Consider \begin{equation*} \dot{x} = Ax + Bu,\quad x \in\mathbb{R}^n,\ u \in\mathbb{R}^m,\quad A \in\mathbb{R}^{n\times n},\ B \in \mathbb{R}^{n\times m} \end{equation*} \begin{equation*} \text{Rank}(\mathcal{C} = [B \ AB \ A^2B \ \cdots \ A^{n-1}B]) = n \iff \text{ (A,B) is Observable } \end{equation*} I came across the following Claim: \begin{equation} Range(\mathcal{C}) = \text{All reachable states} \end{equation} But,
Range($\mathcal{C}$) = $\sum_{i=0}^{n-1}A^iB\cdot v_i, \ \forall v_i \in \mathbb{R}^m$
Reachable states = { $x \in \mathbb{R}^n$ | $x=\int_0^{t_1} e^{A(t_1-\tau)}Bu(\tau)d\tau\ \ \forall u(\tau) \in \text{piecewise continuous }$}, as $e^{At} = \sum_{i=0}^{n-1}A^ic_i(t)$ \begin{equation} x(t_1) = \sum_{i=0}^{n-1}A^iB\int_0^{t_1} c_i(t_1-\tau)u(\tau)d\tau = \sum_{i=0}^{n-1}A^iB\alpha_i(t_1) \end{equation}
In (1) vectors multiplied to $A^iB$ are all the possible vectors whereas in (2) $\alpha_i(t_1) = \int_0^{t_1} c_i(t_1-\tau)u(\tau)d\tau$ may not be all the possible vectors in $\mathbb{R}^m$
So how do we prove that $\alpha_i(t_1)$ could be any possible vector in $\mathbb{R}^m$ by varying $u(t)$?
The idea is actually similar to the Gramian matrix.
First thing to note here is that $c_i(t)$ are linearly independent functions, i.e. there doesn't exist any nonzero constant vector $w$ such that $w^Tc(t)=0$, where $c(t):=[c_1(t)~~\dots~~c_n(t)]^T$. This also means that $$ C(t):=\int_0^t c(\tau)c^T(\tau) d\tau =\int_0^t c(t-\tau)c^T(t-\tau) d\tau $$ is full-rank for all $t$. Assuming single input, we need to find $u(t)$ such that $$ \alpha(t_1) = \int_0^{t_1} c(t_1-\tau) u(\tau) d\tau $$ for any given $\alpha(t) := [\alpha_1(t)~~\dots~~\alpha_n(t)]^T$. For this we can select $$u(t) = c^T(t_1-t) C^{-1}(t_1) \alpha(t_1)$$