Vector space over the complex field 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.
 A: 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$$2\sqrt{\sum_{n=1}^\infty|x_n|^2}\sqrt{\sum_{n=1}^\infty|y_n|^2}.$$

*$\displaystyle\sum_{n=1}^\infty|\lambda x_n|^n=|\lambda|^2\sum_{n=1}^\infty |x_n|^2<\infty.$

