How to find eigenvalues and eigenvectors of Laplace operator? I have a Laplace operator,
$$
L = \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}},
$$
in polynomial space $ p(x, y)$.
How to find its eigenvectors and eigenvalues? I think, that eigenvalues is equal to zero. 
 A: One of the first things you learn about differentiation is that it reduces the degree of a polynomial. Using this idea, you should be able to prove that there's some term in any polynomial $q$ that doesn't appear in $L(q)$. Hence if $L(q) = \lambda q$, compare coefficients to show it must be the case that $\lambda = 0$. Then you only need to find those $q$ such that $L(q) = 0$.
$L$ is a linear operator; what we are really trying to find here is its kernel. The kernel of a linear operator is a vector (sub)space, so we just need to find a basis for it.
It's easy to think of monomials $x^iy^j$ that are zero under both operators – those with both $i$ and $j$ less than 2 – so these will all be eigenvectors, and since they have the same eigenvalue (0), any linear combination of them will be an eigenvector too.
However, more tricky is finding those $p(x,y)$ such that $$\frac{\partial ^2 p}{\partial x^2} = -\frac{\partial^2 p}{\partial y^2}\not= 0$$. To do this, write a polynomial $p(x,y)$ as a polynomial in $x$ that has coefficients polynomials in $y$, then look at the highest power of $x$, say $x^nq(y)$ for some polynomial $q$.
$\partial^2 /\partial x^2 $ is going to produce no terms involving $x^n$, so it follows that $\partial^2 / \partial y^2$ must kill off all these terms itself, i.e. if the term is $x^nq(y)$, it must be the case that $q^{\prime\prime}(y) = 0$. So the highest power of $x$ appears only multiplied by either a constant, or a constant multiple of $y$.
Conversely, if $x^ny$ is in the polynomial you're looking at, you're also going to need a term $-\frac{1}{6}n(n-1)x^{n-2}y^3$ to cancel it out, and then you're going to need another term to cancel that out, and so on. However, it turns out that this works, and so there is a polynomial involving $x^ny$ for each $n$ such that $L$ sends it to zero, let's call it $p_{n,1}(x,y)$. Likewise, there is a $p_{n,0}(x,y)$ that involves $x^n$ alone.
Finding the exact form of these polynomials is a nuisance, but once you've done so, you've found everything: suppose $q(x,y)$ is some polynomial that is mapped to zero, then it must be a linear combination of the polynomials you've found above. If not, then there is a counterexample with the highest power of $x$ minimal, then subtract multiples of $p_{n,0}$ and $p_{n,1}$ until that power of $x$ goes away: by assumption, this result is also not a linear combination of the polynomials found above, but it has smaller highest power of $x$, contradicting minimality. So the polynomials that map to zero (i.e. the eigenvectors) are all linear combinations of the $p_{n,0}$ and $p_{n,1}$.
Phew, that was more work, and more complex, than I anticipated. If it's all too much, just try to understand why the following equations hold (indicating the relevant polynomials are eigenvectors):
\begin{equation*}
L(x^5y - \tfrac{20}{6}x^3y^3 + xy^5) = 0 \\
L(x^6 - 15x^4y^2 + 15x^2y^4 - y^6) = 0 \\
L(x^6 + 6x^5y - 15x^4y^2 - 20x^3y^3 + 15x^2y^4 + 6xy^5 - y^6) = 0
\end{equation*}
A: It would be hard to explain it all here, but using complex analysis you can show that the polynomials whose Laplacian are zero are exactly the polynomials that can be written as the real part of a complex polynomial, i.e. $f(x,y) = {\rm Re}(\sum_{i = 0}^n a_iz^i)$ for some complex numbers $a_0,...,a_n$. So for example $x^2 - y^2 = {\rm Re}(z^2)$ and $xy = {\rm Re}(-{iz^2 \over 2})$ work.
As benmachine mentioned, there are no nonzero eigenvalues. 
