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197
CHAPTER 9
9.1
Assume that
Hence, for
i
=
n
:
Since
β
(
n,n
) = 1, we have
Next, for
i
=
n
1,
or equivalently,
Proceeding in this way, we may thus write
For
β
(
n,i
) to equal
λ
ni
, we must have
This requirement is satisfied by choosing
for all
k
β
ni
,
()
λ
i
()β
1
,
,
=
i
1
…
n
,,
=
β
nn
,
()λ
n
β
1
,
=
λ
1
–
n
β
1
,
=
β
1
,
λ
n
1
β
2
,
=
β
2
,
λ
1
–
n
1
β
1
,
=
λ
1
–
n
1
λ
1
–
n
=
β
,
λ
1
–
i
+1
…
λ
1
–
n
1
λ
1
–
n
=
λ
1
–
k
k
=
i
+1
n
∏
=
λ
1
–
i
+1
…
λ
1
–
n
1
λ
1
–
n
λ
n

i
=
λ
k
λ
1
–
=
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We thus find that
9.2
The matrix inversion lemma states that if
A
=
B
1
+
CD
1
C
(1)
then
A
1
=
B

BC
(
D
+
C
H
BC
)
1
C
H
B
(2)
To prove this lemma, we multiply Eq. (1) by (2):
AA
1
= (
B
1
+
CD
1
C
H
)[
B

BC
(
D
+
C
H
BC
)
1
C
H
B
]
=
B
1
B

B
1
BC
(
D
+
C
H
BC
)
1
C
H
B

CD
1
C
H
BC
(
D
+
C
H
BC
)
1
C
H
B
+
CD
1
C
H
B
We have to show that
AA
1
=
I
. Since (
D
+
C
H
BC
)
1
(
D
+
C
H
BC
)=
I
, and
B
1
B
=
I
,we
may rewrite this result as
AA
1
=
I

C
(
D
+
C
H
BC
)
1
C
H
B
+
CD
1
(
D
+
C
H
BC
)(
D
+
C
H
BC
)
1
C
H
B

CD
1
C
H
BC
(
D
+
C
H
BC
)
1
C
H
B
=
I
 [
C

CD
1
(
D
+
C
H
BC
) 
CD
1
C
H
BC
] · (
D
+
C
H
BC
)
1
C
H
B
= I  (
C

CD
1
D
)(
D
+
C
H
BC
)
1
C
H
B
Since
D
1
D
= I, the second term in this last line is zero; hence,
AA
1
=
I
which is the desired result.
β
ni
,
()
–
terms
λ…λλ
=
λ
n

i
=
199
9.3
We are given
(1)
Let
A
=
B
1
+
CD
1
C
H
(2)
Then, according to the matrix inversion lemma:
A
1
=
B

BC
(
D
+
C
H
BC
)
1
C
H
B
(3)
From Eqs. (1) and (2), we note:
A
=
B
1
=
δ
I
C
=
u
(
n
)
D
=
I
Hence, using Eq. (3):
9.4
From Section 9.6, we have
(1)
Φ
n
()
δ
Iu
n
u
H
n
+
=
Φ
n
Φ
1
–
n
1
δ

I
1
δ
2

u
n
1
1
δ
u
H
n
u
n
+
1
–
u
H
n
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This note was uploaded on 09/11/2010 for the course EE EE254 taught by Professor Ujin during the Spring '10 term at YTI Career Institute.
 Spring '10
 Ujin
 Signal Processing

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