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and
xy
xy
+=
+=
0
0.
In other words, the vectors
v
1
= (1,1) and
v
2
= (1,-1) are eigenvectors corresponding
to eigenvalues 1 and 3, respectively. Let
u
1
= (1/÷2,1/÷2) and
u
2
= (1/÷2,-1/÷2). If P is
the matrix with columns
u
i
, then
1
2
1
2
Ê
ˆ
10
03
=
Ê
Ë
ˆ
¯
Á
Á
˜
˜
-
1
P
=
and
P
AP
.
1
2
1
2
-
Ë
¯
1.8.13. Example.
Let
211
121
112
Ê
ˆ
Á
Á
˜
˜
A =
Ë
¯
We want to find an orthogonal matrix P so that P
-1
AP is a diagonal matrix.
Solution.
The roots of the characteristic polynomial
t
-- -
-
21 1
1 2 1
11 2
Ê
ˆ
(
)
=
2
3
Á
Á
˜
˜
(
)
(
)
det
tI
-
A
det
t
-
-
=-
t
1
t
-
4
Ë
¯
-
-
t
-
are 1 and 4. To find the eigenvectors corresponding to the eigenvalue 1 we need to
solve the equations
-- - =
-- - =
-- - =
xyz
xyz
xyz
0
0
0.
The solution set
X
has the form
{
(
)
}
=-
{
(
)
+-
(
)
}
--
y zyz yz
,,
,
Œ
R
y
11 0
,,
z
1 01
,,
yz
,
Œ
R
.
Applying the Gram-Schmidt algorithm to the basis (-1,1,0), (-1,0,1) produces the
orthonormal basis
u
1
= (-1/÷2,1/÷2,0) and
u
2
= (-1/÷6,-1/÷6,2/÷6) for
X
. Next, to find
the eigenvector for the eigenvalue 4, we need to solve
--=
-+ - =
-- +
2
xyz
xyz
xy z
0
2
0
20
=.
