# Linear algebra tools and demonstrations¶

By Max Kapur. Updated Aug. 23, 2020.

In :
import numpy as np
import matplotlib.pyplot as plt
from matplotlib import cm
import re
# %pip install nfft
from nfft import nfft
from itertools import product, permutations
from scipy.linalg import expm
from scipy.optimize import minimize


## Matrix constructors¶

Handy for homework.

In :
# Yields the n by n Hilbert matrix
def Hilbert(n):
out = np.empty((n,n))
for i, j in product(range(n),repeat=2):
out[i,j] =  1 / (i + j + 1)
return out

g= Hilbert(5)
print(g)
print(np.linalg.cond(g))

[[1.         0.5        0.33333333 0.25       0.2       ]
[0.5        0.33333333 0.25       0.2        0.16666667]
[0.33333333 0.25       0.2        0.16666667 0.14285714]
[0.25       0.2        0.16666667 0.14285714 0.125     ]
[0.2        0.16666667 0.14285714 0.125      0.11111111]]
476607.25024100044

In :
# Checkerboard of 1 and -1
def checkerboard(n):
out = np.empty((n,n))
for i, j in product(range(n),repeat=2):
out[i,j] =  (-1)**(i+j)
return out

print(checkerboard(4))
print((1 - checkerboard(4)) / 2)

[[ 1. -1.  1. -1.]
[-1.  1. -1.  1.]
[ 1. -1.  1. -1.]
[-1.  1. -1.  1.]]
[[0. 1. 0. 1.]
[1. 0. 1. 0.]
[0. 1. 0. 1.]
[1. 0. 1. 0.]]

In :
# A tridiagonal matrix with the given pattern
def tridiag(n,pattern=(-1,2,-1)):
a, b, c = pattern
return a*np.eye(n,k=-1) + b*np.eye(n) + c*np.eye(n, k=1)

print(tridiag(6,pattern=(-1, 4, 2)))

[[ 4.  2.  0.  0.  0.  0.]
[-1.  4.  2.  0.  0.  0.]
[ 0. -1.  4.  2.  0.  0.]
[ 0.  0. -1.  4.  2.  0.]
[ 0.  0.  0. -1.  4.  2.]
[ 0.  0.  0.  0. -1.  4.]]

In :
# Yields all square matrices of a given n whose entries are 0 or 1
return np.array(list(product((0,1),repeat=n**2))).reshape(2**(n**2),n,n)


[[[0 0 0]
[0 1 0]
[1 1 1]]

[[0 0 1]
[0 1 0]
[1 1 1]]

[[0 0 1]
[1 1 1]
[0 0 0]]]

In :
# Yields all permutation matrices of a given n
def perm_mats(n):
perms = list(permutations(range(n)))
out=np.zeros((len(perms), n, n))
for j in range(len(perms)):
for i in range(n):
out[j,i,perms[j][i]]=1
return out

print(perm_mats(4)[[0,8,22]])

[[[1. 0. 0. 0.]
[0. 1. 0. 0.]
[0. 0. 1. 0.]
[0. 0. 0. 1.]]

[[0. 1. 0. 0.]
[0. 0. 1. 0.]
[1. 0. 0. 0.]
[0. 0. 0. 1.]]

[[0. 0. 0. 1.]
[0. 0. 1. 0.]
[1. 0. 0. 0.]
[0. 1. 0. 0.]]]


## Matrix transformations¶

Only one entry so far. cofactor(A) yields the cofactor matrix of $A$ as defined in Strang, Linear Algebra and Its Applications 4e, 4.3:

$C_{ij} = (-1)^{i+j}\det M_{ij}$

Where $M_{ij}$ is the $(n-1)\times(n-1)$ submatrix that results from eliminating the $i$ row and $j$ column of $A$.

Then taking the dot product of any single row of $A$ with any single row of $C$ yields $\det{A}$.

