Q:

how to find eigenvalues of $T_n:R^n\to R^n$ where $T_n (x)=(0,x_1,\frac{x_2}{2},\ldots,\frac{x_{n-1}}{n-1})$, what happen if $n\to\infty$

how to find eigenvalues of $T_n:R^n\to R^n$ where $T_n (x)=(0,x_1,\frac{x_2}{2},\ldots,\frac{x_{n-1}}{n-1})$  what happen if $n\to\infty$, well by definition i do $T_n(x)=\lambda x$ hence
$(0,x_1,\frac{x_2}{2},\ldots,\frac{x_{n-1}}{n-1})=(\lambda x_1,\ldots,\lambda x_n)$ imply
$0=\lambda x_1=\lambda^2 x_2=2\lambda^3 x_3=\ldots =(n-1)!\lambda^n x_n $ so i think that $\lambda$ have to be zero right? and when $n\to\infty$ the operator $T_n$ go to  null operator right? i will apreciate any hint please

A:

It is clear that $0$ is an eigenvalue because $T_n(x) = 0$ iff $x_1 = \ldots = x_{n-1} = 0$. Thus all $(0,\ldots,0,x_n)$ with $x_n \ne 0$ are eigenvectors.
No $\lambda \ne 0$ can be an eigenvalue because then $0=\lambda x_1=\lambda^2 x_2=2\lambda^3 x_3=\ldots =(n-1)!\lambda^n x_n $ implies that $x = 0$.
An alternative way to see this is decribed in Thomas Andrews' comment. We have $(T_n)^n = 0$. If $T_n$ has an eigenvalue $\lambda \ne 0$, then we can take an eigenvector $x \ne 0$ and get $(T_n)^n(x) = \lambda^n x \ne 0$.
Edited:
It is not clear what is meant by $n \to \infty$. There are many possible interpretations, here is one:
Let $\mathbb R^\infty$ denote the vector space of all real sequences $(x_n)$. Define $T_\infty : \mathbb R^\infty \to \mathbb R^\infty, T_\infty(x_1,x_2,\ldots) = (0,x_1,\frac{x_2}{2},\ldots,\frac{x_{n-1}}{n-1},\ldots)$. This map does not have eigenvalues. In fact, it is injective which means that $0$ is no eigenvalue. No $\lambda \ne 0$ can be an eigenvalue because $T_\infty(x) = \lambda x$ implies $0=(n-1)!\lambda^n x_n $ for all $n \in \mathbb N$. This means $x = 0$.

