Eigenvalues of linear operator of the space $\mathbb{R}[x,y]$ 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
$$
\phi(f(x,y)) = y\frac{df}{dx} +x\frac{df}{dy}.
$$


*

*Prove that every integer is an eigenvalue of this operator.

*Find all eigenvalues of this operator. 

*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.
 A: 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 ...).
A: 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.
