# Find all the subspaces of $l_2$ that reduces the right shift operator en $l_2$

Consider the Hilbert Space $$l_2$$.

Let the right and left shift operator $$R$$ and $$L$$ of $$l_2$$ to $$l_2$$ $$R(a_0, a_1, a_2,...)=(0, a_0, a_1, a_2,...).$$ $$L(a_0, a_1, a_2,...)=(a_1, a_2, a_3,...).$$

I need to find all the subspaces $$W$$ of $$l_2$$ that reduces to $$R$$ and $$L$$.

A subspace $$W$$ reduces to operator $$T$$ if $$T(W)\subseteq W$$ and $$T(W^\perp)\subseteq W^\perp$$.

Since that $$R^\ast =L$$ and $$L^\ast=R$$ and

$$W$$ reduces $$T$$ $$\Leftrightarrow$$ $$T(W)\subseteq W$$ and $$T^\ast(W)\subseteq W$$.

Then $$W$$ reduces $$R$$ if and only if $$W$$ reduces $$L$$.

Then I need to find the subspaces $$W$$ of $$l_2$$ so that $$R(W)\subseteq W$$ and $$L(W)\subseteq W$$ but i don´t have idea how to find this subspaces.

Lemma: The algebra of operators on $$\ell^2$$ generated by $$\{I, L, R\}$$ contains all operators whose matrix has finitely many nonzero entries.

Proof: Denoting the canonical basis of $$\ell^2$$ by $$\{e_n\}_{n=0}^\infty$$, notice that $$p:= I-RL$$ is the projection onto the space spanned by $$e_0$$. Moreover $$R^mpL^n$$ is the operator which sends all basis vectors to zero, except for $$e_n$$, which is sent to $$e_m$$. So the matrix of $$R^mpL^n$$ coincides with $$e_{m, n}$$, namely the matrix having all entries zero except for the $$(m,n)$$.th entry which is 1.

This said, any operator whose matrix $$A=(a_{i, j})_{i, j}$$ has finitely many nonzero entries can be written as $$\sum_{i,j=0}^N a_{i,j} e_{i,j} = \sum_{i,j=0}^N a_{i,j} R^ipL^j.$$ QED.

Now, if $$W$$ is a subspace of $$\ell^2$$ reducing $$L$$, it is clear that it is invariant under any operator in the above algebra, and hence also under any finite matrix by the Lemma.

In case $$W$$ is closed, then clearly $$W$$ can only be $$\{0\}$$ or the whole $$\ell^2$$.

On the other hand, I suppose there must be a huge number of non-closed spaces invariant under both $$R$$ and $$L$$, and the characterization of these might be a difficult problem.