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Unformatted text preview: 6.4 THE EIGENVALUE PROBLEM 439 I. Solution Now Ill—AI: file—C039 Sing
—si119 zl—coSQ = £2 — 2icosB+ c0526+ sin29= 12 — 2lcos9+1 So the eigenvalues are the roots of 112  Zlcose+ 1 = 0
that is, 3, : cosBijsinQ
Solving for the eigenvectors as in Example 6.4, we obtain 91 = [1 le and 62 = [1 J'lT
———‘————ﬁ—_________~____ In Example 6.5 we see that eigenvalues can be complex numbers, and that the eigen
vectors may have complex components. This situati on arises when the characteristic
equation has complex (conjugate) roots. \VggVQ‘ \Q ‘ c x, \
your answers using MATLAB. F] 1 F1 2
(a) 1 1] a» 3 2]
the method of Faddeev, obtain the , I L
ristic ol ornials of the matrices —
pm Flo4 F112
3 2 1 (c) 0 5 4 (d) o 2 2
Mi 5 1 b4 4 3 J ﬁ—1 1 3
.2 3 4
.12 112 0 61 l1—1 0
_' ‘1 1 0 (e) 0 11 6 (f) 1 2 1
6 6 *2 —2 l —1
1 1 1 L r L
1 1 0 r4 1 1‘ r1 —4 —2
h
. eigenvalues and corresponding (g) 2 5 4 ( J 0 3 l
' Its of the matrices _—1 ‘1 0.. L1 2 4 5.4.4 Repeated eigenvalues In the examples considered so far the eigenvalues 3.: (i = l, 2, . . . ) of the matrix A have been distinct, and in such cases the corresponding eigenvectors can be found and are
linearly independent. The matrix A is then said to have a full set of linearly independent eigenvectors. It is clear that the roots of the characteristic polynomial C(ﬂ.) may not all
be distinct; and when 0(1) has p S. n distinct roots, 0(1) may be factorized as C(11) =(;1_;1,)'”1(a—a2)’"11.. (ii—aft 5.4 THE EIGENVALUE PROBLEM 443 Adding 3 times cofumn 1 to column 2 followed b y subtracting 2 times column 1 from
column 3 gives ﬁnally 1 0 0
0 0 0
O 0 0 indicating a rank of I. Then from (6.11) the nullity qr2 corresponding to the eigenvalue 2t : 2 there are two line
as found in Example 6.6. In Example 6.7 we again had a repeated eigenvalue )1, = 2 of algebraic multiplicity = 3 — l : 2, continuing that
arly independent eigenvectors, 2.Then
1—2 2 2 —l 2 2
14—21: 0 2—2 1 = 0 0 I
*1 2 2—2 —1 2 0 Performing row and column operations as before produces the matrix y—n CDC 0
1
0 DO this time indicating a rank of 2. From (6.11) the nullity q, = 3 — 2 there is one and only one linearly independent eigenvector associ
value, as found in Example 6.7. = 1, conﬁrming that
ated with this eigen— tajn the eigenvalues and corresponding using the concept of rank, determine how
gmvectors of the matrices many linearly independent eigenvectors 2 2 1 0 2 2 correspond to this value of 3.. Determine a
_ _ ﬁ corresponding set of linearly independent
1 3 1 (b) ‘1 1 2 eigenvectors.
l 2 2 —l —l 2
Given that A = l is a twicerepeated eigenvalue
4 6 6 7 ‘3 ‘4 of the matrix
I 3 2 (d) 3 0 *2
4—5 —2 6—2 —3 2 f—1 that A: l is a threetimes repeated _1 _1 2
value of the matrix 3 #7 _ 5 how many linearly independent eigenvectors
A _ 2 4 3 correspond to this value of 1? Determine a
I — 1 2 2 corresponding set of linearly independent eigenvectors. 446 MATRIX ANALYSIS Use MATLAB th calculations. which in normalized form are
él = [1 2 0]T/J5, £2 2 [0 0 1]T, £3 = [—2 1 UTA/5
Evaluating the inner products, we see that, for example, /
éfé, :§+§+0= 1, é$é3=—§+§+o=o
and that
éléj=5i (1,}: 1, 2, 3) conﬁrming that the eigenvectors form an orthonormal set. . . s; c . . uses1 es s." Wk .
\ \ xx . t s
.\s\\\:\\\\ . .. l . \‘NM roughout to check your Determine the eigenvalues and corresponding eigenvectors of the symmetric matrix Verify Properties 6.1—6.7 of Section 6.4.6. *3 —3 —3
A = —3 1 —1
Given that the eigenvalues of the matrix _3 ﬂ 1 l
4 1 . . I
A _ 2 5 4 and verify that the eigenvectors are mutually
k orthogonal. '
—l *1 0
are 5 3 and 1_ ‘\ _ i The 3 x 3 symmetric matrix A has eigenvalue
’ ' 3 and 2. The eigenvectors corresponding to
(a) conﬁrm Properties 6.1764 of Section 6.4.6; the eigenvalues 6 and 3 are [1 1 2]T and
(b) taking k = 2, conﬁrm Properties 6576.7 of [I I —I]T respectively. Find an eigenvectoii Section 6.4.6. 6.5.1 corresponding to the eigenvalue 2. Numerical methods In practice we may well be dealing with matrices whose elements are decimal
or with matrices of high orders. In order to determine the eigenvalues and cig'en
of such matrices, it is necessary that we have numerical algorithms at our dis The power method Consider a matrix A having :4 distinct eigenvalues 3.1, 12, . . . , 1,, and correspo
linearly independent eigenvectors e], 32, . . . , en. Taking this set of vectors as th‘
we can write any vector x = [x1 x2 . . . xﬂT as a linear combination in the I] H
x: a181+a2£2+ . .+0€nen =2 age;
{:1 ...
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 Fall '10
 ChinC.Lee

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