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Figure 3.10
First example for MSA LUO: eigenvalue estimation.
φ
2
(
1000
)
λ
0744]
T
=
)
=
1
.
1094
w
(
1000
)
=
[0
.
0414, 0
.
7109,
−
0
.
7263, 0
.
2
2
2
(
1000
which are similar to the MSA EXIN results. Figure 3.10 shows that the loss
of step begins after only 1000 iterations. After 4000 iterations,
w
1
and
w
2
,are
parallel. This is also shown in Figure 3.11, which is a plot of
P
12
. Hence, the
loss of step happens much earlier in MSA LUO than in MSA EXIN. Note in
Figure 3.11 that there are no flat portions of the curve before the loss of step (in
MSA EXIN there is a big portion with nearly null gradient); this fact does not
allow us to devise a good stop criterion. Figures 3.12 and 3.13 show the accurate
recovery of the minor subspace. Figures 3.14, 3.15, and 3.16 show the results
for
λ
2
alone, and for
P
12
, respectively, when the true
solutions are taken as initial conditions.
λ
1
and
λ
2
together, for
λ
2
goes to infinity at iteration 20, 168,
λ
1
at iteration 21, 229.
The third example uses, as a benchmark, Problem 6.8 in [121, p. 229]. Here,
2
.
5678
1
.
1626
0
.
9302
−
3
.
5269
"
#
1
.
1626
6
.
1180
4
.
3627
−
3
.
4317
R
=
0
.
9302
4
.
3627
4
.
7218
−
2
.
7970
−
3
.
5269
−
3
.
4317
−
2
.
7970
9
.
0905
has nearly nondistinct eigenvalues,
λ
1
=
1
.
0000 and
λ
2
=
1
.
0001, with corre-
sponding eigenvectors
4433
]
T
z
1
=
[
0
.
.
.
.
9070, 0
0616, 0
0305, 0
1198]
T
z
2
=
[
−
0
.
2866, 0
.
6078,
−
0
.
7308,
−
0
.
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