Solutions for Homework 4 Foundations of Computational
Math 1 Fall 2010
Problem 4.1
Recall that an elementary reflector has the form
Q
=
I
+
αxx
T
∈
R
n
×
n
with
x
2
= 0.
4.1.a
. Show that
Q
is orthogonal if and only if
α
=

2
x
T
x
or
α
= 0
4.1.b
. Given
v
∈
R
n
, let
γ
=
±
v
and
x
=
v
+
γe
1
. Assuming that
x
=
v
show that
x
T
x
x
T
v
= 2
4.1.c
. Using the definitions and results above show that
Qv
=

γe
1
Solution:
We have
Q
T
Q
= (
I
+
αxx
T
)
T
(
I
+
αxx
T
) = (
I
+
αxx
T
)(
I
+
αxx
T
)
=
I
+ 2
αxx
T
+
α
2
x
(
x
T
x
)
x
T
=
I
+ (2
α
+
α
2
x
T
x
)
xx
T
Since
x
is arbitrary we must have
(2
α
+
α
2
x
T
x
) = 0
→
α
=

2
x
T
x
Now taking
x
=
v
+
γe
1
, we have
x
T
v
=
v
T
v
+
γe
T
1
v
=
γ
2
+
γν
1
x
T
x
= (
v
+
γe
1
)
T
(
v
+
γe
1
) =
v
T
v
+ 2
γν
1
+
γ
2
= 2(
γ
2
+
γν
1
)
∴
x
T
x
x
T
v
= 2
Finally,
Qv
= (
I
+
αxx
T
)
v
=
v
+
α
(
x
T
v
)
x
=
v
+
α
(
x
T
v
)
v
+
α
(
x
T
v
)
γe
1
α
(
x
T
v
) =

2
x
T
v
x
T
x
=

1
∴
Qv
=

γe
1
1
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Problem 4.2
4.2.a
This part of the problem concerns the computational complexity question of operation count.
For both
LU
factorization and Householder reflectorbased orthogonal factorization, we
have used elementary transformations,
T
i
, that can be characterized as rank1 updates to
the identity matrix, i.e.,
T
i
=
I
+
x
i
y
T
i
,
x
i
∈
R
n
and
y
i
∈
R
n
Gauss transforms and Householder reflectors differ in the definitions of the vectors
x
i
and
y
i
. Maintaining computational efficiency in terms of a reasonable operation count usually
implies careful application of associativity and distribution when combining matrices and
vectors.
Suppose we are to evaluate
z
=
T
3
T
2
T
1
v
= (
I
+
x
3
y
T
3
)(
I
+
x
2
y
T
2
)(
I
+
x
1
y
T
1
)
v
where
v
∈
R
n
and
z
∈
R
n
. Show that by using the properties of matrixmatrix multiplication
and matrixvector multiplication, the vector
z
can be evaluated in
O
(
n
) computations (a good
choice of version for an algorithm) or
O
(
n
3
) computations (a very bad choice of version for
an algorithm).
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 Spring '11
 Gallivan
 Trigraph, Xt, Qt, Mmin

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