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 $n\in\Bbb{N}$.

| cite | improve this answer | |
  • $\begingroup$ Many thanks! Understood this better with the definition of the new sequence $u_n$ with $u_n = w_{n+1}$ $\endgroup$ – Derp Nov 16 '18 at 18:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.