Valores y vectores propios

Método de Leverrier-Faddeev (algoritmo alterno)

El polinomio caracteristico de un matriz es $$ \begin{equation*} p(\lambda) = c_{0} \lambda^{n} + c_{1} \lambda^{n-1} + c_{2} \lambda^{n-2} + \dots + c_{n-1} \lambda + c_{n} \end{equation*} $$

Algoritmo

$$ \begin{align*} c_{0} &= 1 \\ B_{0} &= I \\ k &= 1,\ldots ,n-1 \\ &\quad c_{k} = -\frac{1}{k} \ \mathrm{tr}(AB_{k-1}) \\ &\quad B_{k} = AB_{k-1} + c_{k} I \\ c_{n} &= -\frac{1}{n} \ \mathrm{tr}(AB_{n-1}) \\ \det A &= (-1)^n c_{n} \\ A^{-1} &= - \frac 1 {c_n} B_{n-1} \end{align*} $$

Seudocódigo

function Leverrier_Faddeev(A)
    I = 1 // matriz identidad
    c[0] = 1
    B[0] = I

    for k=1 to n-1 do
        c[k] = - 1/k * traza(A * B[k-1])
        B[k] = A * B[k-1] + c[k] * I
    end for

    c[n] = - 1/n * traza(A * B[n-1])
    determinante = (-1)^n * c[n]

    if c[0]!=0
        inversa = -1/c[n] * B[n-1]
    else
        inversa = 'No tiene inversa'
    end if

    mostrar c, determinante, inversa
end function

Ejemplo 1

Determinar el polinomio caracteristico de $ A $ $$ \begin{equation*} A = \begin{bmatrix} 3 & 1 & 5 \\ 3 & 3 & 1 \\ 4 & 6 & 4 \end{bmatrix} \end{equation*} $$

Iniciando valores $$ \begin{align*} c_{0} &= 1 \\ B_{0} &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{align*} $$

Bucle 1 $$ \begin{align*} c_{1} &= -\frac{1}{1} \ \mathrm{tr}(A B_{0}) = -\frac{1}{1} (10) = -10 \\ B_{1} &= AB_{0} + c_{1} I = \begin{bmatrix} 3 & 1 & 5 \\ 3 & 3 & 1 \\ 4 & 6 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -7 & 1 & 5 \\ 3 & -7 & 1 \\ 4 & 6 & -6 \end{bmatrix} \end{align*} $$

Bucle 2 $$ \begin{align*} c_{2} &= -\frac{1}{2} \ \mathrm{tr}(A B_{1}) = -\frac{1}{2} (-8) = 4 \\ B_{2} &= AB_{1} + c_{2} I = \begin{bmatrix} 3 & 1 & 5 \\ 3 & 3 & 1 \\ 4 & 6 & 4 \end{bmatrix} \begin{bmatrix} -7 & 1 & 5 \\ 3 & -7 & 1 \\ 4 & 6 & -6 \end{bmatrix} + \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 26 & -14 \\ -8 & -8 & 12 \\ 6 & -14 & 6 \end{bmatrix} \end{align*} $$

Calculando $ c_{3} $ $$ \begin{align*} c_{3} &= -\frac{1}{3} \ \mathrm{tr}(A B_{2}) = -\frac{1}{3} (120) = -40 \end{align*} $$

El polinomio caracteristico es $$ \begin{equation*} p(\lambda) = c_{0} \lambda^{3} + c_{1} \lambda^{2} + c_{2} \lambda + c_{3} = \lambda^{3} - 10 \lambda^{2} + 4 \lambda - 40 \end{equation*} $$

El determinante es $$ \begin{equation*} \det(A) = (-1)^{3} c_{n} = (-1)^{3} (-40) = 40 \end{equation*} $$

Su inversa es $$ \begin{equation*} A^{-1} = -\frac{1}{c_{n}} B_{2} = -\frac{1}{-40} \begin{bmatrix} 6 & 26 & -14 \\ -8 & -8 & 12 \\ 6 & -14 & 6 \end{bmatrix} = \begin{bmatrix} 0.15 & 0.65 & -0.35 \\ -0.20 & -0.20 & 0.30 \\ 0.15 & -0.35 & 0.15 \end{bmatrix} \end{equation*} $$

