Biomedical Engineering Reference
In-Depth Information
þ
þ
;
A
11
A
12
A
21
A
22
A
11
A
13
A
31
A
33
A
22
A
23
A
32
A
33
II
A
¼
(A.56)
A
11
A
12
A
13
A
21
A
22
A
23
A
31
A
32
A
33
III
A
¼
Det
A
¼
:
(A.57)
's that allow the determinant (A.53)
to vanish. We note again that the vanishing of the determinant makes the set of
equations (A.49) linearly dependent. Since the system is linearly dependent, all of
the components of
t
cannot be determined from (A.49). Thus, for each value of
This argument then generates a set of three
l
l
that is a solution to (A.54), we can find only two ratios of the elements of
t
,
t
1
,
t
2
,
and
t
3
. It follows that, for each eigen n-tuple, there will be one scalar unknown.
In this text we will only be interested in the eigenvalues of symmetric matrices.
In Sect. A.7 it is shown that a necessary and sufficient condition for all the
eigenvalues to be real is that that the matrix be symmetric.
Example A.5.5
Find the eigenvalues and construct the ratios of the eigen n-tuples of the matrix
2
4
3
5
:
18 6 6
6 50
601
A
¼
(A.58)
Solution: The cubic equation associated with this matrix is, from (A.54), (A.55),
(A.56), and (A.57),
3
2
l
54
l
þ
891
l
4
;
374
¼
0
;
(A.59)
which has three roots, 27, 18, and 9. The eigen n-tuples are constructed using these
eigenvalues. The first eigen n-tuple is obtained by substitution of (A.58) and
l ¼
27
into (A.49), thus
9
t
1
þ
6
t
2
þ
6
t
3
¼
0
;
6
t
1
12
t
2
¼
0
;
6
t
1
6
t
3
¼
0
:
(A.60)
Note the linear dependence of this system of equations; the first equation is equal
to the second multiplied by (
1). Since
there are only two independent equations, the solution to this system of equations is
t
1
¼
1/2) and added to the third multiplied by (
2
t
2
, leaving an undetermined parameter in the eigen n-tuple
t
.
Similar results are obtained by taking
t
3
and
t
1
¼
l ¼
18 and
l ¼
9.
Problems
A.5.1. Show that the eigenvalues of the matrix
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