1
$\begingroup$

Show that there is a basis in $\mathcal{X}$ in which $A$ and $B$ have matrices of the form $$ A = \begin{bmatrix} A_1 && A_3 \\ 0 && A_2 \end{bmatrix} , B = \begin{bmatrix} B_1 \\ 0\end{bmatrix},$$ where $(A_1,B_1)$ is controllable. Describe $\langle A | \mathcal{B} \rangle$ in this basis.

Notation : $\dot{x}(t) = Ax(t)+Bu(t), \quad x\in \mathcal{X},$ $\quad \mathcal{B} = \mathrm{Im}(B)$ $$\langle A | \mathcal{B} \rangle = \mathcal{B} + A\mathcal{B} + \cdots + A^{n-1}\mathcal{B}$$

I need help to prove the first part.

I begin with assumption that $\exists$ a tranformation $z = T x$ such that $$\dot{z}(t) = T^{-1}ATz(t) + T^{-1}Bu(t)$$

But I am not sure how to proceed further.

$\endgroup$
2
$\begingroup$

Let the controllability matrix $$\mathcal{C}=[\matrix{B & AB & \cdots & A^{n-1}B}]$$ Assume that $rank(\mathcal{C})=r\leq n$. Then, there are $r$ linearly independent column of $\mathcal{C}$ which we may denote as $q_1, q_2,\cdots,q_r$. Augment those columns with $n-r$ columns $q_{r+1},\cdots,q_n$ such that the $n\times n$ matrix $$Q:=\left[\matrix{q_1 & q_2 & \cdots & q_r & q_{r+1} & \cdots & q_n}\right]$$ be invertible. The desired transformation is $$T:=Q^{-1}$$ Proof: Since the columns of $B$ are columns of $\mathcal{C}$ they belong in the subspace spanned by $\{q_1,\cdots,q_r\}$ and therefore $$B=\left[\matrix{\sum_{i=1}^r{b_{1i}q_i} & \cdots & \sum_{i=1}^r{b_{mi}q_i} }\right]\\=\left[\matrix{q_1 & \cdots & q_r & q_{r+1} & \cdots & q_n}\right]\left[\matrix{b_{11} & b_{21} & \cdots & b_{m1}\\ b_{12} & b_{22} & \cdots & b_{m2}\\ \vdots & \vdots & \ddots & \vdots\\ b_{1r} & b_{2r} & \cdots & b_{mr}\\ 0 & 0 &\cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 &\cdots & 0}\right]=Q\left[\matrix{B_1\\0}\right]$$ i.e. $$TB=Q^{-1}B=\left[\matrix{B_1\\0}\right]$$ Also we have that $$AQ=\left[\matrix{Aq_1 & Aq_2 & \cdots & Aq_r & Aq_{r+1} & \cdots & Aq_n}\right]$$ Note that the following property holds:

**P1) The first $r$ columns of $AQ$ defined by $Aq_i$ ($i=1,\cdots,r$) also belong in the subspace spanned by $\{q_1,\cdots,q_r\}$. **

This is due to the special form of $\mathcal{C}$. Since $q_i$ ($i=1,\cdots,r$) are columns of $\mathcal{C}$ then $Aq_i$ are columns of the matrices $A^{j}B$ ($j=0,1,\cdots,n$) and therefore they are either columns of $\mathcal{C}$ or columns of $A^nB$. From the Cayley-Hamilton theorem $A^nB$ is a linear combination of $A^jB$ ($j=0,1,\cdots,n-1$) from which we yield $P1$.

Then, $$AQ= \left[\matrix{\sum_{i=1}^r{a_{1i}q_i} & \cdots & \sum_{i=1}^r{a_{ri}q_i} & \sum_{i=1}^n{a_{r+1,i}q_i} & \cdots & \sum_{i=1}^n{a_{ni}q_i}}\right]\\=\left[\matrix{q_1 & \cdots & q_r & q_{r+1} & \cdots & q_n}\right]\left[\matrix{a_{11} & a_{21} & \cdots & a_{r1} & a_{r+1,1} & \cdots & a_{n1}\\ a_{12} & a_{22} & \cdots & a_{r2} & a_{r+1,2} & \cdots & a_{n2}\\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\ a_{1r} & a_{2r} & \cdots & a_{rr} & a_{r+1,r} & \cdots & a_{nr}\\ 0 & 0 &\cdots & 0 & a_{r+1,r+1} & \cdots & a_{n,r+1}\\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 &\cdots & 0 & a_{r+1,n} & \cdots & a_{n,n}}\right]=Q\left[\matrix{A_1 & A_3\\0 & A_2}\right]$$ or equivalently $$TAT^{-1}=Q^{-1}AQ=\left[\matrix{A_1 & A_3\\0 & A_2}\right]$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.