1.
f ( λ 0 ) = f ( 5 ) = ∣ 1 − 5 1 1 2 1 − 5 1 1 1 3 − 5 ∣ = − 4 ⋅ 7 + 5 − 2 = − 25 f(\lambda_0)=f(5)=\begin{vmatrix}
1-5 & 1&1 \\
2 & 1-5&1\\
1&1&3-5
\end{vmatrix}=-4\cdot7+5-2=-25 f ( λ 0 ) = f ( 5 ) = ∣ ∣ 1 − 5 2 1 1 1 − 5 1 1 1 3 − 5 ∣ ∣ = − 4 ⋅ 7 + 5 − 2 = − 25
f ( λ 1 ) = f ( 4 ) = ∣ 1 − 4 1 1 2 1 − 4 1 1 1 3 − 4 ∣ = − 3 ⋅ 2 + 3 + 5 = 2 f(\lambda_1)=f(4)=\begin{vmatrix}
1-4 & 1&1 \\
2 & 1-4&1\\
1&1&3-4
\end{vmatrix}=-3\cdot2+3+5=2 f ( λ 1 ) = f ( 4 ) = ∣ ∣ 1 − 4 2 1 1 1 − 4 1 1 1 3 − 4 ∣ ∣ = − 3 ⋅ 2 + 3 + 5 = 2
slope = f ( λ 1 ) − f ( λ 0 ) λ 1 − λ 0 = 2 + 25 4 − 5 = − 27 \frac{f(\lambda_1)-f(\lambda_0)}{\lambda_1-\lambda_0}=\frac{2+25}{4-5}=-27 λ 1 − λ 0 f ( λ 1 ) − f ( λ 0 ) = 4 − 5 2 + 25 = − 27
the largest eigenvalue:
λ 2 = λ 1 − f ( λ 1 ) s l o p e = 4 + 2 / 27 = 4.074 \lambda_2=\lambda_1-\frac{f(\lambda_1)}{slope}=4+2/27=4.074 λ 2 = λ 1 − s l o p e f ( λ 1 ) = 4 + 2/27 = 4.074
2.
d 1 = ( 1 2 1 ) d_1=\begin{pmatrix}
1 \\
2 \\
1
\end{pmatrix} d 1 = ⎝ ⎛ 1 2 1 ⎠ ⎞ , d 2 = ( 1 1 1 ) d_2=\begin{pmatrix}
1 \\
1\\
1
\end{pmatrix} d 2 = ⎝ ⎛ 1 1 1 ⎠ ⎞ , d 3 = ( 2 1 3 ) d_3=\begin{pmatrix}
2 \\
1 \\
3
\end{pmatrix} d 3 = ⎝ ⎛ 2 1 3 ⎠ ⎞
∣ ∣ d 1 ∣ ∣ = 1 + 2 2 + 1 = 2.449 ||d_1||=\sqrt{1+2^2+1}=2.449 ∣∣ d 1 ∣∣ = 1 + 2 2 + 1 = 2.449
q 1 = d 1 / ∣ ∣ d 1 ∣ ∣ q_1=d_1/||d_1|| q 1 = d 1 /∣∣ d 1 ∣∣
q 1 T = ( 0.408 0.816 0.408 ) q_1^T=\begin{pmatrix}
0.408 & 0.816&0.408 \\
\end{pmatrix} q 1 T = ( 0.408 0.816 0.408 )
d 2 ′ = d 2 − ( q 1 T d 2 ) q 1 d'_2=d_2-(q_1^Td_2)q_1 d 2 ′ = d 2 − ( q 1 T d 2 ) q 1
d 2 ′ = ( 1 1 1 ) − ( 0.408 0.816 0.408 ) ( 1 1 1 ) ( 0.408 0.816 0.408 ) = ( 0.834 0.334 0.834 ) d'_2=\begin{pmatrix}
1 \\
1\\
1
\end{pmatrix}-\begin{pmatrix}
0.408 & 0.816&0.408 \\
\end{pmatrix}\begin{pmatrix}
1 \\
1\\
1
\end{pmatrix}\begin{pmatrix}
0.408 \\
0.816\\
0.408
\end{pmatrix}=\begin{pmatrix}
0.834 \\
0.334\\
0.834
