Linear System Theory: zero-input response of LTI system

In this note, we consider a $n$-dimensional linear time-invariant (LTI) system described by the state-space equation:

$$\dot{x}(t) = A x(t).$$

It is easy to point out that $x(t) = e^{At}x_0$ is a solution of such LTI system (corresponding to the initial state $x(0)=x_0$) and the set of all the solutions, denoted by $\mathcal{S}$, form a vector space (over $\mathbb{R}$). Now we are going to find a basis for $\mathcal{S}$.

If $A$ had $n$ eigenvectors, say $\{v_1,\ldots,v_n\}$ which corresponding to the eigenvalues $\{\lambda_1,\ldots,\lambda_n\}$, then any initial state of $x$ can be wirtten as a linear combination of $v_i$'s:

$$x_0 = \alpha_1 v_1 + \cdots + \alpha_n v_n$$

which implies that $$\begin{aligned}x(t) &= e^{At}x_0 \\&=e^{At}(\alpha_1 v_1 + \cdots + \alpha_n v_n) \\&= \alpha_1 e^{\lambda_1 t} v_1 + \cdots + \alpha_n e^{\lambda_n t} v_n \\&= \left[\begin{array}{ccc}v_1 & \cdots & v_n  \end{array}\right] \left[\begin{array}{c} \alpha_1 e^{\lambda_1 t}\\ \vdots \\ \alpha_n e^{\lambda_n t}  \end{array}\right] \end{aligned}.$$

Therefore, one may say that $\{v_1, \ldots, v_n\}$ is a basis of $\mathcal{S}$. However, actually $\mathcal{S}$ is a vector space of functions. In other words, the elements or vectors of $\mathcal{S}$ are functions of $t$. Try to rewrite the above equation as

$$x(t) = \left[\begin{array}{ccc} e^{\lambda_1 t} v_1 & \cdots &e^{\lambda_n t} v_n  \end{array}\right] \left[\begin{array}{c} \alpha_1 \\ \vdots \\ \alpha_n  \end{array}\right].$$

We will find that $\{e^{\lambda_1 t} v_1,\ldots, e^{\lambda_n t}v_n\}$ can be considered as a basis of the vector space $\mathcal{S}$ over $\mathbb{R}$.