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?

• $\|Xx\|_2^2 =\langle Xx, Xx\rangle=x^TX^TXx=\langle X^TXx, x\rangle$ and $X^TX$ is symmetric. Jan 28 '21 at 18:18
• @c It is unusual to use $\|X\|$ to refer to the spectral radius. Are you sure that you are interpreting the notation correctly? Jan 28 '21 at 18:42
• @BenGrossmann Thanks, Ben! I have changed to a more commonly used notation. Jan 28 '21 at 19:05
• Thanks, @dan_fulea! So this tells that $\max_{||x||_2 \leq 1} ||Xx||^2_2$ is the largest eigenvalue of $X^T X$. But why it is equal to the square of spectral radius of $X$? Jan 28 '21 at 19:07

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 $$\|X\|_2 = 101 > \rho(X)$$.