# Why the existence and uniqueness of solution of ODE implies existence of $n$ independent solution of $\dot x=A(t)x$ where $A$ is periodic?

Let $$A(t)\in \mathcal M_{n\times n}(\mathbb R)$$ periodic, i.e. $$\exists \quad T>0:\forall t\in \mathbb R, A(t+T)=A(t).$$ Consider the system $$\dot x(t)=A(t)x(t),\quad x(t)\in \mathbb R^n.$$ It's written in my course that if $$t\mapsto A(t)$$ is continuous, the existence and uniqueness theorem for ODE implies the existence of $$n$$ linearly independents solutions.

I don't understand in what this "existence and uniqueness theorem for ODE" implies existence of $$n$$ linearly independents solution. Could someone explain why ?

This theorem says (more or less) :

If $$F:\mathbb R\times \mathbb R^n\to \mathbb R^n$$ is locally Lipschitz w.r.t. the second variable, then the Cauchy problem $$\begin{cases}\dot x(t)=F(t,x(t))\\ x(0)=x_0\end{cases}$$ has a unique (local) solution.

So indeed, we take $$F(t,x)=A(t)x$$ which is obviously globally Lipschitz. But in one case we have a Cauchy problem, and in the other case we have an ODE without initial condition.

A basis for the solution set is given by the $$n$$ functions $$g_k(t)$$ (where $$1 \le k \le n$$), where $$x(t)=g_k(t)$$ is the unique solution of the initital value problem $$\dot x = A x ,\qquad x(0) = e_k ,$$ where $$e_k$$ is the $$k$$th standard basis vector for $$\mathbf{R}^n$$.
Proof: If $$x=z(t)$$ is any solution of $$\dot x = Ax$$, let $$c=z(0)$$. Then $$x=\sum c_k g_k(t)$$ is also a solution, with the same initial value as $$z$$ at $$t=0$$, so by uniqueness $$z=\sum c_k g_k$$. This shows that any solution is a linear combination of the functions $$g_k$$. And if the linear combination $$\sum c_k g_k(t)$$ is the zero function, then in particular it's zero at $$t=0$$, $$\sum c_k e_k = 0$$, so all $$c_k=0$$. This shows that the functions $$g_k$$ are linearly independent. QED.
(By the way, the periodicity of $$A(t)$$ plays no role here, only the continuity.)