Can a set of solutions $(\mathbf{x_1}, \dots, \mathbf{x_k})$ to a linear system $\dot{\mathbf{x}} = \mathbf{Ax}$ be linearly dependent?

Question

Let the set $(\mathbf{x_1}, \dots, \mathbf{x_k})$ be solution to a linear system of differential equations $\dot{\mathbf{x}} = \mathbf{Ax}$. Can we have that for $t=t_1$ that $(\mathbf{x_1}(t_1), \dots, \mathbf{x_k}(t_1))$ are linearly dependent, while for another $t=t_2$ we have that $(\mathbf{x_1}(t_2), \dots, \mathbf{x_k}(t_2))$ is linearly independent, where $t_1 \neq t_2$?

How can I show this? If the solutions are linearly dependent for $t=t_1$ it means that the equation

$$c_1\mathbf{x_1}(t_1) + \dots + c_k\mathbf{x_k}(t_1) = \mathbf{0} $$

has solutions excluding the trivial $c_1 = \dots = c_k = 0$.

If the solutions are linearly independent at $t=t_2$ then the equation

$$c_1\mathbf{x_1}(t_2) + \dots + c_k\mathbf{x_k}(t_2) = \mathbf{0} $$

only has the trivial solution $c_1 = \dots = c_k = 0$.

Answer 1

Let $I \subset \mathbb{R}$ be an interval and $A : I \to \mathbb{K}^{n\times n}$ continuous on $I$. Then for any $t_0 \in I$ and $x_0 \in \mathbb{K}^n$ there exist a unique differentiable $\varphi : I \to \mathbb{K}^n$ so that $\dot{ \varphi }(t) = A(t) \varphi(t)$ for all $t\in I$ and $\varphi(t_0) = x_0$.

Let $ \mathcal{L}$ be the vector space of solutions. For a fixed $t_0 \in I$ consider the initial value map $$i_{t_0} :\mathcal{L} \to \mathbb{K}^n $$ defined by $i_{t_0}(\varphi) = \varphi (t_0) $ for $\varphi \in \mathcal{L}$. Clearly this is a linear map and $i_{t_0}$ is surjective, because there exist a solution for any initial value and injective, because the solution is unique. Therefore $i_{t_0}$ is an isomorphism.

Now let $\varphi_1, \dots, \varphi_k \in \mathcal{L}$ be solutions. Let $ t_1 \in I$ so that $\varphi_1(t_1) , \dots , \varphi_k(t_1) \in \mathbb{K}^n$ are linearly independent, then $\varphi_1, \dots, \varphi_k$ are also linearly independent, because $i_{t_1}$ is an isomorphism and therefore they span a $k$ dimensional subspace of $\mathcal{L}$.

Assume that there exist $t_2 \in I$ so that $\varphi_1(t_2) , \dots, \varphi_k(t_2)$ are linearly dependent, then by the same argument $\varphi_1, \dots , \varphi_k$ would span a subspace of $\mathcal{L}$ with a dimension strictly less than $k$ which is a contradiction.