Finding eigenvalue for a length vector that converges

Question

If $M$ is a linear operator on $\mathbb{R}^3$ with unique and real eigenvalues $\lambda_1 < \lambda_2 < \lambda_3$, such that $\exists x \in \mathbb R^3 \setminus \{0\}$, satisfying the condition $\lim_{n \to \infty} ||M^n x|| = 0$. What are the possible values of $\lambda_1$?

Answer 1

That's not as straightforward of an answer as it seems.

If $x$ is a linear multiple of $\zeta_1$, then we can affirm that $|\lambda_1|<1$.

Otherwise, we have $x=c_1\zeta_1+c_2\zeta_2+c_3\zeta_3$ and that means $M^nx=c_1\lambda_1^{n}\zeta_2+c_2\lambda_2^{n}\zeta_2+c_3\lambda_3^{n}\zeta_3$, implying all $\lambda_i$ should have magnitude less than $1$ to satisfy that property for any arbitrary vector $x$.