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Math 235
Assignment 7 Solutions
1.
For each of the following symmetric matrices, ﬁnd an orthogonal matrix
P
and diagonal
matrix
D
such that
P
T
AP
=
D
.
(a)
A
=
±
1

2

2
1
²
Solution: We have
C
(
λ
) = det(
A

λI
) =
³
³
³
³
1

λ

2

2
1

λ
³
³
³
³
=
λ
2

2
λ

3 = (
λ

3)(
λ
+ 1)
Hence, the eigenvalues are
λ
1
= 3 and
λ
2
=

1.
For
λ
1
= 3 we get
A

λ
1
I
=
±

2

2

2

2
²
∼
±
1 1
0 0
²
. Thus, a basis for
E
λ
1
is
´±

1
1
²µ
.
For
λ
2
=

1 we get
A

λ
2
I
=
±
2

2

2
2
²
∼
±
1

1
0
0
²
. Thus, a basis for
E
λ
2
is
´±
1
1
²µ
.
After normalizing, the basis vectors for the eigenspaces form an orthonormal basis for
R
2
.
Hence,
P
=
±

1
/
√
2 1
/
√
2
1
/
√
2
1
/
√
2
²
is orthogonal and
P
T
AP
=
±
3
0
0

1
²
.
(b)
A
=
0 1 1
1 0 1
1 1 0
Solution: We have
C
(
λ
) =
³
³
³
³
³
³

λ
1
1
1

λ
1
1
1

λ
³
³
³
³
³
³
=
³
³
³
³
³
³

λ
1
2
1

λ
1

λ
0
1 +
λ
0
³
³
³
³
³
³
= (1 +
λ
)(
λ
2

λ

2) = (
λ
+ 1)
2
(
λ

2)
Hence, the eigenvalues are
λ
1
=

1 and
λ
2
= 2.
For
λ
1
=

1 we get
A

λ
1
I
=
1 1 1
1 1 1
1 1 1
∼
1 1 1
0 0 0
0 0 0
Thus, a basis for
E
λ
1
is

1
1
0
,

1
0
1
.
Since the basis is not orthogonal, we need to apply the GramSchmidt procedure. We pick
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 Spring '08
 CELMIN
 Matrices

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