largest singular value of real symmetric matrix

Question

I am trying to find a proof of the following fact: Let $M$ be a real symmetric matrix, then the largest singular value satisfies: $$ \sigma_1(M) = \lim_{k \to \infty} \left[\text{Trace}(M^{2k})\right]^{\frac{1}{2k}} $$

Answer 1

$\newcommand{\diag}{\mathrm{diag}}$ $\newcommand{\tr}{\mathrm{Trace}}$ Since $M$ is symmetric, $M^2 = M'M$, which can then be decomposed as \begin{align*} M'M = O\diag(\sigma_1^2(M), \ldots, \sigma_n^2(M))O', \end{align*} where $O$ is order $n$ orthogonal matrix. It then follows that $M^{2k} = (M'M)^k = O\diag(\sigma_1^{2k}, \ldots, \sigma_n^{2k})O'$, whence $\tr(M^{2k}) = \sigma_1^{2k} + \cdots + \sigma_n^{2k}$. Hence by squeeze principle, it is easy to verify that \begin{align*} \lim_{k \to \infty}(\sigma_1^{2k} + \cdots + \sigma_n^{2k})^{1/2k} = \sigma_1. \end{align*}

Answer 2

There are elementary proofs, as exhibited by the other answer, but the identity in question can also be viewed as a special instance of Gelfand's formula. Since $M$ is real symmetric, we have $\sigma_1(M)=\rho(M)$ and $\operatorname{tr}(M^{2k})= \|M^k\|_F^2$, where $\|\cdot\|_F$ denotes Frobenius norm. Therefore, by Gelfand's formula, $$ \sigma_1(M)=\rho(M) =\lim_{k\to\infty}\|M^k\|_F^{1/k} =\lim_{k\to\infty}\left(\|M^k\|_F^2\right)^{1/2k} =\lim_{k\to\infty}\left(\operatorname{tr}(M^{2k})\right)^{1/2k}. $$