largest singular value of real symmetric matrix I am trying to find a proof of the following fact: Let $M$ be a real symmetric matrix, then the largest singular value satisfies:
$$
\sigma_1(M) = \lim_{k \to \infty} \left[\text{Trace}(M^{2k})\right]^{\frac{1}{2k}}
$$
 A: $\newcommand{\diag}{\mathrm{diag}}$
$\newcommand{\tr}{\mathrm{Trace}}$
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*}
A: 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,
$$
\sigma_1(M)=\rho(M)
=\lim_{k\to\infty}\|M^k\|_F^{1/k}
=\lim_{k\to\infty}\left(\|M^k\|_F^2\right)^{1/2k}
=\lim_{k\to\infty}\left(\operatorname{tr}(M^{2k})\right)^{1/2k}.
$$
