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25
and
±
v
±
2
=(
z
∗
−
α
∗
e
T
1
)(
z
−
α
e
1
)
=
z
∗
z
−
α
∗
z
1
−
αz
∗
1
+
α
∗
α
=
z
∗
z
−
e
−
iθ
±
z
±
e
iθ
ζ
−
e
iθ
±
z
±
e
−
iθ
ζ
+
±
z
±
2
=2
z
∗
z
−
2
±
z
±
ζ.
Then
Qz
=(
I
−
2
uu
∗
)
z
=(
I
−
2
±
v
±
2
vv
∗
)
z
=
z
−
2
v
∗
z
±
v
±
2
v
=
z
−
v
=
α
e
1
.
(b) Let the second column of
A
1
be [
a,
v
1
,...,
v
n
−
1
]
T
. Use the vector
v
to form the
Householder transformation. Then the product
QA
1
leaves the Frst column of
A
1
unchanged and puts zeros below the main diagonal in the second column.
(c) Assume
m>n
. (The other case is similar and left to the reader.)
Initially, let
R
=
A
and
Q
=
I
(dimension
m
).
for
j
=1:
n
,
(1) Let
z
=[
r
jj
,...r
mj
]
T
.
(2) Let the polar coordinate representation of
z
1
be
e
iθ
ζ
.
(3) DeFne
v
=
z
−
α
e
1
where
α
=
−
e
iθ
±
z
±
.
(4) Let
u
=
v
/
±
v
±
, and deFne the Householder transformation by
b
Q
=
I
−
2
uu
∗
.
(5) Apply the transformation to
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.
 Fall '11
 Dr.Robin

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