Linearly Independent Solutions to 1st Order Systems

Theorem
The 1st-order homogeneous linear system of differential equations $$x' = A \left({t}\right) x$$, expressed with the vectors $$x', x: \R \to \R^n$$ and the matrix function $$A: \R \to M_{n \times n} \left({\R}\right)$$, has $$n$$ linearly independent solutions, and if $$ \phi_1, \phi_2, \dots, \phi_n$$ are $$n$$ linearly independent solutions, then $$c_1 \phi_1 + c_2 \phi_2 + \cdots + c_n \phi_n$$, where $$c_i$$ are constants, is a general solution.

Proof
Let $$v_1, v_2, \dots, v_n$$ be linearly independent vectors in $$\R^n$$, and let $$\phi_i$$ be solutions to the IVPs $$x' = A \left({t}\right) x, \, x \left({t_0}\right) = v_i$$ for $$i = 1, 2, \dots, n$$.

Suppose the solutions are not independent, i.e. $$c_1 \phi_1 + c_2 \phi_2 + \cdots + c_n \phi_n = 0$$ for some constants $$c_i$$ not all zero.

Then:
 * $$c_1 \phi_1 \left({t_0}\right) + c_2 \phi_2 \left({t_0}\right) + \cdots c_n \phi_n \left({t_0}\right) = c_1 v_1 + c_2 v_2 + \cdots + c_n v_n = 0$$

meaning the vectors $$v_i$$ are linearly dependent, a contradiction, so the solutions $$\phi_i$$ must be linearly independent.

By linearity of the system, every vector function of the form $$ x = c_1 \phi_1 + \cdots + c_n \phi_n$$ is a solution.

Let $$z$$ be an arbitrary solution of the system.

Since $$\phi_i \left({t_0}\right)$$ are linearly independent and count $$n$$ in number, they form a basis for $$\R^n$$, hence $$z \left({t_0}\right)$$ must be a linear combination of those solutions, and then by uniqueness of solutions $$z$$ is a linear combination of the vector functions $$\phi_i$$.

This proves this is a general solution.