The Jordan Canonical Form of linear operator in two variables polynomial

Question

I've been trying to solve the following exercise,

In the space of bivariate polynomials of the form $f(x,y)=\sum_{n,m=0}^2a_{n,m}x^ny^m$, the lineal operator $T$ is defined by $Tf(x,y)=f(x+1,y+1)$. Obtain a Jordan Canonical Form of T

I think i've made considerable progress, but I'm concerned i am taking the longest path, and i would like to know if there exist an easy way. Let $V$ be the space of bivariate polynomials of degree at most 4, I've done the following,

First, I expressed the operator as $$Tf(x,y)=\sum_{m=0}^2(y+1)^m\left[(a_{0,m}+a_{1,m}+a_{2,m})+ x(a_{1,m}+2a_{2,m})+x^2a_{2,m}\right]$$ Then I observed that whenever $T$ is applied to a function $f$ in $V$, then the non-zero coefficients of the same degree as the polynomial (non-zero coefficients with maximum $n+m$) remain the same. This implies that the only possible eigenvalue is $\lambda=1$. To find the eigenvector i tried to found the set of functions $f\in V$ such that $Tf-\lambda f=Tf-f=0$, using the expressions above it's possible to write the following $$\array{ f(x+1,y+1)-f(x,y)=&\sum_{m=0}^2[(y+1)^m(a_{0,m}+a_{1,m}+a_{2,m})-y^ma_{0,m}]+\\ &x\sum_{m=0}^2[(y+1)^m(a_{1,m}+2a_{2,m})-y^ma_{1,m}]+\\ &x^2\sum_{m=0}^2[(y+1)^ma_{2,m}-y^ma_{2,m}] } $$ Then if this polynomial is $0$ everywhere then if we treat $y$ as a constant the univariate polynomial is $0$ everywhere and then each coeficient is $0$, i.e every sum in the expression above is $0$ and because this is true for every $y$ then this is also a $0$ everywhere polynomial for $y$ where every coefficient should be $0$, after doing this I got a equation system that has a solution only when the following is true $$ \begin{cases} a_{2,2}=a_{2,1}=a_{1,2}=0\\ a_{1,1}+2a_{2_0}=0 \\ a_{0,1}+a_{1,0}=0\\ a_{0,2}=a_{0,2}\\ \end{cases}$$ We can refer to $f\in V$ by it's coefficents, this means $f$ can be expressed as a vector in $\mathbb{R}^9$, in the cannonical basis $f$ can be represented as $f=(a_{0,0},a_{0,1},a_{1,0},a_{0,2},a_{2,0},a_{1,1},a_{2,1},a_{1,2},a_{2,2})$, then a base of the eigenspace can be given by $$\array{ v_1=(1,0,0,0,0,0,0,0,0)\\ v_2=(0,1,-1,0,0,0,0,0,0)\\ v_3=(0,0,0,1,1,-2,0,0,0)\\ }$$

I consider that analyzing the powers of $(T-\lambda I)$, like $(T-\lambda I)^2$ or $(T-\lambda I)^3$ is going to be a struggle, so, how should I construct the Jordan canonical form here?

Answer 1

The matrix of your operator is in fact a Kronecker product, bringing an important conceptual and technical simplification.

Let me take the simplified case considered by Will Jagy where operator $P(x,y) \to P(x+1,y+1)$ is restricted to the space $\mathbb{R}_{1,1}[x,y]$ with bivariate polynomials with degree at most 1 in x and at most one in y. Its matrix with respect to base $(1,x,y,xy)$ is:

$$T_{1,1}=\left(\begin{array}{cc|cc}1&1&1&1\\0&1&0&1\\ \hline0&0&1&1\\0&0&0&1\end{array}\right)$$

can be partitionned in order to recognize in it the Kronecker product of a certain matrix with itself:

$$T_{1,1}=\underbrace{\begin{pmatrix}\color{red}{1}&\color{blue}{1}\\\color{green}{0}&\color{cyan}{1}\end{pmatrix}}_{T_1} \otimes \underbrace{\begin{pmatrix}1&1\\0&1\end{pmatrix}}_{T_1}=\left(\begin{array}{cc|cc}\color{red}{1}\begin{pmatrix}1&1\\0&1\end{pmatrix}&\color{blue}{1}\begin{pmatrix}1&1\\0&1\end{pmatrix}\\ \color{green}{0}\begin{pmatrix}1&1\\0&1\end{pmatrix}&\color{cyan}{1}\begin{pmatrix}1&1\\0&1\end{pmatrix}\end{array}\right) $$

The first matrix $T_1$ represents operator $P(x)\to P(x+1)$ in univariate polynomials of degree at most $1$, the second matrix (of course identical) corresponding to the similar operation $Q(y) \to Q(y+1)$ on the other variable.

