Math 20F  Homeworks 9 & 10  Selected answers
Samuel R. Buss  Winter 2003  UC San Diego
Revision 1.0. – March 13, 2003
Section 6.1, Problem 19.
We did almost exactly this problem as a theorem proved in class.
Section 6.1, Problem 22.
Since
u
T
j
u
i
=
δ
i,j
, we have
A
u
i
=
n
X
j
=1
c
j
u
j
u
T
j
u
i
=
n
X
j
=1
δ
i,j
c
j
u
j
=
c
i
u
i
.
Hence
u
i
is an eigenvector for the eigenvalue
c
i
.
Section 6.3, Problem 1(a).
We did this as an example in class on Wednesday.
Section 6.3, Problem 1(c).
det(
A

λI
)=(2

λ
)(

4

λ
)+8=2
λ
+
λ
2
=(2+
λ
)(
λ
). The
roots of the characteristic polynomial are
λ
1
= 1 and
λ
2
=

2: these are the eigenvalues
of
A
.
Solving
A
x
= 0 for a nontrivial
x
, we ﬁnd that
x
1
=(4
,
1)
T
is a eigenvector corresponding
to the eigenvalue
λ
1
=0.
Solving (
A
+2
I
)
x
= 0, we ﬁnd that
x
2
=(2
,
1)
T
is an eigenvector corresponding to the
eigen value
λ
2
=

2.
The eigenvalues are distinct, hence
x
1
and
x
2
are linearly independent. Therefore,
A
is
diagonalizable in the following form:
A
=
±
42
11
¶
·
±
00
0

2
¶
·
±
¶

1
=
±
¶
·
±
0

2
¶
·
±
1
/
2

1

1
/
22
¶
Section 6.3, Problem 3(c).
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 Winter '03
 BUSS
 Math, Linear Algebra, Algebra, Matrices, Eigenvalue, eigenvector and eigenspace, Singular value decomposition, Orthogonal matrix, linearly independent eigenvectors

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