# How the dimension of a subspace related to the differential operator

I am wondering the link as the title implies. The Spring 87 problem in Berkeley Problems in Mathematics is as follows:

Let $V$ be a finite dimensional linear subspace of $C^{\infty}(\mathbb{R})$. Assume that $V$ is closed under differentiation. Prove that there is a constant coefficient operator $$L=\sum^{n}_{k=0}a_{k}D^{k}$$ such that $V=\{f:Lf=0\}$.

I am wondering why the finite dimensional condition given would imply such a strong result. Because if we assume the required relationship the reverse is not necessarily true(the whole space is obviously closed under any differential operator), I feel something deeper may be buried in this problem I do not know. A hint or an illuminating example would be mostly welcome. I just do not know how to attack this problem.

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Let $\{f_1,\dots,f_N\}$ a basis of $V$. Since $f'_k\in V$ for all $k$, we can write $$f'_k=\sum_{l=1}^Na_{k,l}f_l,$$ and using matrices $$\pmatrix{f'_1\\\vdots\\ f'_N}=A\pmatrix{f_1\\\vdots\\ f_N}.$$ Let $P$ the minimal polynomial of $A$ over $\Bbb R$. We can check that the differential operator $L$ associated to it is such that $V\subset \{f\mid Lf=0\}$. The general theory of differential equations ensures the converse: if $Lf=0$, then $f$ is in the vector space generated by $x^pe^{\lambda x}$, where $\lambda$ are eigenvalues of $A$, $m$ is the multiplicity of the root $\lambda$ in the minimal polynomial and $0\leq p\leq m$. Each of these maps are in $V$.
Note that if $V=\ker L$ for $L$ of the form $\sum_{k=0}^n a_kD^k$, where $a_k$ are constant, then $V$ is necessarily finite dimensional.
For the converse, what I meant is to use the fact that if $Lf=0$, then $f$ is in the vector space generated by $x^me^{i\lambda x}$, where $\lambda$ are eigenvalues of $A$, where $m$ is the multiplicity of the root $\lambda$ in the minimal polynomial. Each of these maps are in $V$. – Davide Giraudo Jul 21 '12 at 11:02
Davide, if this is so, then assuming we have $n-2$ different eigenvalues and 1 eigenvalue of multiplicty 2. By your answer we would have only $n-1$ $f_{i}$s available. Do you mean $x^{i}e^{i\lambda x}$ with $1\le i\le m$? – Bombyx mori Jul 22 '12 at 6:34
OIn fact I don't know why I wrote $i$ in the exponential, and yes we have to take the powers of lower degree. I will edit it. Thanks! – Davide Giraudo Jul 22 '12 at 9:10
Davide, if so then we would have $0,1...m$ together $m+1$ basis vectors. – Bombyx mori Jul 22 '12 at 21:08