I am learning the Perron-Frobenius theorem from some lecture notes. Let $X \in \mathbb{R}^{n}_{++}$ be a square matrix with each element being strictly positive. The theorem says that the spectral radius of matrix $X$, denoted by $\rho(X)$, is an eigenvalue of X, and all other eigenvalues have a strictly smaller absolute value. But why does this implies that
$$
\rho(X) = \sup_{x: ||x||_2 \leq 1} ||Xx||_2 = \sup_{x: ||x||_2 \leq 1} \sqrt{x^T X^T X x}?
$$
If $X$ is further symmetric, I can see it is true. Any idea why this is true in general?
*The same question has also been asked and answered here:
Is spectral radius = operator norm for a positive valued matrix?
Best Answer
The statement that you have made is not true. For example, consider the matrix $$ X = \pmatrix{10 & 100\\1 & 10}. $$ Its eigenvalues are $0,20$, which means that $\rho(X) = 20$. On the other hand, we find that $\|X\|_2 = 101 > \rho(X)$.