Is it possible to find an operator $T:C(\mathbb{R})\rightarrow C(\mathbb{R})$ such that $T=N+D$, where $N$ is nilpotent and $D$ is invertible? Let $C(\mathbb{R})$ is the set of all continuous functions $\mathbb{R}\to\mathbb{R}$.
I want to find a linear operator $T:C(\mathbb{R})\rightarrow C(\mathbb{R})$, proper subspaces $W_{1}, W_{2}$ of $C(\mathbb{R})$ such that $C(\mathbb{R})=W_{1}\oplus W_{2}$ and $T|_{W_{1}}$ is nilpotent $( T|_{W_{1}}\neq 0)$ and $T|_{W_{2}}$ is invertible.
I've tried this one, but it didn't work.
Thanks for your kindly help.
 A: To expand on the other answers, I'll just note that this can be done constructively.
Let $f_1, f_2$ be any continuous functions with $f_1(1) = f_2(2) = 1$, $f_1(2) = f_2(1) = 0$.  (For example, $f_1(x) = 2-x$, $f_2(x)=x-1$.)  Set $W_1 = \mathrm{span}\{f_1, f_2\}$, and $W_2 = \{ f \in C(\mathbb{R}) : f(1) = f(2) = 0\}$.  Then $C(\mathbb{R}) = W_1 \oplus W_2$: if $f \in C(\mathbb{R})$, we can write $$f = (f(1) f_1 + f(2) f_2) + (f - f(1) f_1 - f(2) f_2).$$  We can then define
$$Tf = f - f(1) f_1 - f(2) f_2 + f(1) f_2$$
and it is easy to check that $T$ has the desired properties.
A: Take any non-zero $f_1, f_2\in C(\mathbb{R})$. Consder $W_1=\operatorname{span}\{f_1,f_2\}$. Since $W_1$ is finite dimensional there exist $W_2$ such that $W_1\oplus W_2=C(\mathbb{R})$. Define $N :W_1\to W_1$ by equalities $N(f_1)=f_2$, $N(f_2)=0$ and define $D:W_2\to W_2$ to be identity operator. Then 
$T=N\oplus D$ is desired operator.
A: If by "operator" you mean $\mathbb{R}$-linear endomorphism, the answer is certainly yes.  We may write $C(R) = W_1 \oplus W_2$ where $W_1$ is $2$-dimensional, with basis $e_1, e_2$.  (This may require the Axiom of Choice).  Let $T(e_1) = e_2$, $T(e_2) = 0$ and let $T$ be the identity on $W_2$.
