So I have to find a basis for a polynomial space such that differentiation operator is in Jordan form. I noticed that if I chose a basis to be ${1,x^2,x^3,x^4,...,x^n}$ then the differentiation operator is $$\begin{matrix} 0&1&0&0&\dots&0 \\0&0&2&0&\dots&0 \\0&0&0&3&\dots&0 \\\vdots&&&&\ddots \\0&0&0&0&\dots &n \\0&0&0&0&\dots&0 \end{matrix}$$ This is Jordanish but not Jordan... I wanted to try with a basis where instead of vector $x^n$ I have $\frac{1}{n} x^n$ but of course it does not work. What is the trick here?
Update: That was quick but I think I found it $\{n!,n!x,\frac{n!}{2!}x^2,\dots,\frac{n!}{(n-1)!}x^{n-1},n!x^n\}$ does the job.