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Suppose I have a linear operator defined to be $L={\partial\over\partial x}+{\partial\over\partial y}$ acting on the vector space of polynomials $P(x,y)$ of degree $\leq n$ in $x,y$ respectively.

How do I find its eigenvalues, eigenvectors and the Jordan Canonical matrix representation of $L$?

I guess the JCF of $L$ is a $(n+1)\times (n+1)$ matrix. But how does one convert that sum of partial derivative operators into such a matrix?

Thank you.

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The dimension is wrong. The space of polynomials in $x$ and $y$ of total degree at most $n$ has more than $n+1$ dimensions; if the degree in each of $x$ and $y$ is at most $n$ (so that you can include monomials like $x^ny^n$) then it has even more (it would be an $(n+1)^2\times (n+1)^2$ matrix then). –  Arturo Magidin Nov 20 '11 at 1:32
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1 Answer 1

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There are two methods; one is to translate the information into matrix form and work with that. The other is to work with $L$ directly.

For the first method, the first step is to find a basis for your vector space. In this case, you can take the basis of all monomials of degree at most $n$.

Then apply $L$ to the vectors of the basis in order to find the coordinate matrix of $L$; then work with it.

For example, for $n=2$, a basis would be $\beta=\{1,x,x^2,y,xy, y^2\}$ (note the dimension is not $3$, as you conjectured, but in this case $6$; the dimension is equal to the number of monomials of degree at most $n$).

(Note: I'm interpreting your problem as being with polynomials of total degree at most $n$; if you meant that the degree in each of $x$ and $y$ is at most $n$, then for $n=2$ you would need to add $x^2y$, $xy^2$, and $x^2y^2$ to the basis, and include them in the computations below; the main idea is the same, though.)

Then you apply $L$ to each vector in the basis, and express the answer as a linear combination of the basis: $$\begin{align*} L(1) &= \frac{\partial}{\partial x}1 + \frac{\partial}{\partial y}1= 0 + 0\\ &=0\cdot 1 + 0 x + 0x^2 + 0y + 0xy + 0 y^2,\\ L(x) &= =\frac{\partial}{\partial x}x + \frac{\partial}{\partial y}0 = 1 + 0\\ &=1\cdot 1 + 0 x + 0x^2 + 0y + 0xy + 0 y^2,\\ L(x^2) &= \frac{\partial}{\partial x}x^2 + \frac{\partial}{\partial y}x^2 = 2x + 0\\ &= 0\cdot 1 +2x + 0x^2 + 0y + 0xy + 0y^2,\\ L(y) &= \frac{\partial}{\partial x}y + \frac{\partial}{\partial y}y = 0+1\\ &= 1\cdot 1 + 0x + 0x^2 + 0y + 0xy + 0y^2,\\ L(xy) &= \frac{\partial}{\partial x}xy + \frac{\partial}{\partial y}xy= y+x\\ &= 0\cdot 1 + 1 x + 0x^2 + 1y + 0xy + 0y^2,\\ L(y^2) &= \frac{\partial}{\partial x}y^2 + \frac{\partial}{\partial y}y^2 = 0 + 2y\\ &= 0\cdot 1 + 0x + 0x^2 + 2y + 0xy + 0y^2. \end{align*}$$ Now, relative to the basis, the coordinate matrix of $L$ is: $$\left(\begin{array}{cccccc} 0 & 1 & 0 & 1 & 0 & 0\\ 0 & 0 & 2 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 2\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right)$$ Now that you have a matrix, it is an easy matter to find the characteristic polynomial (it is $t^6$), the eigenvalues (only $\lambda=0$); the eigenspace (there are three free variables, so the dimension of the eigenspace is $3$); the eigenvectors (spanned by $1$, $x-y$, and $x^2 - 2xy + y^2$), the the Jordan canonical form (check the dimension of the nullspaces of $L^2$ (it's dimension is $5$), and of $L^3$ (which is all of the space). This gives you the Jordan canonical form (three blocks total, largest block of size $3$, so it must be one $3\times 3$ block, one $2\times 2$ block, one $1\times 1$ block).

For the other method you would have to work directly with $L$ and polynomials Solve the equation $L(a+bx+cx^2+dy + exy+fy^2) = \alpha(a+bx+cx^2+dy+exy+fy^2)$ to find that the only solutions have $\alpha=0$ (giving you the eigenvalues), and finding the explicit solutions gives the eigenvectors. Note that $L^2$ is not the zero linear transformation, then that $L^3$ is; this gives all the information you need.

(You will find that in the general case something similar occurs: you cannot have any eigenvalues other than $0$, because $L$ always reduces the maximum degree of $p(x)$ by $1$; that you need $n$ applications of $L$ to get the zero linear transformation; and you should find a pretty simple formula for the dimension of the space (hence the size of the matrix) and a nice way of seeing the Jordan canonical form.)

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Very well explained, thanks! –  glinka Nov 20 '11 at 8:55
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