I need to show that the following system is always controllable:

\begin{align}A &= \begin{bmatrix} -\alpha_1I_{k\times k}& -\alpha_2I_{k\times k}& \cdots &-\alpha_{n-1}I_{k\times k}&-\alpha_nI_{k\times k}\\ I_{k\times k}&0_{k\times k}&\cdots&0_{k\times k}&0_{k\times k}\\ 0_{k\times k}&I_{k\times k}&\cdots&0_{k\times k}&0_{k\times k}\\ \vdots & \vdots & \ddots& \vdots&\vdots\\ 0_{k\times k}&0_{k\times k}&\cdots&I_{k\times k}&0_{k\times k}\end{bmatrix}_{nk\times nk}\\ B&=\begin{bmatrix} I_{k\times k}\\ 0_{k\times k}\\ \vdots\\ 0_{k\times k}\end{bmatrix}_{nk\times n} \quad C=\begin{bmatrix}N_1&N_2&\cdots&N_n\end{bmatrix}_{m\times nk}. \end{align}

Now, I think this is actually the controllable canonical form, but I'm really confused about how to show this is always controllable. Can anyone help? Sorry if this is a dumb question.


There's a very simple way to check controllability, indeed if you define the reachability matrix $$ R = \begin{pmatrix}B & AB & \dots & A^{n-1}B\end{pmatrix} $$ then the reachable subspace is the image of R. Hence to check complete controllability you just have to check that $R$ is full rank. First, I think there's an error in the question, $B$ should be $nk\times k$ as you defined it. Anyway, in your case $R$ will be a $nk\times nk$ matrix, the first $k$ columns are given by the columns of $B$, which are linearly independent by definition of $B$ (in fact you have the identity as the first block), thus $rankR\ge k$. The second $k$ columns are given by $AB$ which has the form $$ AB = \begin{pmatrix}\star\\I_k\\0_{k\times k}\\\vdots\\0_{k\times k}\end{pmatrix} $$ where $\star$ means a block you dont care about. Note that the columns of the matrix $$ \begin{pmatrix}B & AB\end{pmatrix} = \begin{pmatrix}I_k & \star \\ 0_{k\times k} & I_k \\ 0_{k\times k} & 0_{k\times k} \\ \vdots & \vdots \\ 0_{k\times k} & 0_{k\times k}\end{pmatrix} $$ are linearly independent reguardless of $\star$. Hence $rankR\ge 2k$. In the same way, for the term $A^k B$ you get the form $$ A^k B = \begin{pmatrix}\star_1 \\ \star_2 \\ \vdots\\ \star_{k}\\ I_k\\0_{k\times k}\\ \vdots \\ 0_{k\times k}\end{pmatrix} $$ where the Identity block is at position $k+1$. hence for any $0\ge k<j\ge n-1$ $A^{k}B$ and $A^jB$ have linearly independent columns, hence you obtain thart $R$ is full rank an the system is completely controllable

  • $\begingroup$ This is a wonderfully illustrative answer. I'd upvote if I could. Thanks for your help! $\endgroup$ – John Alperto Apr 11 '16 at 12:59

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