Find Jordan Basis of triple differentiation operator in $\mathbb{R}_8[x]$

Question

General information and notation at first. Let $A$ be matrix of $T$ and let $T$ be such operator, that $(1, x, x^2, \cdots x^8) \to (0, 0, 0, 6, 24x, 60x^2, 120x^3, 210x^4, 336x^5)$

$T^2$:
$(1, x, x^2 \cdots x^8) \to (0, 0, \cdots, 720, 5040x, 20160x^2)$
Transformation is nilpotent, since $T^3$ leads to zeroes.
Also the only non-zero values are at the diagonal that is above of the main diagonal. Hence, all 9 eigenvalues equal to $0$.

In Jordan normal form of $A$ we have three Jordan blocks of size $3$ with $0$ at diagonal.

Now back to the question. I suspect, there is some pretty interesting way to find Jordan basis. Something that involves playing around with coefficients. Factorials (perhaps, not completely) involved too, since, for example $x^8 \to 336x^5$, where $336 = 8\cdot 7 \cdot 6$. But I can't grasp this idea, unfortunately.

I am acquainted with some general algorithms for finding basis but seems like I lack understanding of this theme. Anyway, what can we do here?

Answer 1

Hint Denote the $3 \times 3$ Jordan block of eigenvalue $0$ by $$J := \pmatrix{\cdot&1&\cdot\\\cdot&\cdot&1\\\cdot&\cdot&\cdot} ,$$ and denote the standard basis of $\Bbb R^3$ by $({\bf E}_1, {\bf E}_2, {\bf E}_3)$. The operator defined on $\Bbb R^3$ by $J$ is characterized by \begin{align*} J{\bf E}_3 &= {\bf E}_2 \\ J{\bf E}_2 &= {\bf E}_1 \\ J{\bf E}_1 &= {\bf 0} . \end{align*} So, to begin defining a Jordan basis $({\bf e}_i)$ of $\Bbb R_8[x]$ with respect to $T$, we choose for ${\bf e}_3$ a vector $p(x)$ such that $T^2 p (x) \neq 0$, i.e., ., $\partial_x^6 p(x) \neq 0$, then take $${\bf e}_2 := T p(x) = \partial_x^3 p(x) \qquad \textrm{and} \qquad {\bf e_1} := T^2 p(x) = \partial_x^6 p(x) ,$$ which ensures exactly that the matrix representation with respect to $\mathcal{B} := ({\bf e}_1, {\bf e}_2, {\bf e}_3)$ of $T\vert_{\mathcal B}$ is $J$. It's convenient, for example, to take ${\bf e}_3 := x^6$, in which case ${\bf e}_2 := T {\bf e_3} = \partial_x^3 (x^6) = \frac{6!}{3!} x^3$ and ${\bf e}_1 = T {\bf e_2} = \partial_x^3 \left(\frac{6!}{3!} x^3\right) = 6!$.

We can then apply the same procedure to find triples $({\bf e}_4, {\bf e}_5, {\bf e}_6)$ and $({\bf e}_7, {\bf e}_8, {\bf e}_9)$ of vectors corresponding to the other copies of the Jordan block $J$.

Additional hint We know we must pick ${\bf e}_6$, ${\bf e}_9$ not in the kernel of $T^2 = \partial_x^6$ and so that $({\bf e}_3, {\bf e}_6, {\bf e}_9)$ is linearly independent, so extending our previous choice we may as well take ${\bf e}_6 = x^7$ and ${\bf e}_9 = x^8$, which determines the rest of the basis.

Answer 2

1. Yes, there is a direct way connected to the concept of partition of an integer (here $9$)

The Jordan decomposition you have found corresponds to this partition where the numbers on the RHS are block's sizes :

$$9 = 3 + 3 + 3$$

$$J=\left(\begin{array}{ccc|ccc|ccc} 0&1&0&0&0&0&0&0&0\\ 0&0&1&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0\\ \hline 0&0&0&0&1&0&0&0&0\\ 0&0&0&0&0&1&0&0&0\\ 0&0&0&0&0&0&0&0&0\\ \hline 0&0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&0&1\\ 0&0&0&0&0&0&0&0&0\\ \hline \end{array}\right)$$

whereas partition :

$$9=4+3+2$$

corresponds to this Jordan decomposition :

$$J=\left(\begin{array}{cccc|ccc|cc} 0&1&0&0&0&0&0&0&0\\ 0&0&1&0&0&0&0&0&0\\ 0&0&0&1&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0\\ \hline 0&0&0&0&0&1&0&0&0\\ 0&0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&0&0&0\\ \hline 0&0&0&0&0&0&0&0&1\\ 0&0&0&0&0&0&0&0&0\\ \hline \end{array}\right)$$

The first one is the only partition compatible with constraints

$$A^2 \ne I \ \ \text{and} \ \ A^3=I$$

Indeed, for a nilpotent operator $A$, to partition :

$$9=s_1 + s_2 + s_3 + \cdots \ \text{where} \ s_1 \ge s_2 \ge s_3...$$

are the resp. sizes of the Jordan blocks, the order of nilpotence of the global Jordan matrix is

$$n=lcm(s_1,s_2,s_3\cdots)\tag{1}$$

ruling out in particular Jordan blocks of size $>3$ like the second decomposition above which has order of nilpotence equal to $12$ ! On the other side, Jordan blocks of size $2$ ar not welcome because if we have some of them, there will remain a Jordan block of size $7$, or $5$ or $3$, and relationship (1) gives an order of nilpotence $\ne 3$...

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2. One can understand it on the equivalent simpler case of the $\color{red}{second}$ differentiation operator in $\mathbb{R}_{\color{red}{3}}[X]$.

Its matrix is

$$A=\pmatrix{0&0&0&0\\0&0&0&0\\6&0&0&0\\0&2&0&0}$$

with $A^2=0$. Its Jordan form is :

$$J=\left(\begin{array}{cc|cc}0&1&0&0\\0&0&0&0\\ \hline 0&0&0&1\\0&0&0&0 \end{array}\right)$$

which is similar to $A$ :

$$V^{-1}AV=J \ \text{with} \ V=\pmatrix{0&1&0&0\\0&0&0&1\\6&0&0&0\\0&0&2&0}$$

This Jordan decomposition being the only possibility for a matrix $A$ such that $A^2=I$ ; indeed, the only challenger would have been :

$$J_1=\pmatrix{0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0}$$

impossible because : $J_1^2 \ne I$.