# Finding eigenvalue for a length vector that converges

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 $$\lambda_1$$?

## 1 Answer

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 $$x$$.

• it can be between -1 and 1 – Saketh Malyala Jun 12 at 19:59