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I was solving this exercise about Jordan normal form:

Find the Jordan normal form $J$ and a string base $S$ for the derivative operator $D : \Bbb C[t]≤3 → \Bbb C[t]≤3$ defined by setting $Dp(t):=p'(t)$.

I easly find the string base $S={\{6,6t,3t^2,t^3}\}$

Because

$t^3-D→3t^2-D→6t-D→6-D→0$

Then I had to find the Jordan normal form and I was very stuck so I read the solution:

$J=J0,4$ or in other words \begin{bmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{bmatrix} Well, I don't get it. I undersand why the $p$ index of the Jordan block is $4$, but I don't understand the $0$ on the main diagonal.

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The diagonal elements are the coefficients $\alpha_{jj}$ in $D(e_i)=\sum_i \alpha_{ij} e_j$, for the scaled basis $e_1,\ldots ,e_4$ of polynomials. But because, say, $D(e_1)=0$, $\alpha_{11}=0$, and because $D(e_2)=e_1+0\cdot e_2$, we have $\alpha_{22}=0$, etc. So all diagonal elements are zero - which was your question.

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