Consider this linear system $\frac{dx}{dt}=Ax+Bu$

Assume that $B\neq 0$ and the system is uncontrollable. It’s easy to show the existence of an invertible state transform $x=Ty$ satisfying $$\frac{dy}{dt}= \begin{pmatrix} A_{1} & A_{2} \\ 0 & A_{3} \end{pmatrix}y+ \begin{pmatrix} B_{1} \\ 0 \end{pmatrix}u$$ by choosing the irrelevant column vector group of the controllability matrix and expand that to a basis of the whole vector space. But I need help to prove that the subsystem $[A_{1},B_{1}]$ is controllable. So could anyone can tell me how to prove it?


The rank of the controllability matrix remains unchanged under such a transformation since $$\bar{\mathcal{C}}=T^{-1}\mathcal{C}$$ where $\mathcal{C}$, $\bar{\mathcal{C}}$ are the controllability matrices of the initial and the transformed system. Thus, $rank(\bar{\mathcal{C}})=r$ with $r$ the dimension of $A_1$. Now if you carry out the calculations to derive $\bar{\mathcal{C}}$ you will obtain $$\bar{\mathcal{C}}=\left[\matrix{B_1 & A_1B_1 &\cdots & A_1^{n-1}B_1\\0 & 0& \cdots &0}\right]\qquad\qquad\qquad\qquad\qquad (1)$$ From $rank(\bar{\mathcal{C}})=r$ we have that $$rank\left[\matrix{B_1 & A_1B_1 &\cdots & A_1^{n-1}B_1}\right]=r\qquad\qquad\qquad\qquad (2)$$ which yields $$rank\left[\matrix{B_1 & A_1B_1 &\cdots & A_1^{r-1}B_1}\right]=r\qquad\qquad\qquad\qquad (3)$$ since all matrices $A_1^{j-1}$ with $j>r$ can be written as a linear combination of $\mathbb{I}_{r}$, $A_1$, $\cdots$, $A_1^{r-1}$ due to Cayley-Hamilton theorem. From (3) we obtain the controllability of $(A_1,B_1)$.

  • $\begingroup$ Very clear and succinct solution, thank you! P.S. the T in the first equation should actually be its inverse, Maybe it’s a typo or maybe you define T as y=Tx, which is different from mine. :D $\endgroup$ – Legolas Hu Nov 8 '18 at 7:32
  • $\begingroup$ @LegolasHu You are right! I was taking $y=Tx$. Now, I noticed you have defined $x=Ty$. I will edit the answer to correct. $\endgroup$ – RTJ Nov 8 '18 at 7:36

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