Determine controllability with big A matrix in linear system

Question

On the linear system below: \begin{aligned} \dot{x} &= Ax + Bu \\ \end{aligned}

where

$$ A = \begin{bmatrix} 1 & 0 &0 &-1 \\ 0 & 1 &2 &-1 \\ 1 & -2 &1 &0 \\ -1 & 0 &0 &2 \\ \end{bmatrix} $$

and

$$ B = \begin{bmatrix} 0 \\ 0 \\ 1\\ 0\\ \end{bmatrix} $$

is there a quick way to check whether it's controllable or not? I know the textbook way by creating the matrix: $$ M = \begin{bmatrix} B &AB & A^2B &A^3B \end{bmatrix} $$

and then finding out if $$ \det{M}$$ is zero or not, but is there any (linear algebra) trick I am missing where the conclusion could be made more easily?

hint: It's not controllable

Answer 1

What you could reason is that the $B$ matrix only 'supplies' terms from the 3rd column in the $A$ matrix. Going one step further, the $AB$ term has only nonzero terms on the 2nd and third row (which is trivial to see). As the first and last row of $A$ have zeros in the second and third column, the $A^iB$th term can never have terms in its first or second row. This reasoning of course only holds for this example, but the idea behind it can be used to quickly see if a controllability (or observability) matrix has full rank (I used this for tests on the subject at uni).

Answer 2

When the eigenvalues of $A$ are given or can relatively easily be obtained one could use the Hautus lemma, which for controllability states that

$$ \text{rank}\begin{bmatrix} \lambda\,I - A & B \end{bmatrix} = n,\,\forall\,\lambda\in\mathbb{C}, $$

or equivalently for all $\lambda$ equal to one of the eigenvalues of $A$.

However, in this case the controllability matrix is probably the easiest route. It can be noted that $M$ can be calculate iteratively (it is not required to calculate the powers of $A$). Namely

$$ M = \begin{bmatrix} L_1 & \cdots & L_n \end{bmatrix}, $$

with $L_1 = B$ and $L_{i+1} = A\,L_i$.

Calculating the controllability matrix in your case yields

$$ M = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 2 & 4 & -2 \\ 1 & 1 & -3 & -11 \\ 0 & 0 & 0 & 0 \end{bmatrix}, $$

which contains two rows of all zeros. However, the rank of a (square) matrix always is reduced by one if it has a row (or column) of all zeros, thus in this case $M$ can't be full rank.