proof of linearly independent for linear maps

Let $L,R: \Bbb R^{\infty} \rightarrow \Bbb R^\infty$ be the linear maps

$$L(a_0, a_1,...) = (a_1, a_2,...)$$ $$R(a_0, a_1,...) = (0, a_0,...)$$

Prove that the set of linear maps $R^kL^l$ is linearly independent.

I have struggled with the problem for a while. I assumed that $c_1R^{k1}L^{l1}+\dots+c_nR^{kn}L^{ln} = 0$ and if $c_1+\dots+c_n \neq 0$, I can conclude that it is contradicted, but I haven't figured out if $c_1 + \dots +c_n =0$.

You can definitely adopt a different method from mine. Any idea would be appreciated!

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I tried to edit but some parts baffled me: what is $\,c_1+...c_n\,$?, and what did you actually mean by "I can conclude they it's contradicted"? –  DonAntonio Oct 21 '12 at 5:25
what's your idea? –  Frank Xu Oct 21 '12 at 5:30
Perhaps you mean $R(a_0,a_1,\ldots) = (0,a_0,a_1,\ldots)$ (i.e. the right shift). The way you have it now, $R^2 = R$. –  Robert Israel Oct 21 '12 at 6:01
exact duplicate. math.stackexchange.com/questions/217700/… –  Julian Kuelshammer Oct 21 '12 at 6:42

$R^kL^l(a_0,a_1,\dots)=(0,\dots,0,a_l,a_{l+1},\dots)$ where the number of zeroes is $k$. Assume that $$\sum_{k=0}^m\sum_{l=0}^n \alpha_{kl}R^kL^l=0,$$ apply this to a sequence $(a_0,a_1,\dots)$ and now look what happens for $k=0$ on the first component of the resulting sequence: we get $\alpha_{00}a_0+\alpha_{01}a_1+\cdots+\alpha_{0n}a_n=0$ because for $k\ge 1$ on the first component of the resulting sequence will be $0$. Now take sequences that have $a_i=0$ for all $i\neq j$ and $a_j=1$, for $j=0,1,\dots,n$. Thus you get $\alpha_{00}=\alpha_{01}=\cdots=\alpha_{0n}=0$.

Now you have $$\sum_{k=1}^m\sum_{l=0}^n \alpha_{kl}R^kL^l=0$$ and look what happens for $k=1$ on the second component of the resulting sequence, and so on.

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