Consider a linear time-invariant system: \[ \begin{align*} \begin{split} \mathbf{\dot{x}} &= \mathbf{Ax+Bu} \\ \mathbf{y} &= \mathbf{Cx+Du} \end{split} \end{align*} \] The closed-form solution of the linear time-invaraint system is well-known: \[ \mathbf{x}(t) = \exp(\mathbf{A}t)\mathbf{x}(0) + \int_{0}^{t}\exp(\mathbf{A}(t-\tau))\mathbf{B}\mathbf{u}(\tau)d\tau \tag{1} \] From this equation, the question of controllability emerges: Given \(\mathbf{x}(t)\in\mathbb{R}^{n}\) which is our desired state to reach, is there any control input signal \(\mathbf{u}(\cdot):[0,t]\rightarrow \mathbb{R}^{m}\) which can reach \(\mathbf{x}(t)\) [1]?
Before discussing this in great detail, we reformulate Equation 1:
\[ \mathbf{x}(t) - \exp(\mathbf{A}t)\mathbf{x}(0) = \int_{0}^{t}\exp(\mathbf{A}(t-\tau))\mathbf{B}\mathbf{u}(\tau)d\tau \tag{2} \]
We introduce an informal (yet correct) derivation of controllability, for single-input-single-output system, and provide a formal derivation for general system in the next section. For that, we use the Cayley-Hamilton Theorem. Given state-matrix \(\mathbf{A}\in\mathbb{R}^{n\times n}\) and its characteristic equation, Cayley-Hamilton theorem shows that: \[ \det(\lambda \mathbf{I_n} - \mathbf{A}) = \lambda^n + a_{n-1}\lambda^{n-1} + \cdots + a_1\lambda + a_0 \] \[ \mathbf{A}^n + a_{n-1}\mathbf{A}^{n-1} + \cdots + a_1\mathbf{A} + a_0\mathbf{I}_n = \mathbf{0}_n \] Hence, \(\exp(\mathbf{A}(t-\tau))\) can be expressed as a sum of \(\mathbf{A}^{n-1}, \cdots \mathbf{A}^{1}, \mathbf{I}_{n}\) with coefficients as a function of \(t\), \(\tau\): \[ \exp(\mathbf{A}(t-\tau)) = \alpha_{n-1}(t,\tau) \mathbf{A}^{n-1} + \alpha_{n-2}(t,\tau) \mathbf{A}^{n-2} + \cdots + \alpha_1(t,\tau) \mathbf{A} + \alpha_0(t,\tau) \mathbf{I}_n \] Hence, the right-hand side of Equation 2 can be formulated as: \[ \begin{align*} \int_{0}^{t}\exp(\mathbf{A}(t-\tau))\mathbf{B}u(\tau)d\tau &= \int_{0}^{t} \alpha_{n-1}(t,\tau) \mathbf{A}^{n-1}\mathbf{B}u(\tau)d\tau + \cdots + \int_{0}^{t}\alpha_0(t,\tau) \mathbf{B}u(\tau)d\tau \\ &= \mathbf{A}^{n-1}\mathbf{B} \int_{0}^{t}\alpha_{n-1}(t,\tau)u(\tau)d\tau + \cdots + \mathbf{B}\int_{0}^{t}\alpha_0(t,\tau)u(\tau)d\tau \\ \int_{0}^{t}\exp(\mathbf{A}(t-\tau))\mathbf{B}u(\tau)d\tau &= \begin{bmatrix} \vert & \vert & & \vert \\ \mathbf{B} & \mathbf{AB} & \cdots & \mathbf{A}^{n-1}\mathbf{B} \\ \vert & \vert & & \vert \end{bmatrix} \begin{bmatrix} \alpha_0' \\ \alpha_1' \\ \vdots \\ \alpha_{n-1}' \end{bmatrix} , ~~~ \forall i \in [0,1,\cdots, n-1]: \alpha_{i}' = \int_{0}^{t} \alpha_{i}(t,\tau)u(\tau)d\tau \end{align*} \] Since \(u(\cdot)\) is defined, it is clear that if the matrix is full rank.