Constant coefficient PDEs have infinite dimensional kernels? I'm looking for exactly what's in the title: a proof of the fact that constant coefficient linear PDEs have infinite dimensional solution spaces (as a subspace of the smooth functions). This should be simple but for some reason I'm not seeing it. Obviously this follows from the $d=2$ case, and I wanted to somehow change variables in such a way that I end up with only one variable (because $d^n/dx^n$ clearly has an infinite dimensional kernel - at least all functions of only $y$) but I'm not really sure what that means.
Edit: I think there should be some sort of correspondence between points in the zero set of the characteristic polynomial and solutions to the equation, but I have no idea how to formalize this.
 A: First: We must rule out the zeroth order operators, since the multiplication operator $\mathbf{c}: f\mapsto cf$ where $c$ is a nonzero constant has only the trivial kernel, and is technically a partial differential operator. 
There is a powerful theorem in analysis (see, e.g. Hormander, Analysis of Linear Partial Differential Operators, volume I, Theorem 7.3.6) that states

All solutions of $P(D)u = 0$ where $P(D)$ is a constant coefficient partial differential operators on $C^\infty(\mathbb{R}^d)$ belong to the closed linear hulls of "exponential solutions", i.e. solutions of the form $$ p(x) e^{i x\cdot \xi} $$
  where $p(x)$ is some polynomial in $x$ and $\xi\in \mathbb{C}^d$ is a generalized frequency. 

We will not use this theorem here, but its validity should be something you can keep in the back of your mind. For the purpose of the present discussion, we can ignore the $p(x)$ factor and concentrate solely on the "plane wave" solutions with generalized frequency $\xi$. 
Note on notation: as in Hormander, we take $D = - i \partial$ to be the differential operator whose Fourier transform is $\xi$. So $P(D)$ is formally a polynomial in the symbol $D$ and can be interpreted as the constant coefficient partial differential operator of degree $n$
$$ P(D) = \sum_{|\beta| \leq n} a_\beta (-i)^{|\beta|} \partial^\beta $$
where $\beta$ ranges over multi-indices and $a_\beta$ take values in $\mathbb{C}$. 
Consider the ansatz $u = \exp (ix\cdot \xi)$. Then we have that
$$ P(D) u = P(\xi) u $$
by a simple computation. So we see that $P(D)u = 0 \iff P(\xi) = 0$. 
The answer to your question then follows from the fact that 

$\{P(\xi) = 0\}$ has infinitely many points, provided that $d > 1$, 

which is in fact a consequence (remembering that $\xi\in \mathbb{C}^d$) of the fundamental theorem of algebra (for each fixed $\xi_2, \ldots, \xi_d\in\mathbb{C}$ the equation $P(\xi_1, \xi_2, \ldots \xi_d)= 0$ has $n$ solutions $\xi_1\in\mathbb{C}$ counting multiplicity; this also explains the case $n = 0$ of zeroth order operators). 
To be slightly more explicit, let $A \subset \{P(\xi) = 0\}$ be any finite subset, then for any mapping $b: A \to \mathbb{C}$ the function 
$$ u_{b,A}(x) = \sum_{\xi\in A} b(\xi) \exp (i x\cdot \xi) $$
is a solution. It is well known that functions of the form $\exp(i x\cdot \xi)$ are linearly independent. 
Note lastly that when $d = 1$ the fundamental theorem of algebra only gives $n$ solutions to the equation $P(\xi) = 0$; this is exactly what we expect in the ODE case (the multiplicities are broken by using the $p(x)$ factors which we've neglected). 
