I am working on the following problem:
Let $k$ be a field, let $d$ be a positive integer, and let $P_d$ be the $k$-vector space of polynomials of degree $\leq d$ in $k[x,y]$. Compute the eigenvalues and eigenvectors of the linear transformation $x\frac{d}{dx} – y\frac{d}{dy} : P_d \longrightarrow P_d$.
So far in my Linear Algebra class, we've only computed eigenvalues and eigenvectors of linear transformations from $\mathbb{R}^n$ to $\mathbb{R}^n$ by considering the associated $n \times n$ matrix so far. Thus, I wasn't sure exactly how to approach this one.
But, I tried approaching this just like the other problems we've done. So, in order to find the eigenvalues, we want to find $\lambda \in k$ such that $Tf = \lambda f$, where $f$ is a (nonzero) function in $k[x,y]$ (which would serve as an eigenvector) and $T$ is the linear transformation specified above. This gives *$x\frac{df}{dx} – y\frac{df}{dy} = \lambda f$.
How can I now solve equation * for eigenvalues and the associated eigenvectors ? Am I supposed to solve a corresponding differential equation? Even this is tough for me, since I'm not used to solving differential equations for functions in two variables. Or is there an easier approach that I'm not seeing?
I wouldn't think that I'm supposed to expand $f$ out to include all of its terms and solve from there — this can get cumbersome since we're dealing with the polynomial ring in two variables $k[x,y]$.
Any help is appreciated. Thanks!
Best Answer
Consider a general $d$ dimensional polynomial $$ f(x,y) = \sum_{i=0}^d \sum_{j=0}^d a_{ij} x^i y^j. $$ Plugging into our linear operator we obtain $$ \sum_{i=0}^d \sum_{j=0}^d (i-j -\lambda) a_{ij} x^i y^j = 0. $$ Since this needs to hold for each coefficient of $x^i y^j$, we require $(i-j -\lambda)a_{ij} = 0$ to hold for all $i,j$. This immediately implies that the eigenvalues will satisfy $\lambda = \pm n$ for $n = 0, 1, \dots, d$. To find the eigenvectors use the relation that $i-j = \lambda$ and all other $i,j$ pairs will have $a_{ij} = 0$. We obtain the eigenvectors $$\sum_{i-j=\lambda} a_{ij} x^i y^j.$$
EDIT: For example $\lambda = 0$, we have the eigenvector $\sum_{i=0}^d a_{ii} (x y)^i$. In the comments you made it should have been $\lambda = 1$ and $a x$ or $\lambda = -1$ and $b y$.