How to divide an uncontrollable LTI system into controllable and uncontrollable parts?

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 $$(A_1,B_1)$$.
• @LegolasHu You are right! I was taking $y=Tx$. Now, I noticed you have defined $x=Ty$. I will edit the answer to correct. – RTJ Nov 8 '18 at 7:36