#!/usr/bin/env python
# coding: utf-8

# # Motivating Power Iteration
# 
# Copyright (C) 2022 Andreas Kloeckner
# 
# <details>
# <summary>MIT License</summary>
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# In[97]:


import numpy as np
import numpy.linalg as la

np.set_printoptions(linewidth=150, precision=3, suppress=True)


# Consider a random diagonal matrix $A$ with entries sorted by magnitude:

# In[98]:


n = 10
eigvals = np.array(sorted(np.random.randn(n), key=abs, reverse=True))

D = np.diag(eigvals)
D


# In[99]:


D @ D @ D


# In[100]:


tmp = D @ D @ D
tmp / tmp[0,0]


# In[101]:


tmp = D @ D @ D @ D @ D @ D @ D @ D @ D @ D @ D @ D
tmp / tmp[0,0]


# This works just as well if the matrix is not diagonalized:

# In[102]:


X = np.random.randn(n, n)
Xinv = la.inv(X)

A = X @ D @ Xinv

A


# In[103]:


A @ A @ A


# At first, it doesn't look like there is much happening, however:

# In[104]:


Apower = A @ A @ A @ A @ A @ A @ A @ A @ A @ A @ A @ A @ A


# In[105]:


tmp = Xinv @ Apower @ X
tmp / tmp[0,0]


# Let $\boldsymbol{y} = A^{13} \boldsymbol{x}$?

# In[106]:


x = np.random.randn(n)

y = Apower @ x


# Anything special about $\boldsymbol{y}$?

# In[107]:


la.norm(A @ y - D[0,0] * y, 2)/la.norm(y, 2)


# Is there a better way to compute $\boldsymbol y$?

# In[109]:


y2 = x
for i in range(13):
    y2 = A @ y2

la.norm(y2-y)


# In[ ]: