# Shift Operator and Invariant Subspace

Let $$V$$ be the set of all infinite real sequences and let $$S$$ be the Shift Operator for the set of all infinite sequences $$(a_n)_{n \in \mathbb N}$$ such that $$S((a_n)_{n \in \mathbb N})=(a_{n+1})_{n \in \mathbb N}$$.

Define a Subspace $$W$$ of $$V$$ such that: $$W = \{(a_n)_{n \in \mathbb N} \in V: a_{n+3} = 2a_{n+2} + a_{n+1} -2a_n\}$$

I need to show that $$W$$ is $$S$$-Invariant, that is $$S((a_n)_{n \in \mathbb N}) \in W$$, however, I am really not sure how to start with this question. I only understood that after the transformation, the first term of the sequence is removed and I will then need to show that the remaining sequence is still in $$W$$. Intuitively $$W$$ seems to be $$S$$-invariant but I am not sure on how to prove this.

Any help and advice will be really appreciated!

Let $$w=(w_n)_{n\in\Bbb{N}}\in W$$ be a sequence. Then for all $$n\in\Bbb{N}$$ you have $$w_{n+3}=2w_{n+2}+w_{n+1}-2w_n.$$ Now consider its shift $$u:=S(w)$$, and denote it by $$u=(u_n)_{n\in\Bbb{N}}$$ so that $$u_n:=w_{n+1}$$ for all $$n\in\Bbb{N}$$. Simply verify that $$u_{n+3}=2u_{n+2}+u_{n+1}-2u_n,$$ for all $$n\in\Bbb{N}$$.
• Many thanks! Understood this better with the definition of the new sequence $u_n$ with $u_n = w_{n+1}$ – Derp Nov 16 '18 at 18:31