Simplest proof of dimension of solution space for linear ODEs

Question

Given a general n-th degree linear ODE, what's the easiest way to prove that there are precisely n linearly independent solutions?

Answer 1

I don't know of an easy/easiest way to prove it for a general $n$th degree linear ODE, but it is worth pointing out that in the constant coefficient case you can get this from elementary linear algebra. The idea is that if $N$ is a positive integer and you have complex numbers $c_1, \dots, c_N$, then the solutions to the differential equation $$ \sum_{n=0}^N c_k y^{(k)} = 0 $$ (here $y^{(k)}$ denotes the $k$th derivative of $y$, interpreted as $y$ when $k=0$) are precisely the elements of the kernel of the operator $$ T = \sum_{n=0}^N c_k D^k $$ where $D$ is differentation, regarded as an operator on a vector space $V$ of functions (there is some freedom in what particular space you choose here; say the set of all infinitely differentiable functions $\mathbb{R} \to \mathbb{C}$). From the fundamental theorem of algebra, you know there are complex numbers $\omega, \omega_1, \dots, \omega_N$ with the property that the polynomial $\sum_{n=0}^N c_k z^k$ factors as $\omega \prod_{n=1}^n (z - \omega_n)$; it follows that your operator $T$ also factors, in the algebra of operators on $V$, as $$ T = \omega \prod_{n=1}^N (D - \omega_n I), $$ where $I$ denotes the identity operator on $V$.

The point is that each of the operators $D - \omega_n I$ has a one-dimensional kernel by basic calculus. (For any $k$, the function $f(t) = \exp(kt)$ is a solution to $y' = k y$, and if $g$ is any other, the quotient rule for derivatives shows that $(g/f)' = 0$. So by a standard argument involving the mean value theorem, $g/f$ is constant; so $\{f\}$ is a basis for $D - kI$.)

And it is a basic linear algebra fact that a product of $n$ operators with one-dimensional kernel, can have kernel of dimension at most $n$. (Follows from the more general assertion that if $S_1: V \to V$ and $S_2: V \to V$ are any operators, the dimension of the kernel of $S_1 S_2$ is at most the dimension of the kernel of $S_1$ plus the dimension of the kernel of $S_2$. This very easy consequence of the rank-nullity theorem--- and does not require $V$ to be finite dimensional.)

Why is the kernel of $T$ exactly $n$-dimensional? Well, just write down $n$ linearly independent elements in it, as they do in textbooks. (Of course, if you have the better sort of textbook, the entire argument just given is in there.)

For non-constant coefficients, factoring the corresponding differential operator is no longer the way you want to approach this. But for a lot of ODE, you can still get reasonably elementary theorems about the dimension of the kernel of the operator by applying some kind of transform (e.g. the Laplace transform) and getting in a position where it is just algebra again.

Answer 2

By the uniqueness property of the Cauchy problem, and linearity.