,1)^T A(1,1)}{2} = 60 $$

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  • $\begingroup$ This answers my question. But how do you know that the $max$ occurs at $(1,1)$? $\endgroup$ – becko Apr 5 '18 at 20:20
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    $\begingroup$ Rayleigh's theorem. Note that $$ x^TAx = x^T\pmatrix{10&50\\50&10}x $$ $\endgroup$ – Omnomnomnom Apr 5 '18 at 22:03
  • $\begingroup$ @becko: Or use Lagrange multipliers on the problem $\max\{ x^T A x | \|x \| =1 \}$ to see that the $\max$ must occur at an eigenvector. Then Perron Frobenius shows that this must be the (unique) $\max$ real eigenvalue and that the result is the same as the stricter problem with $x>0$. $\endgroup$ – copper.hat Apr 6 '18 at 1:57

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