1.
f(λ0)=f(5)=∣∣1−52111−51113−5∣∣=−4⋅7+5−2=−25
f(λ1)=f(4)=∣∣1−42111−41113−4∣∣=−3⋅2+3+5=2
slope = λ1−λ0f(λ1)−f(λ0)=4−52+25=−27
the largest eigenvalue:
λ2=λ1−slopef(λ1)=4+2/27=4.074
2.
d1=⎝⎛121⎠⎞ , d2=⎝⎛111⎠⎞ , d3=⎝⎛213⎠⎞
∣∣d1∣∣=1+22+1=2.449
q1=d1/∣∣d1∣∣
q1T=(0.4080.8160.408)
d2′=d2−(q1Td2)q1
d2′=⎝⎛111⎠⎞−(0.4080.8160.408)⎝⎛111⎠⎞⎝⎛0.4080.8160.408⎠⎞=⎝⎛0.8340.3340.834⎠⎞
∣∣d2′∣∣=1.226
q2T=d2′/∣∣d2′∣∣=(0.6800.2720.680)
d3′=d3−(q1Td3)q1−(q2Td3)q2=⎝⎛213⎠⎞−(0.4080.8160.408)⎝⎛213⎠⎞⎝⎛0.4080.8160.408⎠⎞−−(0.6800.2720.680)⎝⎛213⎠⎞⎝⎛0.6800.2720.680⎠⎞=⎝⎛2−0.333−0.9241−0.666−0.0743−0.499−1.387⎠⎞=⎝⎛0.7430.2601.114⎠⎞
∣∣d3′∣∣=1.364
q3=d3′/∣∣d3′∣∣
q3T=(0.5450.1910.817)
Q(0)=(q1q2q3)=⎝⎛0.4080.8160.4080.6800.2720.6800.5450.1910.817⎠⎞
for R(0) :
r11=∣∣d1′∣∣=2.449,r22=∣∣d2′∣∣=1.226,r33=∣∣d3′∣∣=1.364
r12=q1Td2=1.632,r13=q1Td3=2.856,r23=q2Td3=3.672
R(0)=⎝⎛2.449001.6321.22602.8563.6721.364⎠⎞
D(1)=R(0)Q(0)=⎝⎛2.449001.6321.22602.8563.6721.364⎠⎞⎝⎛0.4080.8160.4080.6800.2720.6800.5450.1910.817⎠⎞=
=⎝⎛3.4962.4990.5574.0512.8300.9284.0003.2341.114⎠⎞
The diagonal elements in matrix D(1) are the first approximation to the eigenvalues of matrix D.
So,
λ1=3.496,λ2=2.830,λ3=1.114
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