Differential operator on C[x,y]

Question

I'm working on the following problem:

By differential operator $D = f\frac{\partial}{\partial{x}} + g\frac{\partial}{\partial{y}}$, we mean a $\mathbb{C}$-linear operator on $\mathbb{C}[x,y]$ such that $D(ab) = D(a)b + aD(b)$ for all $a,b \in \mathbb{C}[x,y]$ and $D(x) = f$ and $D(y) = g$.

a) Let $D = \frac{\partial}{\partial{x}} + x^2\frac{\partial}{\partial{y}}$ be a differential operator on $\mathbb{C}[x,y]$. Prove that for any $a \in \mathbb{C}[x,y]$, there exists a natural number $n = n(a)$ such that $D^n(a) = 0$.

b) Let $A$ be the restriction of the differential operator $D = x\frac{\partial}{\partial{x}} + (x-y)\frac{\partial}{\partial{y}}$ on the space of quadratic polynomials $V = \mathbb{C}x^2 + \mathbb{C}xy + \mathbb{C}y^2$. Find the characteristic polynomial of $A$.

Here is my work so far:

a) Let

$h_1(x) = \alpha_1x + \alpha_2x^2 + ... + \alpha_nx^p$ for some $p \in \mathbb{N}$

$h_2(y) = \beta_1y + \beta_2y^2 + ... + \beta_ny^m$ for some $m \in \mathbb{N}$

$h_3(x,y)$ is the linear combination of terms involving a power of $x$ multiplied by a power of $y$ with coefficients in $\mathbb{C}$

$h_4$ is the linear combination of constant terms in $\mathbb{C}$

Then $a = h_1(x) + h_2(y) + h_3(x,y) + h_4 \in \mathbb{C}[x,y]$. Since the differential operator $D$ is linear (that is, $D(a) = D(h_1(x)) + D(h_2(y)) + D(h_3(x,y)) + D(h_4)$ ), we consider $D$ on $h_1 , h_2, h_3$ and $h_4$ seperately:

$D^{p+1}(h_1(x)) = 0$.

$D^{m+1}(h_2(y)) = 0$.

$D^{q}(h_3(x,y)) = 0$ for some $q \in \mathbb{N}$.

$D(h_4) = 0$.

Thus, $D^n(a) = 0$ for $n = max(p+1, m+1, q, 1) \in \mathbb{N}$.

Is this a sound proof for part (a) ? How could my proof be improved further ? Did I skip over any important details?

b) This one introduces a new concept for me -- I'm used to finding characteristic polynomials of matrices. In the past, I've found the eigenvalues of a differential operator, but I'm not sure how to find the corresponding characteristic polynomial of the operator. I suppose this just boils down to finding the multiplicity of the eigenvalues found, but even this was difficult for me here. Still, I tried to approach it to the best of my abilities. I attempted to find the eigenvalues first.

Let $h(x,y) = ax^2 + bxy + cy^2 \in V$, where $a, b , c \in \mathbb{C}$ .

Then $A(h(x,y)) = \lambda h(x,y)$

$\Rightarrow$ $x(2ax + by) + (x-y)(bx + 2cy) = \lambda ax^2 + \lambda bxy + \lambda cy^2$

$\Rightarrow$ $(2a + b)x^2 + (2c)xy + (-2c)y^2 = \lambda ax^2 + \lambda bxy + \lambda cy^2$

$\Rightarrow$ $\lambda a = 2a + b$, $2c = \lambda b$, $2c = -\lambda c$.

The last two equations give $\lambda = 0$ or $b + c = 0$.

If $\lambda = 0$, then $c = 0$, $b$ can be any complex number, and $a = -\frac{b}{2}$. Thus, it seems to me that $\lambda = 0$ is an eigenvalue with associated eigenvectors $h(x,y) = \frac{-b}{2}x^2 + bxy$. Isn't this an eigenvalue with an infinite number of eigenvectors, then? How can I find the multiplicity of such an eigenvalue, which will help me build the characteristic polynomial of $A$ ?

If $b + c = 0$, then, I'm struggling to find the eigenvalues and associated eigenvectors in this case. It seems the eigenvalue $\lambda$ would depend on the coefficients $a,b,c$, and so again, we get an infinite number of eigenvectors, and I'm not sure how to determine the eigenvalue's multiplicity.

Thanks!

Answer 1

Part a: Order monomials first by power of $y$ then by power of $x$. Observe that $D$ takes constants to $0$ and $$ D \sum_{i=0}^M c_i x^i = \sum_{i=0}^{M-1} c_i i x^{i-1} \text{,} $$ so any element of $\Bbb{C}[x]$ of degree $M$ is sent to zero by $D^{M+1}$. Also,

$$ D \sum_{j=0}^M c_j y^j = \sum_{i=0}^{M-1} c_j x^2 y^{j-1} \text{,} $$ so any element of $\Bbb{C}[y]$ of degree $M$ is sent to an element of $\Bbb{C}[x]$ having degree $2M$ by $D^M$, so is sent to zero by an application of $D^{3M + 1}$. Finally, $$ D \sum_{i,j=0}^M c_{ij} x^i y^j = \sum_{i,j=0}^{M} \left( c_{ij} i x^{i-1}y^j + c_{ij} jx^{2+i}y^{j-1}\right) \text{.} $$ After at most $M$ applications of $D$, the leading monomial is a constant multiple of $y^M$, so after at most $M+1$ applications of $D$, the largest power of $y$ appearing in any monomial is $M-1$. By a simple induction, after at most $(M+1)M/2$ applications of $D$, $y$ no longer appears in any monomial and the resulting element of $\Bbb{C}[x]$ is reduced to zero by a finite number of applications of $D$.

For part b: I think you're overcomplicating this. $V$ is a three $\Bbb{C}$ dimensional vector space with basis $\{x^2, xy, y^2\}$ (which we take, in that order, for our representations with complex $3$-vectors), so we study the linear operator by seeing where it sends each element of the basis. \begin{align*} D(x^2) &= 2 x^2, &\text{so}&& A\begin{pmatrix}1\\0\\0\end{pmatrix} &= \begin{pmatrix}2\\0\\0\end{pmatrix} \text{,} \\ D(xy) &= x^2, &\text{so}&& A\begin{pmatrix}0\\1\\0\end{pmatrix} &= \begin{pmatrix}1\\0\\0\end{pmatrix} \text{, and} \\ D(xy) &= 2xy - y^2, &\text{so}&& A\begin{pmatrix}0\\0\\1\end{pmatrix} &= \begin{pmatrix}0\\2\\-1\end{pmatrix} \text{.} \end{align*} So (in the chosen basis) $A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & -1 \end{pmatrix}$ and $\det(A - I\lambda) = (-1-\lambda)(2-\lambda)(-\lambda) = -\lambda^3 + \lambda^2 + 2\lambda$.