The University of Texas at Austin
Department of Electrical and Computer Engineering
EE362K: Introduction to Automatic Control—Fall 2009
Solutions for Problem Set Two
C. Caramanis
Due: Monday, September 21, 2009.
This problem set is intended to get us started thinking about linear algebra, and also to continue
giving us some practice with Matlab.
1. The ﬁrst column of
A
1
is a multiple of the second column: (1
,

2)
*
2 = (

2
,
4). Therefore,
any vector of the form (2
x,

x
) will be in the null space of
A
1
. Generically, the range of any
matrix is the span of the columns. Since the columns of
A
1
are linearly dependent, the range
is just onedimensional, and equal to Span(1
,

2).
For
A
2
, you can check that any vector of the form: (2
x,x,

5
x/
2) is in the null space.
Moreover, there are two linearly independent columns, which means that the pan of the
columns is two dimensional, and that therefore the dimension of the range is 2. Since the
dimension of the null space plus the dimension of the range equals
n
, or in this case 3, we
know that indeed the null space is the onedimensional subspace: Span(2
,
1
,

5
/
2). The range
will be the entire (
x,y
)plane, or, Span
{
(1
,
0
,
0)
,
(0
,
1
,
0)
}
.
2. As mentioned in the book, the consensus algorithm is realizable by the update law:
x
[
k
+ 1] =
x
[
k
]

γ
(
D

A
)
x
[
k
]
where
A
is the adjacency matrix of the graph and
D
is a diagonal matrix with entries cor
responding to the number of neighbors of each node. The constant
γ
describes the rate at
which the estimate of the average is updated based on information from neighboring nodes.
(a)
A
and
D
for
K
10
graph are given by:
A
=
0 1 1 1 1 1 1 1 1 1
1 0 1 1 1 1 1 1 1 1
1 1 0 1 1 1 1 1 1 1
1 1 1 0 1 1 1 1 1 1
1 1 1 1 0 1 1 1 1 1
1 1 1 1 1 0 1 1 1 1
1 1 1 1 1 1 0 1 1 1
1 1 1 1 1 1 1 0 1 1
1 1 1 1 1 1 1 1 0 1
1 1 1 1 1 1 1 1 1 0
,D
=
9 0 0 0 0 0 0 0 0 0