Implementación

import numpy as np

def Leverrier_Faddeev(A):
    m, n = A.shape
    I = np.eye(n)
    B = np.zeros((n+1,n,n))
    c = np.zeros(n+1)

    c[0] = 1.0
    B[0] = I

    for k in range(1,n):
        c[k] = - 1/k * np.trace(np.dot(A,B[k-1]))
        B[k] = np.dot(A,B[k-1]) + c[k]*I

    c[n] = - 1/n * np.trace(np.dot(A,B[n-1]))
    determinante = (-1)**n * c[n]

    if c[0]!=0:
        inversa = -1/c[n] * B[n-1]
    else:
        inversa = np.zeros((n,n)) * np.nan

    print('p(x) =', c)
    print('determinante =', determinante)
    print('inversa =')
    print(inversa)

A = np.array([[3,1,5],
              [3,3,1],
              [4,6,4]], float)
Leverrier_Faddeev(A)

p(x) = [  1. -10.   4. -40.]
determinante = 40.0
inversa =
[[ 0.15  0.65 -0.35]
 [-0.2  -0.2   0.3 ]
 [ 0.15 -0.35  0.15]]

#revisando
print('p(x) =', np.poly(A))
print('determinante =', np.linalg.det(A))
print('inversa =')
print(np.linalg.inv(A))

p(x) = [  1. -10.   4. -40.]
determinante = 40.0
inversa =
[[ 0.15  0.65 -0.35]
 [-0.2  -0.2   0.3 ]
 [ 0.15 -0.35  0.15]]

B = np.array([[1,-4,-1,-4],
              [2,0,5,-4],
              [-1,1,-2,3],
              [-1,4,-1,6]],float)
Leverrier_Faddeev(B)

p(x) = [ 1. -5.  9. -7.  2.]
determinante = 2.0
inversa =
[[  1.   -1.   -4.    2. ]
 [ -0.5  -4.  -11.    2.5]
 [ -0.    3.    8.   -2. ]
 [  0.5   3.    8.   -1.5]]

#revisando
print('p(x) =', np.poly(B))
print('determinante =', np.linalg.det(B))
print('inversa =')
print(np.linalg.inv(B))

p(x) = [ 1. -5.  9. -7.  2.]
determinante = 2.0
inversa =
[[  1.   -1.   -4.    2. ]
 [ -0.5  -4.  -11.    2.5]
 [ -0.    3.    8.   -2. ]
 [  0.5   3.    8.   -1.5]]

C = np.array([[0.01,0,0,0,0],
              [0,0.01,0,0,0],
              [0,0,0.99,0,0],
              [0,0,0,100,0],
              [0,0,0,0,10000]],float)
Leverrier_Faddeev(C)

p(x) = [  1.00000000e+00 -1.01010100e+04  1.01020102e+06  -1.01020099e+06  1.99010181e+04  1.05845976e+01]
determinante = -10.5845976024
inversa =
[[ -9.35323058e+02  -0.00000000e+00  -0.00000000e+00  -0.00000000e+00  -0.00000000e+00]
 [ -0.00000000e+00  -9.35323058e+02  -0.00000000e+00  -0.00000000e+00  -0.00000000e+00]
 [ -0.00000000e+00  -0.00000000e+00  -9.44941026e+00  -0.00000000e+00  -0.00000000e+00]
 [ -0.00000000e+00  -0.00000000e+00  -0.00000000e+00  -9.52868576e-02  -0.00000000e+00]
 [ -0.00000000e+00  -0.00000000e+00  -0.00000000e+00  -0.00000000e+00   4.25900631e-03]]

#revisando
print('p(x) =', np.poly(C))
print('determinante =', np.linalg.det(C))
print('inversa =')
print(np.linalg.inv(C))

p(x) = [  1.00000000e+00  -1.01010100e+04  1.01020102e+06  -1.01020099e+06  1.99009999e+04  -9.90000000e+01]
determinante = 99.0
inversa =
[[  1.00000000e+02  0.00000000e+00  0.00000000e+00  0.00000000e+00  0.00000000e+00]
 [  0.00000000e+00  1.00000000e+02  0.00000000e+00  0.00000000e+00  0.00000000e+00]
 [  0.00000000e+00  0.00000000e+00  1.01010101e+00  0.00000000e+00  0.00000000e+00]
 [  0.00000000e+00  0.00000000e+00  0.00000000e+00  1.00000000e-02  0.00000000e+00]
 [  0.00000000e+00  0.00000000e+00  0.00000000e+00  0.00000000e+00  1.00000000e-04]]