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 $x = 0$.
A: Notice that $T^n = 0$ so by the spectral mapping theorem we have
$$\sigma(T)^n = \sigma(T^n) = \{0\} \implies \sigma(T) = \{0\}$$
so $0$ is the only eigenvalue.
For the case $n\to \infty$, have a look at $\ell^\infty$, the space of all bounded sequences equipped with the supremum norm $\|(x_n)_n\|_\infty= \sup_{n\in\Bbb{N}}|x_n|$. Then we can look at $T : \ell^\infty \to \ell^\infty$ given by
$$T(x_1,x_2,x_3, \ldots) = \left(0,x_1,\frac{x_2}{2},\ldots,\frac{x_{k-1}}{k-1},\ldots\right).$$
Notice that $T^n$ acts as
$$T^n(x_1,x_2,x_3, \ldots) = \left(\underbrace{0,\ldots, 0}_{n}, \frac{x_1}{1\cdot 2 \cdots  n},\frac{x_2}{2\cdot 3 \cdots  (n+1)},\ldots,\frac{x_{k-1}}{(k-1)\cdot k \cdots (k-1+n)},\ldots\right)$$
so
$$\|T^n(x_n)_n\|_\infty = \sup_{k\in\Bbb{N}} \frac{|x_{k-1}|}{(k-1)\cdot k \cdots (k-1+n)} \le \frac1{n!} \sup_{k\in\Bbb{N}}|x_k| = \frac1{n!}\|x\|_\infty$$
and therefore $\|T^n\| \le \frac1{n!}$. We have a formula for spectral radius:
$$\sup_{\lambda \in \sigma(T)} |\lambda| = r(T) = \lim_{n\to\infty} \|T^n\|^{\frac1n} \le \lim_{n\to\infty} \frac1{\sqrt[n]{n!}} = 0$$
so again $\sigma(T) = \{0\}$.
