Can RowReduce work in this matrix?

Question

The matrix $Q$ with dimensions $n\times2*n*m$ is structured by $$Q=[B|AB|\cdots|A^{2*n-1}B]$$ where $Q$ is an augmented matrix built from a $3\times3$ matrix, $A$, and a $3\times2$ matrix, $B$.

I need $\DeclareMathOperator{\rank}{rank}\rank(Q)<n$ but I get $\rank(Q)=n$.

If I RowReduce matrices $A$ and $B$ before that calculation the $Q$ matrix has $\rank<n$ as I want, but its elements aren't the ones I need to continue, but 0's and 1's.

Any Idea on how to get the desired $Q$ matrix?

If I multiply $Q$ by $Q_\text{reduced}$ (with dimensions $2*n*m\times n$) I get an $n\times n$ matrix, $W$ with $\rank(W)=n$, so that's not what I want.

I've no Idea how to implement the Mathematica code here. Any suggestions would be helpful.

CellGroupData[{
  A0 = RandomInteger[{1, 10}, {4, 4}],
  B0 = RandomInteger[{1, 10}, {4, 3}],
  A1 = RowReduce[A0],
  B1 = RowReduce[B0],
  m = 3,
  n = 4,

  (*Original Matrices Proc *)
  Do[Lnn[j] = MatrixPower[A0, j - 1].B0, {j, 2*n}],
  Do[Flatten[Lnn[j][[i]], m], {i, n}, {j, 2*n}],    
  M222 = Table[Lnn[j][[i]], {i, n}, {j, 2*n}],
  R222 = Flatten[M222],
  Qnn = Partition[R222, 2*n*m],
  η1 = MatrixRank[Qnn],
  {p1, r1} = Dimensions[Qnn],
  QN11 = RowReduce[Qnn],

  (*RowReduced Matrices used now *)
  Do[Lnn1[j] = MatrixPower[A1, j - 1].B1, {j, 2*n}],
  Do[Flatten[Lnn1[j][[i]], m], {i, n}, {j, 2*n}],    
  M22 = Table[Lnn1[j][[i]], {i, n}, {j, 2*n}],
  R22 = Flatten[M22],
  Qn = Partition[R22, 2*n*m],
  η = MatrixRank[Qn],
  {p, r} = Dimensions[Qn],
  QN = RowReduce[Qn],

  Print[MatrixForm[Qnn], MatrixForm[QN11]],
  Print[MatrixForm[Qn], MatrixForm[QN]]}];

Answer 1

The Answer is no it can't!

Random Integer sets as Linear Independent the Columns B and the first column of the product AB, and linear dependent among these three all the others.

If you're an engineer and want to test this augmented Controllability Crit your system HAS to be NOT Controllable and implement the Static Feedback that makes it controllable.

Answer 2

Whether the Q matrix has rank $<n$ or not depends on the specific A and B. For instance, if B is a matrix of all zeroes, then Q is a matrix of all zeroes and so has rank $< n$.

You can test for the rank of Q using the MatrixRank[ ] function.

You can build Q from A and B using the function ControllabilityMatrix[ ].

For example, if a={{-1, 1, 0}, {0, -1, 0}, {0, 0, 0}} and b={{1, 0}, {1, 1}, {0, 0}}, the Q matrix is

q = ControllabilityMatrix[StateSpaceModel[{a, b}]]

and the rank is

 MatrixRank[q]

which is 2.

Of course, Aristarchus is correct, if you build A and B from random integers (or random reals) then Q is full rank with probability one.