MFDS-NM-05: Eigenvalue dan Eigenvector Numerik

Outline:

• Referensi/Materi
• Referensi Video
• Code/Module
• Forum Diskusi

Referensi/Materi

• https://www.math.tamu.edu/~dallen/linear_algebra/chpt6.pdf
• https://www.math.upenn.edu/~deturck/m320/text/part8.pdf
• Materi Bahasa Indonesia

Referensi Video

• https://www.youtube.com/watch?v=hrDgjfc-RJk
• https://www.youtube.com/watch?v=-2bBukw4az8

Open Code in Google Colaboratory

Metode Numerik
EigenValue dan EigenVektor Numerik

(C) Taufik Sutanto - 2020

tau-data Indonesia ~ https://tau-data.id/mfds-nm-05/

Eigenvalue (Eigenvector) Numerik:¶

• Pendahuluan Eigenvalue dan eigen vektor
• Aplikasi & Intuisi
• Karakteristik Polinomial
• Power Method
• Rayleigh Quotient
• Deflated Matrix Method
• Jacobi Method

Karektiristik Polynomial ~ Permasalahan Mencari Akar¶

Keterbatasan - Companion Matrix

Contoh lain permasalahan numerik menggunakan AKP¶

In [1]:

# Python implementation
import numpy as np

A = np.array([[ 1, 1 ],
              [ 0, 2 ]])
A

Out[1]:

array([[1, 1],
       [0, 2]])

In [2]:

# Operasi Matrix
print('A \n',A)
print('A*A \n',A.dot(A))
print('A+A \n',A+A)
print("A' \n",A.transpose())

A 
 [[1 1]
 [0 2]]
A*A 
 [[1 3]
 [0 4]]
A+A 
 [[2 2]
 [0 4]]
A' 
 [[1 0]
 [1 2]]

In [3]:

# Eigenvalue dan eigenvector

A = np.array([[ 3, -1 ],[ -1, 3 ]])
print(A)
eVa, eVe = np.linalg.eig(A)
print('Eigenvalues \n',eVa)
print('Eigenvectors \n',eVe)

[[ 3 -1]
 [-1  3]]
Eigenvalues 
 [4. 2.]
Eigenvectors 
 [[ 0.70710678  0.70710678]
 [-0.70710678  0.70710678]]

In [4]:

A = [[1,2,3],[0,4,5],[0,0,6]]
A = np.array(A)
eVa, eVe = np.linalg.eig(A)
print('Eigenvalues \n',eVa)
print('Eigenvectors \n',eVe)

Eigenvalues 
 [1. 4. 6.]
Eigenvectors 
 [[1.         0.5547002  0.51084069]
 [0.         0.83205029 0.79818857]
 [0.         0.         0.31927543]]

In [5]:

eVe[1]/eVe[1].max()

Out[5]:

array([0.        , 1.        , 0.95930327])

Latihan 1:¶

• Diberikan matrix sbb: $$ \begin{bmatrix} -2 & -3 \\ 6 & 7 \end{bmatrix} $$
• Tentukan Error Relatif aproksimasi eigenvalue terkait matrix tersebut
• jika persamaan karakteristik polinomial-nya diselesaikan dengan metode Newton
• tiga iterasi ($\lambda_3$) dengan $\lambda_0 = 1$

In [6]:

def F(x):
    return x**2-5*x+4
def f(x):
    return 2*x-5
# Newton (L3, Lo=1)
Lo =1
L1 = Lo - F(Lo)/f(Lo)
L2 = L1 - F(L1)/f(L1)
L3 = L2 - F(L2)/f(L2)
L3
eksak = 1
abs(L3-eksak)/abs(eksak) # Error_Relatif

Out[6]:

0.0

Power Method - Contoh

In [7]:

# Contoh
A = np.array([[ 2, 9 ],[ 9, 1 ]])
x = np.array([1,1]).transpose()
N = 5
for i in range(N):
    xo = x
    x = A.dot(x)
    x1 = x
    eigen = max(abs(x1))/max(abs(xo))
print('Eigenvalue = ', eigen)
print('Eigenvector = ', x)

Eigenvalue =  10.63820909893284
Eigenvector =  [132584 124129]

In [8]:

eVa, eVe = np.linalg.eig(A)
print('Eigenvalues \n',eVa)
print('Eigenvectors \n',eVe)

Eigenvalues 
 [10.51387819 -7.51387819]
Eigenvectors 
 [[ 0.72645372 -0.68721539]
 [ 0.68721539  0.72645372]]

In [9]:

# Seringnya Power method menggunakan normalisasi vector di setiap langkahnya
x = np.array([1,1]).transpose()
N = 6
for i in range(N):
    x = x/x.max()
    xo = x
    x = A.dot(x)
    x1 = x
    eigen = max(abs(x1))/max(abs(xo))
print('Eigenvalue = ', eigen)
print('Eigenvector = ', x/x.max())

Eigenvalue =  10.426061968261632
Eigenvector =  [1.         0.95301842]

Latihan 2:¶

• Diberikan matrix sbb: $$ \begin{bmatrix} 1 & 1 \\ 4 & 1 \end{bmatrix} $$
• Tentukan Error Relatif aproksimasi eigenvalue
• terkait matrix tersebut jika eigenvalue-nya didekati
• dengan metode Rayleigh power Method dua iterasi ($\lambda_2$) dengan $x_0 = [1,1]'$

In [10]:

# Contoh
A = np.array([[ 1, 1 ],[ 4, 1 ]])
x = np.array([1,1]).transpose()
N = 4
for i in range(N):
    xo = x
    x = A.dot(x)
    x1 = x
    eigen = max(abs(x1))/max(abs(xo))
print('Eigenvalue = ', eigen)
print('Error Relatif = ', abs(3-eigen)/3)

Eigenvalue =  2.951219512195122
Error Relatif =  0.016260162601626032

Rayleigh Quotient

• Secara umum Power method Lambat, methode Rayleigh mempercepat iterasi Power Method
• 

In [11]:

# Contoh (Sebelumnya)
import numpy as np
A = np.array([[ 1, 0, 2 ],[ 0, 1, 1 ], [2, 1, 2]])
A = np.linalg.inv(A)
x = np.array([1,1, 1]).transpose()
N = 2
eksak = -1
for i in range(N):
    xo = x
    x = A.dot(x)
    x1 = x
    eigen = max(abs(x1))/max(abs(xo))
    eigenRayleigh = x.transpose().dot(A).dot(x)/x.transpose().dot(x)
    
print('EigenValue Menurut metode Rayleigh 2 iterasi =', eigenRayleigh)    
    
print('EA Eigenvalue (Power Method)= ', abs(eigen-eksak))
print('EA Eigenvalue (Rayleigh)= ', abs(eigenRayleigh-eksak))
print('Eigenvector = ', x)

EigenValue Menurut metode Rayleigh 2 iterasi = -0.5555555555555552
EA Eigenvalue (Power Method)=  2.0
EA Eigenvalue (Rayleigh)=  0.44444444444444475
Eigenvector =  [ 0.33333333  0.66666667 -0.33333333]

End of Module

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Referensi:

1. Johansson, R. (2018). Numerical Python: Scientific Computing and Data Science Applications with Numpy, SciPy and Matplotlib. Apress.
2. John H. Mathews, Numerical Methods for Mathematics, Prentice Hall, 1992 [Refernsi Utama]
3. Heath, M. T. (2018). Scientific computing: an introductory survey (Vol. 80). SIAM.
4. Conte, S. D., u0026amp; De Boor, C. (2017). Elementary numerical analysis: an algorithmic approach (Vol. 78). SIAM.