Im trying to understand a proof from my professor about the dimension of the solution space of an nth order homogeneous linear equation, namely that it is n-dimensional given the conditions below. In that proof, im supposed to understand that that conclusion follows from the following lemma.
Lemma: Let $t_0 \in I$ and $x_1(t),…,x_r(t)$ be solutions of
$x^{(n)}(t)+p_{n-1}(t)x^{(n-1)}(t)+…..+p_0(t)x^{(0)}(t) = 0$
where the coefficients $p_i(t),1\leq i \leq n,$ are continuous on the interval $I$.
Then $\{x_1(t),…,x_r(t)\}$ is linearly independent if and only if ${\bf \bar x_1}(t_0),…, {\bf \bar x_r}(t_0)$ are linearly independent vectors in $\Re^n$, where ${\bf\bar x_i}(t_0)=(x_i(t_0),x_i'(t_0),…,x_i^{(n-1)}(t_0))$
I understand the lemma, but i fail to make the connection to the theorem, any hints?
//More specifically, what guarantees that we can find $n$ independent solutions? Existence and uniqueness theorem says that there exists a unique solution, but what implies that that solution consists of $n$ independent solutions in general?
Best Answer
This is guaranteed by Liouville's Formula which says that any matrix $X(t)$ satisfying an equation $X' = A(t) X$, where $A$ is a continuous matrix-valued function on $I$ has the following property:
$$ \det(X(t)) = \det(X(t_o)) \exp\left( \int_{t_o}^t \mathrm{tr}(A(t')) \mathrm{d}t'\right) $$
thus, if the initial conditions are chosen to be linearly independent, then the columns of x are always linearly independent (determinant non-zero).