Let $A$ be an $n \times n$ invertible matrix with real entries. I know that its largest singular value, $\sigma_{\max}$ is equal to its operator norm. Moreover, there are singular vectors $u, v \in \mathbb{R}^n$ such that
\begin{equation} Au=\sigma_{\max}v, \qquad A^Tv=\sigma_{\max}u \end{equation}
by definition of the singular value. However, generally $u$ and $v$ are not the same, so that $\sigma_{\max}$ is not an eigenvalue of $A$. I wonder when $\sigma_{\max}$ becomes an eigenvalue of $A$, so that there is nonzero $x \in \mathbb{R}^n$ such that $Ax=\lVert A \rVert x$.