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 $$egin{aligned}x(t) &= e^{At}x_0 \&=e^{At}(alpha_1 v_1 + cdots + alpha_n v_n) \&= alpha_1 e^{lambda_1 t} v_1 + cdots + alpha_n e^{lambda_n t} v_n \&= left[egin{array}{ccc}v_1 & cdots & v_n end{array} ight] left[egin{array}{c} alpha_1 e^{lambda_1 t}\ vdots \ alpha_n e^{lambda_n t} end{array} ight] end{aligned}.$$
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
$$x(t) = left[egin{array}{ccc} e^{lambda_1 t} v_1 & cdots &e^{lambda_n t} v_n end{array} ight] left[egin{array}{c} alpha_1 \ vdots \ alpha_n end{array} ight].$$
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 $mathbb{R}$.