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Let $l_2$ denote the set of all complex sequences $(x_1,x_2,\dots)$ such that $\sum_{i=1}^\infty |x_i|^2<\infty$. Show that $l_2$ is a vector space over $\mathbb{C}$ if for $x=(x_1,x_2,\dots)$ and $y=(y_1,y_2,\dots)$ in $l_2$ we define $\langle x,y\rangle=\sum_{i=1}^\infty x_i\bar y_i$, then $\langle.,.\rangle$ is a inner product in $l_2$. If we define $T:l_2\longrightarrow l_2$ by $(x_1,x_2,\dots)\mapsto(0,x_1,x_2,\dots)$, then show that $T$ has no eigenvalue.

For the second part, if we consider that $\lambda$ is an eigenvalue, then $Tx=\lambda x$. Which gives $(0,x_1,x_2,\dots)=\lambda (x_1,x_2,x_3,\dots)$ and this gives us absurd as $(x_1,x_2,\dots)=(0,0,\dots)$ and get no value of $\lambda$. Am I correct? If not please help me where I made mistake? And for the first one to prove $l_2$ is a vector space over $\mathbb{C}$, help me. Thanks in advance.

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    $\begingroup$ Your proof that $T$ has not eigenvalues is correct. For the first part just use the axioms of a vector space. $\endgroup$
    – Marios Gretsas
    Sep 12, 2017 at 22:09

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I don't know whether you are correct or not when you state that “this gives us absurd as $(x_1,x_2,\ldots)=(0,0,\ldots)$”, because you don't tell us how you reched that absurd.

If $\lambda=0$, then it follows from$$(0,x_1,x_2,\ldots)=\lambda(x_1,x_2,x_3,\ldots)\tag{1}$$than the sequence $(x_n)_{n\in\mathbb N}$ is the null sequence. Therefore, $0$ is not an eigenvalue. If $\lambda\neq0$, it follows from $(1)$ that $\lambda x_1=0$ and therefore $x_1=0$. Since $x_1=0$, it follow from $(1)$ that $\lambda x_2=0$ and therefore that $x_2=0$. And so on. So, again, the sequence $(x_n)_{n\in\mathbb N}$ is the null sequence and therefore, $\lambda$ is not an eigenvalue.

The space $l_2$ is a vector space because, if $(x_n)_{n\in\mathbb N},(y_n)_{n\in\mathbb N}\in l_2$ and $\lambda\in\mathbb C$, then:

  • $\displaystyle\sum_{n=1}^\infty|x_n+y_n|^2\leqslant\sum_{n=1}^\infty|x_n|^2+2\sum_{n=1}^\infty|x_n|.|y_n|+\sum_{n=1}^\infty|y_n|^2$. The first and the third sums are finte, by assumption. The second one is finite too, by Cauchy-Schwarz: it is less that or equal to$$2\sqrt{\sum_{n=1}^\infty|x_n|^2}\sqrt{\sum_{n=1}^\infty|y_n|^2}.$$
  • $\displaystyle\sum_{n=1}^\infty|\lambda x_n|^n=|\lambda|^2\sum_{n=1}^\infty |x_n|^2<\infty.$
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  • $\begingroup$ yah sir I do the same as you. But for the second one I have to show the equality, but you shows the inequality above. Is it ok? $\endgroup$
    – abcdmath
    Sep 12, 2017 at 22:37
  • $\begingroup$ @abcdmath Which inequality? Do you mean $\displaystyle|\lambda|^2\sum_{n=1}^\infty|x_n|^2<\infty$? $\endgroup$
    – José Carlos Santos
    Sep 12, 2017 at 22:39
  • $\begingroup$ No no the first one, $\sum_{n=1}^\infty |x_n+y_n|²≤ \sum_{n=1}^\infty |x_n|²+\sum_{n=1}^\infty |y_n|²+2\sum_{n=1}^\infty |x_n|.|y_n|$ $\endgroup$
    – abcdmath
    Sep 12, 2017 at 22:49
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    $\begingroup$ @abcdmath If $z,w\in\mathbb C$, $|z+w|^2=(z+w).\overline{(z+w)}=|z|^2+2\operatorname{Re}(z.\overline{w})+|w|^2\leqslant|z|^2+2|z|.|w|+|w|^2$. $\endgroup$
    – José Carlos Santos
    Sep 12, 2017 at 22:52
  • $\begingroup$ Thank you sir I get it, it's too easy, but where the norm function will required in the above case? @José Carlos Santos $\endgroup$
    – abcdmath
    Sep 12, 2017 at 22:58

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