# Jordan Canonical Form set of polynomials over complex field

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.

• What are $P_4(C)$, $p''$ and $p'$? – Aqua Aug 15 '17 at 20:57
• 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. – Frank Drebin Aug 15 '17 at 22:56

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 $\{N^4(X^4),N^3(X^4),N^2(X^4),N(X^4),X^4\}$.
• $N(X^4)=12X^2-12X^3$; then the $5^{th}$ column of $N$ is $[0,0,12,-12,0]^T$. – loup blanc Aug 15 '17 at 23:08