We can continue this process on the (
N

1)
×
(
N

1) submatrix
on the lower right. We take
x
2
=
0
0
˜
a
32
˜
a
42
.
.
.
˜
a
N
2
,
y
2
=
0
0
r
2
0
.
.
.
0
,
with
r
2
=
k
x
2
k
2
,
and
H
2
=
I

2
u
2
u
T
2
,
u
2
=
x
2

y
2
k
x
2

y
2
k
2
.
Then
A
3
=
H
2
A
2
H
2
=
a
11
r
1
0
0
· · ·
0
r
1
˜
a
22
r
2
0
· · ·
0
0
r
2
b
a
33
b
a
34
· · ·
b
a
3
N
0
0
b
a
43
b
a
44
· · ·
b
a
4
N
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0
0
b
a
N
3
b
a
N
4
· · ·
b
a
NN
Repeating this procedure
N

1 times gives us a tridiagonal matrix
T
=
A
N
=
H
N
H
N

1
· · ·
H
1
AH
1
H
2
· · ·
H
N
=
Q
T
AQ
.
where
Q
=
H
1
H
2
· · ·
H
N
.
Note that since all of the
H
n
are orthonormal, so is
Q
.
23
Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 13:22, November 13, 2019
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Technical Details: Solving Toeplitz systems and
the LevinsonDurbin algorithm
We start by looking at how to solve
Hv
=
y
for a very particular
righthand side. Consider
h
0
h
1
· · ·
h
N

1
h
1
h
0
h
1
· · ·
h
N

2
h
2
h
1
h
0
· · ·
h
N

3
.
.
.
h
N

1
· · ·
· · ·
h
0
v
[1]
v
[2]
.
.
.
.
.
.
v
[
N
]
=
h
1
h
2
.
.
.
h
N

1
h
N
.
(7)
Here the first
N

1 entries of the “observation” vector
y
match the
last
N

1 entries in the first column of
H
. This system is special
ized, but not at all contrived — (
7
) are called the
YuleWalker
equations
, and appear in many different places in statistical signal
processing. Moreover, solving systems with general righthand sides
solve systems of the form (
7
) as an intermediate step.
The crux of the LevinsonDurbin algorithm relies upon a seemingly
innocuous fact. Let
J
be the
N
×
N
exchange matrix
(also called
the
counter identity
):
J
=
0 0
· · ·
0 0 1
0 0
· · ·
0 1 0
0 0
· · ·
1 0 0
.
.
.
.
.
.
1 0
· · ·
0
.
Applying
J
to a vector
x
reverses the entries in
x
. For example,
0 0 1
0 1 0
1 0 0
5

1
7
=
7

1
5
.
24
Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 13:22, November 13, 2019
It should be clear that
J
T
=
J
and
J
2
=
JJ
=
I
.
Applying
J
to the left of a matrix reverses all of its columns, while
applying
J
to the right reverses the rows. In particular, if
H
is a
symmetric Toeplitz matrix, then
J
H
=
0 0
· · ·
0 0 1
0 0
· · ·
0 1 0
0 0
· · ·
1 0 0
.
.
.
.
.
.
1 0
· · ·
0
h
0
h
1
· · ·
h
N

1
h
1
h
0
h
1
· · ·
h
N

2
h
2
h
1
h
0
· · ·
h
N

3
.
.
.
h
N

1
· · ·
· · ·
h
0
=
h
N

1
h
N

2
· · ·
h
0
h
N

2
h
N

3
h
N

4
· · ·
h
1
h
N

3
h
N

4
h
N

5
· · ·
h
2
.
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 Staff