For your case, you just have to consider the $9 \times 9$ matrix:

$$T_{2,2}=\underbrace{\begin{pmatrix}1&1&1\\0&1&2\\0&0&1\end{pmatrix}}_{T_2} \otimes \underbrace{\begin{pmatrix}1&1&1\\0&1&2\\0&0&1\end{pmatrix}}_{T_2}=\left(\begin{smallmatrix}1&1&1&1&1&1&1&1&1\\ 0&1&2&0&1&2&0&1&2\\ 0&0&1&0&0&1&0&0&1\\ 0&0&0&1&1&1&2&2&2\\ 0&0&0&0&1&2&0&2&4\\ 0&0&0&0&0&1&0&0&2\\ 0&0&0&0&0&0&1&1&1\\ 0&0&0&0&0&0&0&1&2\\ 0&0&0&0&0&0&0&0&1\end{smallmatrix}\right)$$

Matrices that can be expressed into the form of Kronecker products have many properties. For example their eigenvalues/vectors are very easy to get using the eigenvalues/vectors of their constituents. Dealing with the Jordan form of a Kronecker product, we have a theorem recalled [here] (Jordan Block of Kronecker Product) (coming from Horn and Johnson classical book) giving the number of Jordan blocks: we must expect blocks of size

$$5=3+3-1, \ \ 3=3+3-3, \ \ 1=3+3-5$$

which is indeed the case. One can verify that $$V^{-1}T_{2,2}V=J$$

with

$$J=\left(\begin{smallmatrix} 1& 1& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 1& 1& 0\\ 0& 0& 0& 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1\end{smallmatrix}\right) \& V=\left(\begin{smallmatrix}24& 0& 0& 0& 0& 0& 0& 0& 0\\ 0&12&-12&12&-6& 1& 0&-2&-4\\ 0& 0& 2&-5&11& 0& 0.5&-1.25&-2\\ 0&12& 0&-4& 0&-1& 0& 2& 4\\ 0& 0& 8&-8& 0& 0& 0& 2& 4\\ 0& 0& 0& 2&-6& 0& 0& 0.5& 0\\ 0& 0& 2& 1& 0& 0&-0.5&-0.75&-2\\ 0& 0& 0& 2& 0& 0& 0&-0.5& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0\end{smallmatrix}\right)$$

Results obtained with the following Matlab program:

 T2=[1 1 1;
     0 1 2;
     0 0 1]
 jordan(T2), % gives matrix [1 1 0;0 1 1;0 0 1]
 T22=kron(T2,T2)
 [V,J]=jordan(T22)
 inv(V)*T22*V-J, % gives 0 matrix

Another reference for Jordan decomposition of Kronecker products: Theorem 56 p. 48 of this MSc. thesis.

Answer 2

A more brute-force approach: clearly $1,\,x-y$ and $(x-y)^2$ are eigenvectors of $T$. Therefore we expect the matrix of $T$ to look simpler if the ordered basis contains them as members, with probably some other members divisible by $x-y$. When the characteristic of the underlying field is $0$, we may try the following: $$ \{1,\,x+y,\,xy,\,(x-y)^2,\,xy(x+y),\,x^2y^2,\,x-y,\,(x+y)(x-y),\,xy(x-y)\}. $$ The matrix of $T$ with respect to this ordered basis is $$ \pmatrix{1&2&1&0&2&1\\ &1&1&0&3&2\\ &&1&0&6&6\\ &&&1&1&1\\ &&&&1&2\\ &&&&&1\\ &&&&&&1&2&1\\ &&&&&&&1&1\\ &&&&&&&&1}. $$ Let $B=A(1:6,1:6)-I_6$. It is straightforward to verify that $\operatorname{rank}(B)=4$ and $B^4\ne0=B^5$. Therefore the Jordan form of $B$ is $J_5(0)\oplus0$. Also, the Jordan form of $A(7:9,7:9)-I_3$ is obviously $J_3(0)$. Hence the Jordan form of $T$ is $J_5(1)\oplus J_3(1)\oplus1$.