Graphics Programs Reference
In-Depth Information
h=h/2;
if num
eVals<k;eVal=eVal+h;
elseif num
_
_
eVals>k;eVal=eVal-h;
else; break
end
end
% If eigenvalue located, change upper limit of
% search and record result in
{
r
}
ValMax = eVal; r(k+1) = eVal;
end
EXAMPLE 9.11
Bracket each eigenvalue of the matrix in Example 9.10.
Solution
In Example 9.10 wefound that all the eigenvalues lie in (0
,
8).We nowbisect
this interval and use the Sturm sequence to determine the number of eigenvalues in
(0
,
4).With
λ
=
4, the sequence is—see Eqs. (9.49)
P
0
(4)
=
1
=
−
=
P
1
(4)
4
4
0
2
2
(1)
P
2
(4)
=
(4
−
4)(0)
−
=−
4
2
2
(0)
P
3
(4)
=
(5
−
4)(
−
4)
−
=−
4
Since a zero value is assigned the sign opposite to that of the preceding member, the
signs in this sequence are(
+
,
−
,
−
,
−
). The onesign change shows the presence of
oneeigenvalue in (0
4).
Next we bisect the interval(4
,
,
8) and compute the Sturm sequence with
λ
=
6:
P
0
(6)
=
1
P
1
(6)
=
4
−
6
=−
2
2
2
(1)
P
2
(6)
=
(4
−
6)(
−
2)
−
=
0
2
2
(
P
3
(6)
=
(5
−
6)(0)
−
−
2)
=
8
In this sequence the signs are(
+
,
−
,
+
,
+
), indicating twoeigenvalues in (0
,
6).
Therefore
0
≤
λ
1
≤
44
≤
λ
2
≤
66
≤
λ
3
≤
8
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