Show The Jordan Normal Form Of $\varphi$.

Question

Fix a nonnegative integer $n$, and consider the linear space $$\mathbb{R}_n\left [x,y \right ] := \left\{ \sum_{\substack{ i_1,i_2;\\ i_1+i_2\leq n}}a_{i_1i_2}x^{i_1}y^{i_2}\quad\Big|{}_{\quad}a_{i_1i_2}\in \mathbb{R} ; \ i_1,i_2 \text{ are non-negative integers}\right \}$$ over $\mathbb{R}$ where two operations, addition and scalar multiplication, are defined as usual. $\\$

A linear map $\varphi$ from $\mathbb{R}_n\left [x,y \right ]$ to $\mathbb{R}_n\left[x,y \right ]$ defined as following:

$$\forall f(x,y)\in \mathbb{R}_n\left [x,y \right ],\quad\varphi(f):=2\frac{\partial f }{\partial x}+ \frac{\partial f }{\partial y}.\quad$$ $\\$ Show the jordan normal form of $\varphi$.

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When $n=1,$ $$span\{2,x,-\frac{1}{2}x+y\}=\mathbb{R}_1\left [x,y \right ], $$$$\varphi(2,x,-\frac{1}{2}x+y)=(2,x,-\frac{1}{2}x+y)\left(\begin{array}{cc|cc} 0 & 1& 0\\ 0& 0& 0\\ \hline 0& 0& 0 \end{array}\right).$$

When $n=2,$ $$span\{1,x,y,xy,x^2,y^2\}=\mathbb{R}_2\left [x,y \right ], $$it is not difficult to calculate the jordan normal form of $\varphi$ is

$$\left(\begin{array}{ccc|cc|c} 0& 1& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ \hline0& 0& 0 & 0& 1& 0\\ 0& 0& 0 & 0& 0& 0\\ \hline 0& 0& 0 & 0& 0& 0\\ \end{array}\right).$$

But how to generalize it to any integer $n$ and prove the generalization is correct ？

Answer 1

I would have posted the followig train of thoughts as a comment to my question,but the symbols failed to display on mobile devices. $\\$

By using the mathematical inductive method,we only need to prove：

For any fixed integer $n$, $\textbf{(1)}$ the order of jordan normal form of $\varphi$ is $\frac{(n+1)(n+2)}{2} $; $\textbf{(2)} $ the minimal polynomial of $\varphi$ is $\lambda^{n+1};$ $\textbf{(3)}$ $\dim\ker(\varphi)$ is $n+1$$\left(i. e.rank(\varphi)=\frac{n(n+1)}{2}\right).$

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It is not difficult to get $\textbf{(1)}$ and $\textbf{(2)}$.With regard to $\textbf{(3)}$，we just observe that $$span \left \{ \varphi(1)\arrowvert\varphi(x),\varphi(y)\right \}=span\left\{1\right\};$$$$span \left \{ \varphi(1)\arrowvert\varphi(x),\varphi(y)\arrowvert\varphi(xy),\varphi(x^2),\varphi(y^2)\right \}=span\left\{1\right\}\oplus span\left \{ x,y \right \}; $$$$span \left \{ \varphi(1)\arrowvert\varphi(x),\varphi(y)\arrowvert\varphi(xy),\varphi(x^2),\varphi(y^2)\arrowvert\varphi(x^2y),\varphi(xy^2),\varphi(x^3),\varphi(y^3)\right \}=span\left\{1\right\}\oplus span\left \{ x,y \right \}\oplus span\left\{xy,x^2,y^2\right\};$$$$\cdots\quad\cdots\quad\cdots\quad\cdots\quad.$$

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$\\$And then，$\textbf{(1)}+\textbf{(2)}+\textbf{(3)} \Longrightarrow \textbf{(4)}$ For any fixed integer $n$, the jordan normal form of $\varphi$ is a nilpotent matrix its order is $\frac{(n+1)(n+2)}{2}$ and the total number of jordan blocks is $n+1$. Those are

$(0)_{1\times1},\begin{pmatrix} 0& 1 \\ 0& 0\\ \end{pmatrix}_{2\times 2},\begin{pmatrix} 0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\\ \end{pmatrix}_{3\times3},\cdots \cdots,\begin{pmatrix} 0 & 1 & & & \\ & 0 & 1& & \\ & & \ddots& \ddots& \\ & & & 0 & 1\\ & & & & 0 \end{pmatrix}_{(n+1)\times(n+1)}.$

Answer 2

There is an anther approach.

Let $f^{k}_{1}=(x-2y)^k (k=0，1，\cdots，n)$$, \varphi(f^{k}_1)=0.$For a fixed $ k(\geq 1),$We only need to find $f^{k}_{t}(t=1，2\cdots，k),$such that $\varphi (f^{k}_{t+1})=f^{k}_{t},$ $$\varphi(f^{k}_{1},f^{k}_{2},\cdots,f^{k}_{k+1})$$$$=(f^{k}_{1},f^{k}_{2},\cdots,f^{k}_{k+1})\begin{pmatrix} 0 & 1 & & & \\ & 0 & 1& & \\ & & \ddots& \ddots& \\ & & & 0 & 1\\ & & & & 0 \end{pmatrix}_{(k+1)\times(k+1)}.$$ Using some knowledge of First order linear partial differential equation, it is not difficult to find those $f^{k}_{t}$s.