Kalman rank condition

Remarks

The proof uses the controllability Gramian $W_T = \int_0^T e^{-As} B B^T e^{-A^T s} \, ds$, built from the matrix exponential. A vector $v$ lies in the kernel of $W_T$ exactly when $v^T e^{-As} B = 0$ for all $s \in [0, T]$, and by Taylor expansion and the Cayley–Hamilton theorem this holds exactly when $v^T A^k B = 0$ for $k = 0, \ldots, n - 1$. So $W_T$ is invertible if and only if $\rank \mathcal{C} = n$. If $W_T$ is invertible, the control $u(s) = B^T e^{-A^T s} W_T^{-1} (e^{-AT} x_1 - x_0)$ steers $x_0$ to $x_1$ in time $T$. If $\rank \mathcal{C} < n$, every trajectory from $0$ stays in the column space of $\mathcal{C}$, so the states outside it are not reachable from $0$. Example: for the double integrator, $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ give $\mathcal{C} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, of rank $2$, so the system is controllable.

Going deeper

For a student

The condition is purely algebraic and easy to check by hand or by computer. By duality, the system $x' = A x$, $y = C x$ is observable if and only if the matrix with block rows $C, CA, \ldots, CA^{n-1}$ has rank $n$. This is the Kalman condition for the pair $(A^T, C^T)$. When the condition fails, a linear change of coordinates splits the state into a controllable part and a part on which the control has no effect. Controllability of $(A, B)$ is also equivalent to the possibility of placing the eigenvalues of $A + BK$ anywhere, symmetric under complex conjugation, by a choice of the feedback matrix $K$.