Eigenvalues of linear operator of the space $\mathbb{R}[x,y]$

Question

Let $\mathbb{R}[x,y]$ be a space of real polynomials of two variables. Define a linear operator $\phi$ on this space by the formula $$ \phi(f(x,y)) = y\frac{df}{dx} +x\frac{df}{dy}. $$

1. Prove that every integer is an eigenvalue of this operator.
2. Find all eigenvalues of this operator.
3. Is this operator diagonazible?

For the first item. It is not hard to show by induction that $(x+y)^n$ is a vector( of the corresponding vector space $\mathbb{R}[x,y]$) belonging to the eigenvalue $n$ and $(x-y)^n$ to the eigenvalue $-n$. Every constant polynomial belongs to the zero eigenvalue.

But I don't know how to solve the second and third item. I want to show that there are no any other (non-integer) eigenvalues.

Also we can see that the system $$ \left((x\pm y)^n| n\in \mathbb{Z} \right) $$ is linearly independent.

I will be gratefull for hints and ideas.

Answer 1

The operator $D = x\partial_{y} + y\partial_{x}$ is a derivation, meaning that it satisfies the Leibniz (product) rule $D(PQ) = D(P)Q + PD(Q)$. This has the consequence that the product of eigenvectors is again an eigenvector, with eigenvalue equal to the sum of the eigenvalues of the eigenvectors. For example $(x + y)^{p}(x-y)^{q}$ is an eigenvector of $D$ with eigenvalue $p - q$.

To check that all the eigenvalues of $D$ are integers and to decide whether $D$ is diagonalizable, you need to find a basis of $\mathbb{R}[x, y]$. One knows the monomial basis $x^{p}y^{q}$. In this exercise it seems more convenient to work with powers of $x + y$ and $x - y$ (or to make a change of variables ...).

Answer 2

Solving

$$ y f_x+x f_y = \lambda f $$

using the method of characteristics

$$ \frac{dx}{y}=\frac{dy}{x}\Rightarrow x^2-y^2 = C_1 $$

$$ \frac{d(x+y)}{x+y} = \frac{df}{\lambda u}\Rightarrow f = C_2(x+y)^{\lambda} =\phi(x^2-y^2)(x+y)^{\lambda} $$

so $\lambda \in \mathbb{N}$ is an eigenvalue because $f(x,y)$ should be a polynomial.