\end{pmatrix} d 2 ′ = ⎝ ⎛ 1 1 1 ⎠ ⎞ − ( 0.408 0.816 0.408 ) ⎝ ⎛ 1 1 1 ⎠ ⎞ ⎝ ⎛ 0.408 0.816 0.408 ⎠ ⎞ = ⎝ ⎛ 0.834 0.334 0.834 ⎠ ⎞
∣ ∣ d 2 ′ ∣ ∣ = 1.226 ||d'_2||=1.226 ∣∣ d 2 ′ ∣∣ = 1.226
q 2 T = d 2 ′ / ∣ ∣ d 2 ′ ∣ ∣ = ( 0.680 0.272 0.680 ) q_2^T=d'_2/||d'_2||=\begin{pmatrix}
0.680 & 0.272&0.680 \\
\end{pmatrix} q 2 T = d 2 ′ /∣∣ d 2 ′ ∣∣ = ( 0.680 0.272 0.680 )
d 3 ′ = d 3 − ( q 1 T d 3 ) q 1 − ( q 2 T d 3 ) q 2 = ( 2 1 3 ) − ( 0.408 0.816 0.408 ) ( 2 1 3 ) ( 0.408 0.816 0.408 ) − − ( 0.680 0.272 0.680 ) ( 2 1 3 ) ( 0.680 0.272 0.680 ) = ( 2 − 0.333 − 0.924 1 − 0.666 − 0.074 3 − 0.499 − 1.387 ) = ( 0.743 0.260 1.114 ) d'_3=d_3-(q_1^Td_3)q_1-(q_2^Td_3)q_2=\begin{pmatrix}
2 \\
1\\
3
\end{pmatrix}-\begin{pmatrix}
0.408 & 0.816&0.408 \\
\end{pmatrix}\begin{pmatrix}
2 \\
1\\
3
\end{pmatrix}\begin{pmatrix}
0.408 \\
0.816\\
0.408
\end{pmatrix}--\begin{pmatrix}
0.680 & 0.272&0.680 \\
\end{pmatrix}\begin{pmatrix}
2 \\
1\\
3
\end{pmatrix}\begin{pmatrix}
0.680 \\
0.272\\
0.680
\end{pmatrix}=\begin{pmatrix}
2- 0.333-0.924\\
1-0.666-0.074\\
3-0.499-1.387
\end{pmatrix}=\begin{pmatrix}
0.743\\
0.260\\
1.114
\end{pmatrix} d 3 ′ = d 3 − ( q 1 T d 3 ) q 1 − ( q 2 T d 3 ) q 2 = ⎝ ⎛ 2 1 3 ⎠ ⎞ − ( 0.408 0.816 0.408 ) ⎝ ⎛ 2 1 3 ⎠ ⎞ ⎝ ⎛ 0.408 0.816 0.408 ⎠ ⎞ − − ( 0.680 0.272 0.680 ) ⎝ ⎛ 2 1 3 ⎠ ⎞ ⎝ ⎛ 0.680 0.272 0.680 ⎠ ⎞ = ⎝ ⎛ 2 − 0.333 − 0.924 1 − 0.666 − 0.074 3 − 0.499 − 1.387 ⎠ ⎞ = ⎝ ⎛ 0.743 0.260 1.114 ⎠ ⎞
∣ ∣ d 3 ′ ∣ ∣ = 1.364 ||d'_3||=1.364 ∣∣ d 3 ′ ∣∣ = 1.364
q 3 = d 3 ′ / ∣ ∣ d 3 ′ ∣ ∣ q_3=d'_3/||d'_3|| q 3 = d 3 ′ /∣∣ d 3 ′ ∣∣
q 3 T = ( 0.545 0.191 0.817 ) q_3^T=\begin{pmatrix}
0.545 & 0.191&0.817 \\
\end{pmatrix} q 3 T = ( 0.545 0.191 0.817 )
Q ( 0 ) = ( q 1 q 2 q 3 ) = ( 0.408 0.680 0.545 0.816 0.272 0.191 0.408 0.680 0.817 ) Q^{(0)}=\begin{pmatrix}
q_1 & q_2&q_3 \\
\end{pmatrix}=\begin{pmatrix}
0.408 & 0.680&0.545 \\
