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Let $V$ be the vector space of the real sequences. Let $S:V\to V$ be a linear operator such that $S(a_1,a_2,\dots)=(a_2,a_3,\dots)$.

What are the eigenvectores of the map $S$?

Prove that the subspace $W=\{X_n \in V\mid X_{n+2}=X_{n+1} + X_n\}$ is invariant by $S$, $\dim W = 2$ and find a basis for $W$ formed by eigenvectors of $S$.

I found that the eigenvector was v=a1(1,¶,¶²,¶³,...) and i applied S in W and found S(X1, X2, X1+X2,...)=(X2, X1+X2, X1+2X2,...) (All of the terms of this sequence is a linear combination of X1 and X2, and hence dim$W$=2, right?) But i couldn't find a basis with these eigenvectores... PS: ¶ is the eigenvalue.

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    Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 2017-02-10
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    Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. – 2017-02-10
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    Okay, no problem. I found that the eigenvector was v=a1(1,¶,¶²,¶³,...) and i applied S in W and found S(X1, X2, X1+X2,...)=(X2, X1+X2, X1+2X2,...) (This is an element of W, right?) But i couldn't find a basis with these eigenvectores... PS: ¶ is the eigenvalue. – 2017-02-10
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    Better to add that into the question... – 2017-02-10
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    "All the terms of this sequence is a linear combination of X1 and X2, and hence dim W=2, right?" No, you are confusing the first two entries of a sequence $(x_1,x_2,x_3,\ldots)$ with taking a linear combination of two vectors (in order to span $W$). Keep in mind that the vectors of $V$ are *real sequences*. – 2017-02-10
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    I have seem my mistake and i solved the problem a few minutes ago, Thanks! – 2017-02-11

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Hint for the eigenvalus: let $\mu \in \mathbb R$ and $x:=(1, \mu, \mu^2,...)$

Then $Sx= \mu x$

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    This post does not answer the question. – 2017-02-10