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Math
2940
Solutions, Fall
2011
Section
6
.
4
2
) We have (
A
T
CA
)
T
=
A
T
C
T
(
A
T
)
T
=
A
T
CA
. When
A
is 6
×
3 we have
A
T
is 3
×
6. So
C
must be 6
×
6 and
A
T
CA
is 3
×
3.
5
) We need to ﬁnd the unit eigenvectors of
A
. They will be our columns.
The characteristic polynomial of
A
=
1
0
2
0

1

2
2

2
0
is (1

λ
)(

1

λ
)(

λ
)

4(1

λ
) + 4(1 +
λ
) =
λ
(9

λ
2
). The eigenvalues are 0
,
3
,

3 wih eigenvectors (2
,
2
,
1), (2
,

1
,
2)
and (1
,

2
,

2) . So our matrix is
2
/
3
2
/
3
1
/
3
2
/
3

1
/
3

2
/
3
1
/
3

2
/
3

2
/
3
.
7
) (a)
±
1
b
b
1
²
has characteristic polynomial
λ
2

2
λ
+ 1

b
2
which has roots 1
±
b
. So as
long as

b

>
1 one eigenvalue will be negative.
(b) The pivots have the same signs as the eigenvalues.
(c) The sum of the eigenvalues is the trace of the matrix, 2. So both eigenvalues cannot be
negative.
15
)
±
i
1
1

i
²
has characteristic polynomial
λ
2
which has 0 as a double root. If there were
independent vectors with 0 eigenvalue, our matrix would be the zero matrix!
23
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