0.816 & 0.272&0.191\\
0.408&0.680&0.817
\end{pmatrix} Q ( 0 ) = ( q 1 q 2 q 3 ) = ⎝ ⎛ 0.408 0.816 0.408 0.680 0.272 0.680 0.545 0.191 0.817 ⎠ ⎞
for R ( 0 ) R^{(0)} R ( 0 ) :
r 11 = ∣ ∣ d 1 ′ ∣ ∣ = 2.449 , r 22 = ∣ ∣ d 2 ′ ∣ ∣ = 1.226 , r 33 = ∣ ∣ d 3 ′ ∣ ∣ = 1.364 r_{11}=||d'_1||=2.449,r_{22}=||d'_2||=1.226,r_{33}=||d'_3||=1.364 r 11 = ∣∣ d 1 ′ ∣∣ = 2.449 , r 22 = ∣∣ d 2 ′ ∣∣ = 1.226 , r 33 = ∣∣ d 3 ′ ∣∣ = 1.364
r 12 = q 1 T d 2 = 1.632 , r 13 = q 1 T d 3 = 2.856 , r 23 = q 2 T d 3 = 3.672 r_{12}=q_1^Td_2=1.632,r_{13}=q_1^Td_3=2.856,r_{23}=q_2^Td_3=3.672 r 12 = q 1 T d 2 = 1.632 , r 13 = q 1 T d 3 = 2.856 , r 23 = q 2 T d 3 = 3.672
R ( 0 ) = ( 2.449 1.632 2.856 0 1.226 3.672 0 0 1.364 ) R^{(0)}=\begin{pmatrix}
2.449 & 1.632&2.856 \\
0 & 1.226&3.672\\
0&0&1.364
\end{pmatrix} R ( 0 ) = ⎝ ⎛ 2.449 0 0 1.632 1.226 0 2.856 3.672 1.364 ⎠ ⎞
D ( 1 ) = R ( 0 ) Q ( 0 ) = ( 2.449 1.632 2.856 0 1.226 3.672 0 0 1.364 ) ( 0.408 0.680 0.545 0.816 0.272 0.191 0.408 0.680 0.817 ) = D^{(1)}=R^{(0)}Q^{(0)}=\begin{pmatrix}
2.449 & 1.632&2.856 \\
0 & 1.226&3.672\\
0&0&1.364
\end{pmatrix}\begin{pmatrix}
0.408 & 0.680&0.545 \\
0.816 & 0.272&0.191\\
0.408&0.680&0.817
\end{pmatrix}= D ( 1 ) = R ( 0 ) Q ( 0 ) = ⎝ ⎛ 2.449 0 0 1.632 1.226 0 2.856 3.672 1.364 ⎠ ⎞ ⎝ ⎛ 0.408 0.816 0.408 0.680 0.272 0.680 0.545 0.191 0.817 ⎠ ⎞ =
= ( 3.496 4.051 4.000 2.499 2.830 3.234 0.557 0.928 1.114 ) =\begin{pmatrix}
3.496 & 4.051&4.000 \\
2.499 & 2.830&3.234\\
0.557&0.928&1.114
\end{pmatrix} = ⎝ ⎛ 3.496 2.499 0.557 4.051 2.830 0.928 4.000 3.234 1.114 ⎠ ⎞
The diagonal elements in matrix D ( 1 ) D^{(1)} D ( 1 ) are the first approximation to the eigenvalues of matrix D D D .
So,
λ 1 = 3.496 , λ 2 = 2.830 , λ 3 = 1.114 \lambda_1=3.496,\lambda_2=2.830,\lambda_3=1.114 λ 1 = 3.496 , λ 2 = 2.830 , λ 3 = 1.114
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