$\lambda = \max_{\mathbf{x}}\frac{\mathbf{x}^{T}\mathbf{A}\mathbf{x}}{\mathbf{x}^{T}\mathbf{x}}$ for non-negative matrices $A$?

Question

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 $?

Answer 1

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, $$ \max_x \frac{x^TAx}{x^Tx} = \frac{(1,1)^T A(1,1)}{2} = 60 $$