536 Chapter 19The matrix eigenvalue problem
with eigenvalues λ
11 = 1 − 1 , λ
21 = 11 , and λ
31 = 12. The corresponding solutions of the
secular equations are obtained by replacing λin the equations by each root in turn.
We use equations (1) and (2) to solve for xand yin terms of (arbitrary) z:
We see that equation (3) is identical to (1) and gives no further information. Similarly,
The three eigenvectors are therefore
0 Exercises 4 –9
When the eigenvalues,λ
k(k 1 = 11 , 2 ,=,n), are distinct (with no two having the same
value, as in Example 19.3) then there exist ndistinct (linearly independent) eigenvectors.
When two or more eigenvalues are equal (multipleor degenerate eigenvalues) there
may exist fewerthan ndistinct eigenvectors. But symmetric(and Hermitian) matrices
always have the full complement of ndistinct eigenvectors.
EXAMPLES 19.4Multiple eigenvalues
(i) The characteristic equation of the non-symmetric matrix
and the single eigenvalueλ 1 = 13 is doubly-degenerate. Substitution of this value
of λin either secular equation gives just one eigenvector. Thus,
and the single eigenvector is ,yarbitrary.
x=
y
1
1
()20 0
3−+=→−+= → =
=λ
λxy xy xy
21
14
21
14
30
2−
−
−−
is =− =
λ
λ
()λ
xxx
1230
1
1
1
2
1
=−
,=
zz, ==
z
1
3
1
λλ==: −
−
=
=
→=
32 1
2
4
11 2 5
0
0
()
()
x
x
y
y
z
z
xz,, yz= 3
λλ==:
−−
=
=
→=
21
12
3
11 3 5
0
0
()
()
x
x
y
y
z
z
xz,,yz= 2
λλ==−:
−−
=
=
→=
11
12 11 5 5
0
0
0
()
()
x
x
y
y
z
z
x ,,
()
yz
xyz
=−
30 −+ + =