Graphics Programs Reference
In-Depth Information
The characteristicequationis
1
−
λ
−
1
0
2
3
|
A
−
λ
I
| =
−
12
−
λ
−
1
=−
3
λ
+
4
λ
−
λ
=
0
(b)
0
−
11
−
λ
The roots of thisequationare
λ
1
=
0,
λ
2
=
1,
λ
3
=
3. To c ompute the eigenvector
λ
3
, wesubstitute
λ
=
λ
3
into Eq. (9.2), obtaining
corresponding the
⎡
⎣
⎤
⎦
⎡
⎣
⎤
⎦
=
⎡
⎣
⎤
⎦
−
2
−
1
0
x
1
x
2
x
3
0
0
0
−
1
−
1
−
1
(c)
−
−
0
1
2
Weknow that the determinant of the coefficient matrix iszero, so that the equations
are not linearly independent. Therefore, wecan assign an arbitrary valuetoany one
componentof
x
and use two of the equationstocompute the other twocomponents.
Choosing
x
1
=
1, the first equation of Eq. (c) yields
x
2
=−
2and fromthe third equation
we get
x
3
=
1. Thus the eigenvectorassociatedwith
λ
3
is
⎡
⎣
⎤
⎦
1
x
3
=
−
2
1
The other twoeigenvectors
⎡
⎣
⎤
⎦
⎡
⎣
⎤
⎦
1
0
1
1
1
x
2
=
x
1
=
−
1
can beobtainedinthe same manner.
It issometimes convenienttodisplay the eigenvectors as columnsofamatrix
X
.
For the problemathand,this matrix is
⎡
⎣
⎤
⎦
111
1
x
1
x
3
X
=
x
2
=
0
−
2
1
−
11
It is clear from the aboveexample that the magnitudeofan eigenvectoris indeter-
minate; onlyits direction canbecomputed fromEq. (9.2). It iscustomary to
normalize
the eigenvectors by assigning a unit magnitudetoeach vector. Thus the normalized
eigenvectors in our example are
⎡
⎣
/
√
3
1
/
√
21
/
√
6
⎤
⎦
1
/
√
3
/
√
6
X
=
1
0
−
2
/
√
3
/
√
21
/
√
6
1
−
1
Throughoutthischapterwe assumethat the eigenvectors are normalized.
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