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Solving the following exercise of a list I have: "$H$ is a complex Hilbert space admitting an orthonormal basis $\{e_n\}, n\in \mathbb{N}$ ; $\{\lambda_n\}\subset \mathbb{C}\setminus \{0\}$ is a sequence such that $\{|\lambda_n|\}$ is decreasing and tends to zero. Prove that $T(e_n) = \lambda_ne_n$ defines a continuous, one-to-one linear function $T$ of $H$ in itself ". I have solved this and I am interested now in the related question:

Find the solutions of the equation $T(x) - \alpha x = y$, where $\alpha$ is a complex cube root of 1 and $y$ is fixed in $H$.

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  • $\begingroup$ I have made some formatting changes to your post; please take not of what I did so that you know how to make your questions more readable in the future. $\endgroup$ – Omnomnomnom May 8 '15 at 18:44
  • $\begingroup$ It may help to rewrite the equation as $$ (T - \alpha \,\text{id})(x) = y $$ What we're looking for is an eigenvector of $T$ associated with $\alpha$. $\endgroup$ – Omnomnomnom May 8 '15 at 18:48
  • $\begingroup$ @Omnomnomnom: wouldn't an eigenvector satisfy $(T - \alpha \;\text{id})(x) = 0$, so any two solutions to $(T - \alpha \;\text{id})(x) = y$ differ by an eigenvector (in this case)? $\endgroup$ – Robert Lewis May 8 '15 at 18:52
  • $\begingroup$ @Omnomnomnom Thanks. By the way, I am going to learning writing in English and Tex Commands. It is hard for me, I beg you all people of math.stackexchange for your tolerance with my bad English $\endgroup$ – Piquito May 8 '15 at 18:57
  • $\begingroup$ @RobertLewis ahhhh whoops! Don't know what I was thinking there. ahem I mean, that's what I meant. Yep. $\endgroup$ – Omnomnomnom May 8 '15 at 18:57
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Note that if $x = \sum_{n \in \Bbb N} {x_n e_n}$, then $$ T x - \alpha x = \sum_{n \in \Bbb N} (\lambda_n - \alpha)x_n e_n $$ Now, if $y = \sum_{n \in \Bbb N} y_n e_n$, use the above to solve for each $x_n$.

That is, we have $$ (\lambda_n - \alpha)x_n = y_n $$ If $\lambda_n \neq \alpha$ for every $n$, then we can define $x$ by $$ x_n = \frac{y_n}{\lambda_n - \alpha} $$ If $\lambda_n = \alpha$ but $y_n \neq 0$, then there is no solution for $x$. If $\lambda_n = \alpha$ but $y_n = 0$, then we'll have infinitely many solutions.

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  • $\begingroup$ So x is unique provide that any $\lambda_n$ is equal to $\alpha$ ? $\endgroup$ – Piquito May 8 '15 at 19:08
  • $\begingroup$ $x$ exists and is unique provided that none of the $\lambda_n$ is equal to $\alpha$. $\endgroup$ – Omnomnomnom May 8 '15 at 19:12
  • $\begingroup$ @sasha Are you suggesting that T - $\alpha$I is not 1-1? $\endgroup$ – Piquito May 8 '15 at 19:13
  • $\begingroup$ @LuisGomezSanchez it is not 1-1 if $\lambda_n = \alpha$ for some $n$. $\endgroup$ – Omnomnomnom May 8 '15 at 19:14
  • $\begingroup$ @LuisGomezSanchez I made a arithmetic mistake edited it $\endgroup$ – sashas May 8 '15 at 19:15

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