Understanding controllability indices I'm teaching myself linear control systems through various online materials and the book Linear Systems Theory and Design by Chen. I'm trying to understand controllability indices. Chen says that looking at the controllability matrix $C$ as follows
$$ C = \begin{bmatrix} B & AB & A^2 B & \cdots & A^n B \end{bmatrix} = \begin{bmatrix} b_1 & \cdots & b_p & | & A b_1 & \cdots & A b_p & | & \cdots & | & A^{n-1} b_1 & \cdots & A^{n-1} b_p \end{bmatrix}$$
that reading left to right, the columns are linearly independent until some column $A^ib_m$, which is dependent on the columns to the left - this makes sense - but that subsequent columns associated with $b_m$ (e.g. $A^{i+k}b_m$) are also linearly dependent on their proceeding columns, or that "once a column associated with $b_m$ becomes linearly dependent, then all columns associated with $b_m$ thereafter are linearly dependent". Unfortunately this last point isn't intuitive to me at all - if anyone can shed some light on this or help me see it I'd be very grateful!
Thanks in advance for any help.
Luke
 A: Say $A b_1$ is linearly dependent on the $b_i$, namely $A b_1 = \lambda_i b_i$ (summation implicit). Then $A^2 b_1 = A \lambda_i b_i = \lambda_i A b_i$, so $A^2 b_1$ depends linearly on the columns $A b_i$ which are on its left.
A: I am studying the same subject now, from the same book and had exactly the same question you had.
The answer by @Pait doesn't show that the resulting set $\mathit{\lambda_iAb_i}$ is a subset of the LHS vectors relative to $\mathit{A^2b_1}$ in the controllability matrix. However I had understood it myself, and I will post it here for reference.
Say $\pmb{A^ib_m}$ is the first vector associated with $\pmb{b_m}$ that is dependent on some of the vectors $\pmb{v_k}$ at its left hand side (LHS).
That is: $$\pmb{A^ib_m}= \sum_{k=1}^{i*m-1}\alpha_k\pmb{v_k} \ \ \ (1)$$
Now, for the next vector associated with $\pmb{b_m}$, i.e. $\pmb{A^{i+1}b_m}$, we have: $$\pmb{A^{i+1}b_m}= \sum_{k=1}^{i*m-1}\alpha_k\pmb{Av_k} \ \ \ (2)$$
Now we show (or clarify) that the set $\{\pmb{Av_k}\ |\ k\in\Bbb{Z},\ 1\le k \le (i*m-1)\}$ is a subset of the vectors at LHS of $\pmb{A^{i+1}b_m}$ in the controllability matrix.
The sum in (1) can be split into:
$$\pmb{A^ib_m} = \underbrace{\sum_{k=0}^{i-1}\alpha_{mk}^{'}\pmb{A^kb_m}}_{Vectors\ which\ are\\ associated\ with\ \pmb{b_m}\\ but\ are\ on\ LHS} + \underbrace{(\sum_{k=0}^i\alpha_{1k}^{'}\pmb{A^kb_1} +\sum_{k=0}^i\alpha_{2k}^{'}\pmb{A^kb_2}+\cdots+\sum_{k=0}^i\alpha_{(m-1)k}^{'}\pmb{A^kb_{m-1}})}_{Vectors\ associated\ with\ \pmb{b_j}\ columns\\ which\ are\ before\ \pmb{b_m}\ in\ the\ \pmb{B}\ matrix}$$
Now by multiplying by $\pmb{A}$ on both sides:
$$\pmb{A^{i+1}b_m} = \sum_{k=1}^{i}\alpha_{mk}^{'}\pmb{A^kb_m} + (\sum_{k=1}^{i+1}\alpha_{1k}^{'}\pmb{A^kb_1} +\sum_{k=1}^{i+1}\alpha_{2k}^{'}\pmb{A^kb_2}+\cdots+\sum_{k=1}^{i+1}\alpha_{(m-1)k}^{'}\pmb{A^kb_{m-1}})$$
Clearly, the vectors in the summations are subset of the vectors that are at LHS of $\pmb{A^{i+1}b_m}$ in the controllability matrix.
