Eigenvalues of the hyperbolic Laplacian I am working on the following Linear Algebra 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 $k[x,y]$. The hyperbolic Laplacian on polynomials of degree $d$ is the differential operator $D_d = -y^2(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2})$. For each $d \leq 5$, compute the eigenvalues of $D_d$, and, for each eigenvalue, compute the dimension of the eigenspace. (Your answer will depend on $d$.)
Here is my work so far: 
Let $f(x,y) = \sum_{i = 0}^d \sum_{j = 0}^d a_{i,j}x^iy^j \in k[x,y]$ be a $d$-dimensional polynomial, where $a_{i,j} \in k$. Then $D_d(f) = \lambda f$ 
$\Rightarrow -y^2(\frac{\partial^2f}{\partial x^2} + \frac{\partial^2f}{\partial y^2}) = \lambda f$
$\Rightarrow \sum_{i = 0}^d \sum_{j = 0}^d i(1-i)a_{i,j}x^{i-2}y^{j+2} + \sum_{i = 0}^d \sum_{j = 0}^d j(1-j)a_{i,j}x^iy^j = \sum_{i = 0}^d \sum_{j = 0}^d a_{i,j} \lambda x^iy^j$
$\Rightarrow \sum_{i = 0}^d \sum_{j = 0}^d a_{i,j}[i(1-i)x^{i-2}y^{j+2} + (j(1-j)-\lambda)x^iy^j] = 0$
From the above equation, I'm struggling with finding the solutions for the eigenvalues $\lambda$ and their corresponding eigenspace dimensions. How do I deconstruct the above equation for my solutions for $\lambda$ ? It seems a bit overwhelming. 
Thanks! 
 A: Let $Q_d$ denote the space of polynomials of total degree exactly $d$ in
$k[x,y]$. Then we have the direct sum decomposition
$P_d=\bigoplus_{k=0}^d Q_k$ and further each $Q_d$ is stable by $D$.
So, it suffices to study the eigenproperties of $D$ on each $Q_d$.
Next, if we put $b_j=y^jx^{d-j}$, notice that ${\cal B}=(b_{d},b_{d-1},b_{d-2},\ldots,b_{0})$ is a basis of $Q_d$ (in particular, $Q_d$ has dimension $d+1$).
Also, we have $D(b_{d})=-d(d-1)b_{d}, D(b_{d-1})=-(d-1)(d-2)b_{d-1}$ and $D(b_j)=-j(j-1)b_j-(d-j)(d-j-1)b_{d-j}$ for $j\lt d-1$. So the matrix of $D$ with respect to ${\cal B}$ is upper triangular, and this immediately gives us all the eigenvalues : they are the $f(j)=-j(j-1)$ for $0\leq j\leq d$ (note that some of those values may not be distinct).
The first values of $f$ are $f(0)=f(1)=0$, $f(2)=-2$, $f(3)=-6$, $f(4)=-12$ and
$f(5)=-20$.
So on $Q_d$, the eigenvalues of $D$ are $0$ with multiplicity $2$ (except for $d=0$ where it is $1$), and all the $f(k)$ for $k\leq d$, with multiplicity $1$.
Adding up, on $P_d$ the eigenvalues of $D$ are $0$ with multiplicity $2d+1$ (except for $d=0$ where it is $1$), and all the $f(k)$ for $k\leq d$, where $f(k)$ has multiplicity $d-k+1$.
