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N: $P_4(C) \to P_4(C)$ defined by $N(p) = p''- 3p'$

Find the canonical form and a canonical basis for the mapping of $N$.

I am not sure how to compute this when it is over $\mathbb{C}$ instead of over $\mathbb{R}$. I tried looking on the internet for set of polynomials over a field but could not find anything helpful.

A full solution would be lovely to show how you would even compute $N^2$, $N^3$, etc.

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  • $\begingroup$ What are $P_4(C)$, $p''$ and $p'$? $\endgroup$ – Aqua Aug 15 '17 at 20:57
  • $\begingroup$ P4(C) is the vector space of the polynomials of the complex variable X of degree ≤4 p′,p′′,⋯ are the derivatives of the polynomial p. $\endgroup$ – Frank Drebin Aug 15 '17 at 22:56
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I assume that $P_4(\mathbb{C})$ is the vector space of the polynomials of the complex variable $X$ of degree $\leq 4$ and $p',p'',\cdots$ are the derivatives of the polynomial $p$.

The matrix of $N$ is $5\times 5$ and is nilpotent. Indeed, $N^5=0$ and $N^4\not= 0$ (because $N^4(X^4)\not= 0$).

Then the jordan form of $N$ is $J_5$, the nilpotent jordan block of dimension $5$. A basis that "Jordanizes" $N$ is $\{N^4(X^4),N^3(X^4),N^2(X^4),N(X^4),X^4\}$.

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  • $\begingroup$ So to actually compute N and N(x^4) we would just put X degree 4 in matrix form and then square it to get N^2, multiply by N again to get N^3 and so on. Correct? $\endgroup$ – Frank Drebin Aug 15 '17 at 22:59
  • $\begingroup$ A better question would be what does the N matrix look like? That's where I'm confused. $\endgroup$ – Frank Drebin Aug 15 '17 at 23:02
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    $\begingroup$ $N(X^4)=12X^2-12X^3$; then the $5^{th}$ column of $N$ is $[0,0,12,-12,0]^T$. $\endgroup$ – loup blanc Aug 15 '17 at 23:08

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