Solutions for Homework 4 Numerical Linear Algebra 1
Fall 2011
Problem 4.1
Given that we know the SVD exists for any complex matrix
A
∈
C
m
×
n
, assume that
A
∈
R
m
×
n
has rank
k
with
k
≤
n
, i.e.,
A
is real and it may be rank deficient, and show that the
SVD of
A
is all real and has the form
A
=
U
parenleftbigg
S
0
parenrightbigg
V
T
=
U
k
Σ
k
V
T
k
where
S
∈
R
n
×
n
is diagonal with nonnegative entries,
U
=
(
U
k
U
m

k
)
,
U
T
U
=
I
m
V
=
(
V
k
V
n

k
)
,
V
T
V
=
I
n
U
k
∈
R
m
×
k
,
and
V
k
∈
R
n
×
k
Solution:
The result follows simply from the relationship of the SVD to the symmetric eigenvalue
decomposition.
A
T
A
and
AA
T
are both real symmetric positive semidefinite and therefore
have real orthogonal eigenvectors that define orthogonal
U
and
V
as needed. The rank is
reflected in the nonzero eigenvalues which are the squares of the nonzero singular values and
therefore define Σ
k
and
S
.
U
k
and
V
k
follow immediately from the appropriate partitionings
of
U
and
V
.
Problem 4.2
Let
T
∈
R
n
×
n
be a symmetric tridiagonal matrix, i.e.,
e
T
i
Te
j
=
e
T
j
Te
i
and
e
T
i
Te
j
= 0 if
j < i
−
1 or
j > i
+ 1. Consider
T
=
QR
where
R
∈
R
n
×
n
is an upper triangular matrix and
Q
∈
R
n
×
n
is an orthogonal matrix.
Recall, the nonzero structure of
R
was derived in class and shown to be
e
T
i
Re
j
= 0 if
j < i
(upper triangular assumption) or if
j > i
+ 2, i.e, nonzeros are restricted to the main
diagonal and the first two superdiagonals.
(4.2.a) Show that
Q
has nonzero structure such that
e
T
i
Qe
j
= 0 if
j < i
−
1, i.e.,
Q
is
upper Hessenberg.
(4.2.b) Show that
T
+
=
RQ
is a symmetric triagonal matrix.
(4.2.c) Prove the Lemma in the class notes that states that choosing the shift
μ
=
λ
,
where
λ
is an eigenvalue of
T
, results in a reduced
T
+
with known eigenvector
and eigenvalue.
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Solutions:
To show that
Q
is upper Hessenberg we can analyze the structure of the product
of Given’s rotations used to produce it,
Q
=
G
1
G
2
· · ·
G
n

1
. We have the column structure
G
i
e
i
=
γ
i
e
i
−
σ
i
e
i
+1
G
i
e
i
+1
=
σ
i
e
i
+
γ
i
e
i
+1
G
i
e
j
=
e
j
,
j < i,
j > i
+ 1
and the row structure
e
T
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 Fall '06
 gallivan
 Matrices, Diagonal matrix, Orthogonal matrix, symmetric tridiagonal matrix, Qi ΛQT, Van Loan Problem

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