Eigenvalues of double right shift operator

Question

Consider the Operator S/ $S: l_2 → l_2$ , $S(x_1,x_2,x_3,....)=(0,0,x_1,x_2,x_3,...)$ I know the right shift operator has no eigenvalues,but does this operator either? I this for the same reason as for the right shift operator? Thanks!

Answer 1

Consider the right-shift operator $(x_{1},x_{2},\ldots)\mapsto(0,x_{1},x_{2},\ldots)$. Suppose there exists an eigenvector $\mathbf{x}\neq0$ of the shift operator. By definition, we can find a scalar $\lambda$ such that $$ \lambda(x_{1},x_{2},\ldots)=(0,x_{1},x_{2},\ldots). $$ This implies that $x_{1}=0$, $\lambda x_{2}=x_{1}=0$, etc., so that $\mathbf{x}=0$, a contradiction.

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Consider now the double-shift $(x_{1},x_{2},\ldots)\mapsto(0,0,x_{1},x_{2},\ldots)$. As above, suppose there exists an eigenvector $\mathbf{x}\neq0$ of the double-shift operator. By definition, we can find a scalar $\lambda$ such that $$ \lambda(x_{1},x_{2},\ldots)=(0,0,x_{1},x_{2},\ldots). $$ What can you conclude?

Answer 2

If $S$ is the right shift, then you want to know if there exists $\lambda\in\mathbb{C}$ and $x\ne 0$ such that $S^2 x = \lambda x$. If there were such an $x$, then $$ (S+\sqrt{\lambda}I)(S-\sqrt{\lambda}I)x = 0. $$ Because $(S-\sqrt{\lambda}I)x \ne 0$ (you know $S$ has no eigenvalues,) you conclude that $y=(S-\sqrt{\lambda}I)x \ne 0$ and $Sy = -\sqrt{\lambda}y$, which is a contradiction. Hence $S^2$ has no eigenvalues.