# On uniqueness of approximate eigen sequence of linear operator on Hilbert spaces

Assume there is no eigenvalues of $$T$$. Let $$\lambda$$ is an approximate eigenvalue but not eigenvalue of a self-adjoint operator $$T$$. That means there exist sequences $$\{x_{n}\}$$ unit vectors in $$\mathcal{H}$$ such that $$\|(T-\lambda I)x_{n}\| \to 0$$. Can there always exist $$y_{n}$$ orthogonal to $$x_{n}$$ i.e $$\langle y_{n},x_{n}\rangle=0$$., such that $$\|(T-\lambda I)y_{n}\| \to 0$$, with $$\|y_{n}\|=1$$.

• If $\lambda\neq 0$ then these will imply $|\lambda|\|x_{n}-y_{n}\| \to 0$, that says $\lambda\sqrt{2} \to 0$, a contradiction. – mathlover Feb 11 at 10:07

You already noted that it is not possible if $$\lambda \ne 0$$. I believe the answer is yes in the case $$\lambda = 0$$.

Assume $$Tx_n \to 0$$ where $$\|x_n\| = 1$$.

Notice that $$|\langle x_{n+1}, x_n\rangle| \le 1$$ and it is equal to $$1$$ if and only if there exists $$q_n \in \mathbb{C}$$, $$|q_n| = 1$$ such that $$x_{n+1} = q_nx_n$$.

Let $$y_1$$ be any vector such that $$y_1 \perp x_1$$ and $$\|y_1\| = 1$$. Define inductively $$y_n = \begin{cases} \frac{x_{n+1}-\langle x_n, x_{n+1}\rangle x_n}{\sqrt{2\left(1-|\langle x_n, x_{n+1}\rangle|^2\right)}} &\text{if } |\langle x_{n+1}, x_n\rangle| < 1\\ y_{n-1} &\text{if } |\langle x_{n+1}, x_n\rangle| = 1 \end{cases}$$

Then $$y_n \perp x_n$$, $$\|y_n\| = 1$$, and $$Ty_n \to 0$$ since it is not possible that $$y_n = y_1$$ for all $$n \in \mathbb{N}$$. In that case we would have $$q_n(Tx_1) = Tx_n \to 0$$ so $$x_1 = 0$$ since $$|q_n| = 1$$. But this is a contradiction with $$\|x_1\| = 1$$.

The question was modified after I posted my answer. The requirement that $$T$$ has no eigen values has been added.

On $$\ell^{2}$$ let $$Te_n=\frac 1 n e_n$$ where $$(e_n)$$ is the usual basis. Then $$(T-0I)e_n \to 0$$ and $$(T-0I)e_{n+1} \to 0$$; of course $$\langle e_n, e_{n+1} \rangle =0$$ for all $$n$$. Also, $$0$$ is not an eigen value and $$T$$ is self adjoint.

• But if $\sigma_{eig}(T)=\emptyset$ then?? – mathlover Feb 11 at 9:17
• If $T(a_n)=0 a_n$ then $\frac 1 n a_n =0$ for all $n$, This implies $a_n=0$ for all $n$. Hence $0$ is not an eigen value. – Kavi Rama Murthy Feb 11 at 9:21
• @mathlover I have modified my answer. – Kavi Rama Murthy Feb 11 at 9:30
• Yes, ok I see. Does it always exist? – mathlover Feb 11 at 9:32
• @mathlover If you want t make major changes to a question you should ask a new question. If that is not possible, you should at least display the edited part of the question. – Kavi Rama Murthy Feb 11 at 9:48