NumLinAlg Problem 2-8

# Numeric linear Algebra, Project 2, Problem 8
# Nicholas Lamicela
import numpy as np

# Input
A = np.matrix(((1, 2, 4),(1, 3, 3),(0, 0, 1)))

# Apparently, numpy returns U, the diagonal of s, and V transpose,
# so that A = U * diag(s) * Vt.
U, s, Vt = np.linalg.svd(A)

# Rank 1 and 2 approximation:
# This line truncates s to r elements then pads the matrix to nxn with 0s.
n = 3
r = 1
S1 = np.pad(np.diag(s[0:r]), (0,n-r), mode='constant')
r = 2
S2 = np.pad(np.diag(s[0:r]), (0,n-r), mode='constant')

# Output
print("Original Matrix:")
print(A)

print("Rank-1 Approximation:")
print(U*S1*Vt)

print("Rank-2 Approximation:")
print(U*S2*Vt)

==================== Output ====================
Original Matrix:
[[1 2 4]
 [1 3 3]
 [0 0 1]]
Rank-1 Approximation:
[[ 1.01071168  2.51252097  3.64361002]
 [ 0.95524374  2.37463362  3.44364841]
 [ 0.17868048  0.44418052  0.64414212]]
Rank-2 Approximation:
[[ 0.9420343   2.013099    4.00704661]
 [ 1.03999729  2.99096147  2.99513772]
 [ 0.11405527 -0.02577403  0.98613485]]