Suppose we are given a matrix $B \in \mathbb{R}^{n\times n}$, and $A \in \mathbb{R}^n$
Then form a matrix $M$
$M = \begin{bmatrix} A & BA & B^2A & \ldots & B^n A \end{bmatrix}$
How can you partition the columns this matrix into linearly dependent and independent sets?
Attempt:
going from the left, suppose $B^{k+1}A$ is the first vector that is linearly dependent on the previous columns, then
$$B^{k+1}A = c_0 A + c_1 BA + \ldots + c_{k}B^kA$$
Then
$$BB^{k+1}A = B^{k+2}A = c_0 AB + c_1 B^2A + \ldots + c_{k}B^{k+1}A$$
But $$B^{k+1}A = A = c_0 A + c_1 BA + \ldots + c_{k}B^kA$$
Therefore $$B^{k+2}A = c_0 AB + c_1 B^2A + \ldots + c_{k}(c_0 A + c_1 BA + \ldots + c_{k}B^kA)$$
Continue this way, we can partition the matrix into one set of $k$ linearly independent columns and $n - k$ linearly dependent columns
Do I have this correctly? Is there a general theorem governing this result?
I think the problem arises from Linear Control Theory:observability. The matrix $M$ is called observability matrix. And there are some criteria on whether its rank is n.