Finding $T^n$, where $T: l^{1} \rightarrow l^{1}$

Question

Let $T: l^{1} \rightarrow l^{1}$ be a bounded map, where each $j$, $T\left( \varepsilon _{j}\right)= \varepsilon _{j}+2 \varepsilon _{j+2}$ where $\left\{\varepsilon _{j}\right\}$ is the standard basis in $l^{1}$.

1. Find $T(x)$ and $T^{n}(x)$ for $n=1,2,3, \ldots$ where $x=$ $\left(x_{n}\right) \in l^{1}$.
2. Determine $\operatorname{Ker}(T-I)$. Show that if $\lambda=1$ an eigenvalue of $T$. If it is so, find are the regarding eigenvectors?

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Now, for the first part the question, \begin{aligned} &T(e_k)=(0,0, \ldots, \underbrace{1}_{k^ {th}}, 0, \ldots)+2(0,0, \ldots, \underbrace{1}_{{k+2}^{th}}, 0, \ldots)\\ &\begin{gathered} T(e_k)=\left(0,0, \ldots, 1, 0, 2, 0, \ldots\right) \\ \end{gathered}\\ \end{aligned} And then, let's take any $x \in \ell ^1$, and consider $T(x)$: \begin{aligned} T x=T\left(\sum_{n=1}^{\infty} x_{n} e_{n}\right)=& \sum_{n=1}^{\infty} x_{n} T\left(e_{n}\right) \\ =&\left(x_{1}, 0,2 x_{1}, 0, \ldots\right)+\left(0, x_{2}, 0,2 x_{2}, 0, \ldots\right) \\ &+\left(0,0, x_{3}, 0,2 x_{3}, 0, \ldots\right)+\ldots \\ &=\left(x_{1}, x_{2}, 2 x_{1}+x_{3}, 2 x_{2}+x_{4}, 2 x_{3}+x_{5}, \ldots\right) \end{aligned}

Now, I thought of using the formula given in Kreyszig (pg. 423) -using the Binomial theorem shortly- for $\lambda =1$: \begin{aligned} T_{\lambda}^{n}=(T-\lambda I)^{n} &=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) T^{k}(-\lambda)^{n-k} \\ &=(-\lambda)^{n} I+T \sum_{k=1}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) T^{k-1}(-\lambda)^{n-k} \end{aligned}

This is what I have in my mind so far for the first part of this question. Am I allowed to use the Binomial theorem directly? Am I on the right track? And, I am open to any suggestions for both parts (I haven't started solving (2), but it seems easy, I am just stuck at part 1).

Edit: For the second part, I have come up with the following:

Now, consider $T=I+2 S ^2$,

$(T-\lambda I) \lambda=0=\left(I+2 S^2-\lambda I\right) x=0$

$2\left(S^{{2}}-\left(\frac{\lambda-1}{2}\right) I\right) x=0$

$2\left(S+\sqrt{\frac{(\lambda-1)}{2}}\right)\left(S-\sqrt{\frac{(\lambda-1)}{2}} I\right) x=0$

$S$ nas no eigenvalues, so $\quad S x \neq \sqrt{\frac{\lambda-1}{2}} x$. So, $S x =-\sqrt{\frac{\lambda-1}{2}} x $ but we have a conradiction. So, $T$ has no eigenvalue!

Answer 1

I prefer the following explanation. Assume $Tx=\lambda x$ for $x\neq 0.$ Then $S^2x=(\lambda -1)x.$ As $S^2$ is an isometry, then $\lambda\neq 1.$ Denote $\mu=\lambda-1\neq 0.$ Then $\mu^{-1}S^2x=x$ i.e. $$\mu^{-1}(0,0,x_1,x_2,\ldots , x_{n-2},\ldots )=(x_1,x_2,\ldots, x_n,\ldots )$$ This implies $$x_1=x_2=0,\qquad \mu^{-1}x_{n-2}=x_n, \ n\ge 3$$ Therefore $x_n=0$ for any $n\ge 1.$

Answer 2

It is useful to embed $\ell^1$ into the ring of formal power series $K[[t]]$ via the map $(x_0, x_1, \dots) \mapsto \sum_m x_mt^m$ where $K$ is the coefficient field (e.g. $K = \Bbb R$).

The map $T$ is then the restriction to $\ell^1$ of the map $f \mapsto (1 + t^2)f$ on $K[[t]]$.

It follows that $T^n$ sends $f$ to $(1 + t^2)^nf$. Applying binomial theorem gives $T^n(f) = \left(\sum_i \binom n i t^{2i}\right) f$ and hence $(x_0, x_1, \dots)$ is mapped via $T^n$ to $(y_0, y_1, \dots)$ where $y_m = \sum_i \binom n i x_{m - 2i}$.

For the second part, we consider the equation $(T - I)f = 0$ on $K[[t]]$, which translates to $t^2 f = 0$. Since $K[[t]]$ is an integral ring, this implies $f = 0$. Therefore the kernel of $T - I$ is zero.

The same argument shows that $T - k$ has zero kernel for any $k \in K$.