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# 30.3: El poder de una matriz

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• Para una matriz diagonalizable$$A$$, tenemos$$C^{-1}AC = D$$. Entonces tenemos$$A = CDC^{-1}$$
• Nosotros tenemos$$A^2 = CDC^{-1}CDC^{-1} = CD^{2}C^{-1}A^{n} = CDC^{-1} \ldots CDC^{-1} = CD^{n}C^{-1}$$.
• Porque las columnas de$$C$$ son vectores propios, entonces podemos decir que los vectores propios para$$A$$ y$$A^n$$ son los mismos si$$A$$ es diagonalizable.
• Si$$x$$ es un vector propio de$$A$$ con el valor propio correspondiente$$\lambda$$, entonces también$$x$$ es un vector propio de$$A^n$$ con el valor propio correspondiente$$\lambda^n$$.
# Here are some libraries you may need to use
%matplotlib inline
import numpy as np
import sympy as sym
import networkx as nx
import matplotlib.pyplot as plt
from urllib.request import urlretrieve

sym.init_printing(use_unicode=True)
'answercheck.py');

## Gráfica Caminata Aleatoria

• Defina las siguientes matrices:
• $$I$$es la matriz de identidad
• $$A$$es la matriz de adyacencia
• $$D$$es una matriz diagonal de grados (número de aristas conectadas a cada nodo)

$W=\frac{1}{2}(I + AD^{-1}) \nonumber$

• La matriz de caminata aleatoria perezosa$$W$$,, toma un vector de distribución de cosas$$p_t$$, y lo difunde a sus vecinos:

$p_{t+1}=Wp_{t} \nonumber$

• Para alguna distribución inicial de cosas$$p_0$$,, podemos calcular cuánto estaría en cada nodo a la vez,$$t$$, alimentando de la$$W$$ siguiente manera:

$p_{t}=W^{t}p_{0} \nonumber$

• Al enchufar la expresión anterior se obtiene:

$p_{t}=\left( \frac{1}{2}(I+AD^{-1}) \right)^t p_{0} \nonumber$

##### Hacer esto

Utilizando álgebra matricial, mostrar que$$\frac{1}{2}(I + AD^{-1})$$ es similar a$$I-\frac{1}{2}N$$, donde$$N=D^{-\frac{1}{2}}(D-A)D^{-\frac{1}{2}}$$ está la gráfica normalizada Laplaciana.

## Gráfico de Caminata Aleatoria sobre Barbell

Para generar el gráfico de barra, ejecute la siguiente celda.

n = 60 # number of nodes
B = nx.Graph() # initialize graph

## initialize empty edge lists
edge_list_complete_1 = []
edge_list_complete_2 = []
edge_list_path = []

## generate node lists
node_list_complete_1 = np.arange(int(n/3))
node_list_complete_2 = np.arange(int(2*n/3),n)
node_list_path = np.arange(int(n/3)-1,int(2*n/3))

## generate edge sets for barbell graph
for u in node_list_complete_1:
for v in np.arange(u+1,int(n/3)):
edge_list_complete_1.append((u,v))

for u in node_list_complete_2:
for v in np.arange(u+1,n):
edge_list_complete_2.append((u,v))

for u in node_list_path:
edge_list_path.append((u,u+1))

# G.remove_edges_from([(3,0),(5,7),(0,7),(3,5)])

## draw graph
pos=nx.spring_layout(B) # positions for all nodes

### nodes
nx.draw_networkx_nodes(B,pos,
nodelist=list(node_list_complete_1),
node_color='c',
node_size=400,
alpha=0.8)
nx.draw_networkx_nodes(B,pos,
nodelist=list(node_list_path),
node_color='g',
node_size=200,
alpha=0.8)
nx.draw_networkx_nodes(B,pos,
nodelist=list(node_list_complete_2),
node_color='b',
node_size=400,
alpha=0.8)

### edges
nx.draw_networkx_edges(B,pos,
edgelist=edge_list_complete_1,
width=2,alpha=0.5,edge_color='c')
nx.draw_networkx_edges(B,pos,
edgelist=edge_list_path,
width=3,alpha=0.5,edge_color='g')
nx.draw_networkx_edges(B,pos,
edgelist=edge_list_complete_2,
width=2,alpha=0.5,edge_color='b')

plt.axis('off')
plt.show() # display
##### Hacer esto

Generar la matriz de caminata aleatoria perezosa$$W$$,, para la gráfica anterior.

A = nx.adjacency_matrix(B)
A = A.todense()

d = np.sum(A,0) # Make a vector of the sums.
D = np.diag(np.asarray(d)[0])
#Put your answer to the above question here.
from answercheck import checkanswer
checkanswer.matrix(W, "7af4a5b11892da6e1a605c8239b62093")
##### Hacer esto

Calcular los valores propios y vectores propios de$$W$$. Hacer una matriz diagonal$$J$$ con los valores propios en la diagonal. Nombra la matriz de vectores propios$$V$$ (cada columna es un vector propio).

#Put your answer to the above question here.

Ahora nos aseguramos de que construimos$$V$$ y$$A$$ correctamente comprobando que$$W = VJV^{-1}$$

np.allclose(W, V*J*np.linalg.inv(V))
##### Hacer esto

Deja que tu$$p_0 = [1,0,0, \ldots, 0]$$. Calcular$$p_t$$ para$$t = 1,2,\ldots,100$$, y trazar$$||v_1 - p_t||_1$$ versus$$t$$, donde$$v_1$$ está el vector propio asociado con el valor propio más grande$$\lambda_1 = 1$$ y cuya suma es igual a 1. (Nota:$$||\cdot||_1$$ puede calcularse usando np.linalg.norm (v_1-p_t, 1).)

#Put your answer to the above question here.

### Comparar con Gráfica Completa

Si completa lo anterior, haga lo mismo para una gráfica completa en el mismo número de nodos.

##### Pregunta

¿Qué notas de la gráfica que es diferente de la anterior?

This page titled 30.3: El poder de una matriz is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Dirk Colbry via source content that was edited to the style and standards of the LibreTexts platform.