Suppose $\bf b$ is an $n$-dimensional column vector that can be decomposed as ${\bf b} = {\bf b}_1 + {\bf b}_2$. In this decomposition, ${\bf b}_1$ and ${\bf b}_2$ hold the non-zero components of $\bf b$ respectively (e.g. if $\bf b$= $\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}^\top$, then this decomposition can be $\bf b_{1}$= $\begin{bmatrix} 1 & 0 & 3 \end{bmatrix}^\top$ and $\bf b{2}$=$\begin{bmatrix} 0 & 2 & 0 \end{bmatrix}^\top$).
Suppose $\bf A$ is a real matrix of order n and let $\bf C$=[$\bf b$,$\bf Ab$, ... , $\bf A^{n-1}b$], $\bf C_{1}$=[$\bf b_{1}$,$\bf Ab_{1}$, ... , $\bf A^{n-1}b_{1}$] and $\bf C_{2}$=[$\bf b_{2}$,$\bf Ab_{2}$, ... , $\bf A^{n-1}b_{2}$]. If we have rank([$\bf C_{1}$,$ \bf C_{2} $])=n, can we prove rank($\bf C$)=n?
The answer is no. As an example, consider $$ A = \pmatrix{1&1\\1&1}, \quad b = \pmatrix{1\\-1}, \quad b_1 = \pmatrix{1\\0}, \quad b_2 = \pmatrix{0\\-1}. $$ We find that $C_1 = [b_1 \ \ Ab_1]$ and $C_2 = [b_2 \ \ Ab_2]$ each have full row rank $n=2$, as does the block matrix $[C_1 \ \ C_2]$. However, $C = [b\ \ Ab]$ has a zero column and therefore does not have full rank.