# Transpose of the Shift operators

Let $X = \ell^2$ The operators $\textbf{L}$ and $\textbf{R}$ are defined as

$$\textbf{R}x = (0, a_0, a_1...) \;\; \textbf{L}x = (a_1, a_2, a_3...)$$
show that they are the transposes of one another $\textbf{L}' = \textbf{R},\;\; \textbf{R}' = \textbf{L}.$

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What have you tried? –  Cameron Buie Dec 2 '12 at 14:50
I have not been able to come much longer then the definition of transpose. Any general hints on how to find transposes like this? –  Johan Dec 2 '12 at 14:51
Is the transpose uniquely determined? If so I guess it could follow by putting $\ell = (\ell_0, \ell_1, \ell_2...)$ where $\ell_i$ is the projection in the ith coordinate. –  Johan Dec 2 '12 at 14:57

By definition of transpose, $$\langle R'x,y\rangle = \langle x,Ry\rangle = \sum_{j=2}^\infty x_jy_{j-1}=\sum_{j=1}^\infty x_{j+1}y_{j}=\langle Lx,y\rangle.$$ As this equality holds for any $y\in X$, we deduce that $R'x=Lx$; as this holds for any $x$, we have $R'=L$.