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Math 235
Assignment 6 Solutions
1.
Let
A
=
1
2

4
2

2

2

4

2
1
. Find an orthogonal matrix
P
that
diagonalizes
A
and the corresponding diagonal matrix.
Solution: The characteristic polynomails is
C
(
λ
) = det(
A

λI
) = det
1

λ
2

4
2

2

λ

2

4

2
1

λ
= det
1

λ
2

4
2

2

λ

2

3

λ
0

3

λ
= det
5

λ
2

4
4

2

λ

2
0
0

3

λ
=

(
λ
+ 3)(
λ
2

3
λ

18) =

(
λ
+ 3)
2
(
λ

6)
.
Hence, the eigenvalues of
A
are
λ
1
=

3 and
λ
2
= 6.
For
λ
1
=

3 we have
A

λ
1
I
=
4
2

4
2
1

2

4

2
4
∼
1 1
/
2

1
0
0
0
0
0
0
.
Thus, linearly independent eigenvectors corresonding to
λ
1
are
~v
1
=

1
2
0
, and
~v
2
=
1
0
1
.
For
λ
2
= 6 we have
A

λ
2
I
=

5
2

4
2

8

2

4

2

5
∼
1 0
1
0 1 1
/
2
0 0
0
.
Thus, an eigenvector corresonding to
λ
2
is

2

1
2
.
We ﬁrst observe that eigenvectors for
λ
1
are not orthogonal. So, applying the GramSchmidt
procedure to
{
~v
1
,~v
2
}
: Let
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This note was uploaded on 11/30/2010 for the course MATH 235/237 taught by Professor Wilkie during the Spring '10 term at Waterloo.
 Spring '10
 WILKIE
 Math

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