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.

  • $\begingroup$ Neither is it a square matrix. $\endgroup$
    – amd
    Jul 2 '18 at 20:53
  • $\begingroup$ @amd please,clarify what you mean $\endgroup$
    – Kran
    Jul 2 '18 at 20:55
  • $\begingroup$ The matrix that you’ve got in your question has dimension $(n+1)\times n$. It makes no sense to speak of its Jordan normal form. You need a row of zeros at the bottom. $\endgroup$
    – amd
    Jul 2 '18 at 20:56
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    $\begingroup$ Have you figured out what the Jordan normal form of the differentiation operator is? That might be a good starting point. You’re on the right track with the basis that you’re trying, at any rate. $\endgroup$
    – amd
    Jul 2 '18 at 20:56
  • $\begingroup$ @amd now I got what you meant I will updeate the matrix. I found the basis I think. And do you mean that I should know the normal form since the matrix Ihave have characteristic polynomial $\lambda^n=0$ what implies the only eigenvalue is 0? $\endgroup$
    – Kran
    Jul 2 '18 at 21:05

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