Solutions for Homework 2 Numerical Linear Algebra 1
Fall 2011
Problem 2.1
Let
x
∈
C
n
and
y
∈
C
n
be two arbitrary vectors. Consider determining a circulant matrix
C
∈
C
n
×
n
such that
y
=
Cx
2.1.a
. Assume that
C
exists for a given pair (
x, y
), show how to construct it.
2.1.b
. When is
C
unique for a given pair (
x, y
)?
2.1.c
. When does
C
not exist
for a given pair (
x, y
)?
Solution:
There are two equivalent lines of reasoning. The first is in the “spatial” domain. Denote
C
=
C
(
c
) to indicate that it is generated by its first row
c
T
. We have
Cx
=
y
C
(
c
)
x
=
y
H
(
x
)
c
=
y
C
(
x
r
)
c
r
=
y
where
H
(
x
)
∈
C
n
×
n
.
H
(
x
) is not however circulant in the same fashion as
C
(
c
). It is a Hankel matrix not a
Toeplitz matrix. A Hankel matrix is constant along its antidiagonals. It is therefore circulant
with respect to the first row using a circulant shift to the left (not to the right as with
C
(
c
))
and with respect to the first column using a circulant shift up (not down as with
C
(
c
)).
The Hankel/left circulant form of the matrix vector product can be transformed into
the Toeplitz/right circulant form of the matrix vector product.
This is done by using a
Toeplitz/right circulant matrix and a righthand side vector
x
r
and
c
r
that are the vectors
x
and
c
with their elements in reverse order respectively, i.e.,
P
T
=
(
e
n
e
n

1
. . .
e
1
)
,
x
r
=
Px,
c
r
=
Pc
Since
x
and
y
are given, one way to find
c
is to solve the system with the matrix,
H
(
x
),
defined by
x
. Therefore, to have a solution we must have
y
∈R
(
H
(
x
)) or
y
∈R
(
C
(
x
r
)).
1
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These facts are easily seen by example with
n
= 4.
x
=
ξ
0
ξ
1
ξ
2
ξ
3
c
=
γ
0
γ
1
γ
2
γ
3
y
=
C
(
c
)
x
=
γ
0
ξ
0
+
γ
1
ξ
1
+
γ
2
ξ
2
+
γ
3
ξ
3
γ
3
ξ
0
+
γ
0
ξ
1
+
γ
1
ξ
2
+
γ
2
ξ
3
γ
2
ξ
0
+
γ
3
ξ
1
+
γ
0
ξ
2
+
γ
1
ξ
3
γ
1
ξ
0
+
γ
2
ξ
1
+
γ
3
ξ
2
+
γ
0
ξ
3
=
γ
0
ξ
0
+
γ
1
ξ
1
+
γ
2
ξ
2
+
γ
3
ξ
3
γ
0
ξ
1
+
γ
1
ξ
2
+
γ
2
ξ
3
+
γ
3
ξ
0
γ
0
ξ
2
+
γ
1
ξ
3
+
γ
2
ξ
0
+
γ
3
ξ
1
γ
0
ξ
3
+
γ
1
ξ
0
+
γ
2
ξ
1
+
γ
3
ξ
2
=
H
(
§
)
c
=
γ
3
ξ
3
+
γ
2
ξ
2
+
γ
1
ξ
1
+
γ
0
ξ
0
γ
3
ξ
0
+
γ
2
ξ
3
+
γ
1
ξ
2
+
γ
0
ξ
1
γ
3
ξ
1
+
γ
2
ξ
0
+
γ
1
ξ
3
+
γ
0
ξ
2
γ
3
ξ
2
+
γ
2
ξ
1
+
γ
1
ξ
0
+
γ
0
ξ
3
=
C
(
x
r
)
c
r
The system of course can be solved via the DFT which gives the more elegant view of
the problem in the Fourier domain (essentially the eigendomain).
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 Fall '06
 gallivan
 Matrices, Block matrix, matrix vector product

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