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with('linalg'):
Define the system matrices
A
,
B
A:=diag(2,2,2,3,3);
d
2
0
0
0
0
0
2
0
0
0
0
0
2
0
0
0
0
0
K
3
0
0
0
0
0
K
3
A[1,2]:=1;
1, 2
d
1
print(A);
2
1
0
0
0
0
2
0
0
0
0
0
2
0
0
0
0
0
K
3
0
0
0
0
0
K
3
B:=transpose(matrix(2,5,[1,1,2,0,0,0,1,1,1,1]));
d
1
0
1
1
2
1
0
1
0
1
Compute the controllability matrix
R:=evalm(augment(B,A&*B,A&*A&*B,A&*A&*A&*B,A&*A&*A&*A&*B));
R
d
1
0
3
1
8
4
20
12
48
32
1
1
2
2
4
4
8
8
16
16
2
1
4
2
8
4
16
8
32
16
0
1
0
K
3
0
9
0
K
27
0
81
0
1
0
K
3
0
9
0
K
27
0
81
rank(R);
4
The pair (A,B) is not controllable. Let us compute the controllable subspace
colspace(R);
1
0
0
0
0
,
0
0
0
1
1
,
0
0
1
0
0
,
0
1
0
0
0
T1:=transpose(stackmatrix(Matrix(1, 5, {(1, 1) = 1, (1, 2) = 0,
(1, 3) = 0, (1, 4) = 0, (1, 5) = 0}), Matrix(1, 5, {(1, 1) = 0,
(1, 2) = 0, (1, 3) = 0, (1, 4) = 1, (1, 5) = 1}), Matrix(1, 5, {
Problem 3, Exercise set 6
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(1, 1) = 0, (1, 2) = 0, (1, 3) = 1, (1, 4) = 0, (1, 5) = 0}),
Matrix(1, 5, {(1, 1) = 0, (1, 2) = 1, (1, 3) = 0, (1, 4) = 0,
(1, 5) = 0})));
T1
d
1
0
0
0
0
0
0
1
0
0
1
0
0
1
0
0
0
1
0
0
The controllable subspace has dimension 4, and it is spanned by the columns of the matrix T1. Let us
find a complementary subspace of the controllable subspace. For simplicity, we compute the orthogonal
complement:
nullspace(transpose(T1));
0
0
0
K
1
1
T2:=matrix(5,1,[0,0,0,1,1]);
T2
d
0
0
0
K
1
1
Collecting all the vectors, we form a change of basis adapted to the controllable subspace. The
corresponding transformation matrix T is
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 Fall '10
 Serrani

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