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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?

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  • $\begingroup$ $\|Xx\|_2^2 =\langle Xx, Xx\rangle=x^TX^TXx=\langle X^TXx, x\rangle$ and $X^TX$ is symmetric. $\endgroup$
    – dan_fulea
    Commented Jan 28, 2021 at 18:18
  • $\begingroup$ @c It is unusual to use $\|X\|$ to refer to the spectral radius. Are you sure that you are interpreting the notation correctly? $\endgroup$ Commented Jan 28, 2021 at 18:42
  • $\begingroup$ @BenGrossmann Thanks, Ben! I have changed to a more commonly used notation. $\endgroup$
    – zxzx179
    Commented Jan 28, 2021 at 19:05
  • $\begingroup$ 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 $X$? $\endgroup$
    – zxzx179
    Commented Jan 28, 2021 at 19:07

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

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  • $\begingroup$ Thanks, Ben!! I'm happy (for it is now resolved) and sad (for I had followed a bad note) to hear this... $\endgroup$
    – zxzx179
    Commented Jan 28, 2021 at 20:50

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