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

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.

Answer 2

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 3

I think the key point you are missing is that RowReduce must be done on the transpose of $Q$.

Using the same example as bill s and the nifty way he pointed out to construct $Q$ using ControllabilityMatrix, we have:

a = {{-1, 1, 0}, {0, -1, 0}, {0, 0, 0}};
b = {{1, 0}, {1, 1}, {0, 0}};
Q = ControllabilityMatrix[StateSpaceModel[{a, b}]];

If we want to compute the span of the controllable space using RowReduce, this is what needs to be done.

span = Transpose[RowReduce[Transpose[Q]]]
(* {{1, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}} *)

The controllable space is the span of the columns.

Of course, the dimension of the space will be the same whether we evaluate it using $Q$ or its span.

 MatrixRank /@ {Q, span}
 (* {2, 2} *)

(If we are working with the ObservabilityMatrix, then we need not go through Transpose to determine the span of the observable space.)