In :
def cofactor(A):
n = len(A)
out = np.zeros((n,n))
for i in range(n):
for j in range(n):
M = A
M = np.delete(M, i, axis=0)
M = np.delete(M, j, axis=1)
out[i,j]=np.linalg.det(M)*(-1)**(i+j)
return out

g = tridiag(4,pattern=(-1,4,3))
print(g)
print(cofactor(g))
# Use the dot product of any two rows to get the determinant
print(g @ cofactor(g))
# Numpy's evaluation of the same
print(np.linalg.det(g))

[[ 4.  3.  0.  0.]
[-1.  4.  3.  0.]
[ 0. -1.  4.  3.]
[ 0.  0. -1.  4.]]
[[ 88.  19.   4.   1.]
[-57.  76.  16.   4.]
[ 36. -48.  76.  19.]
[-27.  36. -57.  88.]]
409.0
409.00000000000017


## Demonstrations¶

### Visualizing the convergence of discrete and continuous Markov processes¶

Reference: Gilbert Strang, Linear Algebra and Its Applications 4e, sections 5.3 and 5.4.

arr or $A$ gives a random discrete Markov matrix indicating the relationship of next year’s population to this one. It satisfies the defining property of a discrete Markov matrix:

$\sum_{i=1}^n A_{ij} = 1$ for all $j$

That is, the sum of each column is one.

arr_cont or $A'$ gives the continuously-compounded form of the same, which satisfies a similar equation:

$\sum_{i=1}^n A'_{ij} = 0$ for all $j$

arr_cont equals simply $A - I$. This is easy to see by analogy with $(1+r)^{t}$ and $e^{rt}$, respectively the discrete and continuous forms of the one-dimensional exponential growth equation. In matrix form, the equations become:

• Discrete: $\qquad u_{t} = A^t u_0$
• Continuous: $\quad u(t) = \exp(tA')u_0$

Both matrices have the same eigenvectors (and the eigenvalues are simply reduced by one in arr_cont). The graph below suggests that the continuous form approaches its eigenvectors slightly faster.

In :
n = 6
np.random.seed(1987)

# A random n-by-n Markov matrix
arr = np.random.random_sample((n,n))
arr = arr / arr.sum(axis=0)

# and its continuous form (just subtract I)
arr_cont = arr - np.eye(n)

# n random population start values
u0 = np.random.randint(51, size=n)

print(arr)
print(arr.sum(axis=0))

[[0.11181654 0.02909357 0.2179892  0.29565424 0.17371954 0.0716327 ]
[0.11727295 0.1519251  0.15913957 0.06874233 0.15073924 0.52364584]
[0.13912152 0.30975534 0.14804066 0.14492082 0.19171334 0.0938279 ]
[0.4303738  0.286293   0.19879452 0.16772347 0.05583325 0.10939849]
[0.14950011 0.08155211 0.05743246 0.20054737 0.28280312 0.16920595]
[0.05191508 0.14138088 0.2186036  0.12241177 0.1451915  0.03228911]]
[1. 1. 1. 1. 1. 1.]


arr has its largest eigenvalue $\lambda_1 = 1$. Thus, we expect the matrix to converge on the corresponding eigenvector $\eta_1$. For our matrix $\eta_1$ is complex, so there will be oscillatory behavior as $t \to \infty$, but in this case the oscillations are gentle enough that they do not show on the plot.

In :
# Eigenvalues of arr
print(np.linalg.eigvals(arr))
# First eigenvector of arr
print(np.linalg.eig(arr))

[ 1.        +0.j          0.17975222+0.07357139j  0.17975222-0.07357139j
-0.1769291 +0.1599462j  -0.1769291 -0.1599462j  -0.11104826+0.j        ]
[-0.38536198+0.j          0.22279832+0.27868169j  0.22279832-0.27868169j
-0.16825134+0.43297401j -0.16825134-0.43297401j -0.47484858+0.j        ]

In :
t = 6

# Discrete results from time 0 to t
results_discrete = np.stack([np.linalg.matrix_power(arr, i) @ u0 for i in range(t+1)])
x_discrete = range(t+1)

# Continuous results from time 0 to t
results_cont = np.stack([expm(t*arr_cont) @ u0 for t in np.linspace(0,t,101)])
x_cont = np.linspace(0,t,101)

# Plot
plt.figure(figsize=(8,10))
plt.ylabel('population')
plt.xlabel('time')
plt.title('Comparison of discrete and continuous Markov processes')

plt.plot(x_discrete, results_discrete, color='yellowgreen')
plt.plot(x_cont, results_cont, color='seagreen')

# Legend
plt.plot([],[], color='yellowgreen', label='discrete')
plt.plot([],[], color='seagreen', label='continuous')
plt.legend()

Out:
<matplotlib.legend.Legend at 0x26f70c8d7b8> ### Image compression using SVD¶

References:

Public-domain image from the Biodiversity Heritage Library.

Singular value decomposition (SVD) factors a matrix into three components:

$$A = U S V^H$$

The middle $S$ is a diagonal matrix containing the square roots of the eigenvalues of $A^H A$ (or $A A^H$). Because $S$ is sorted from greatest $\sigma$ to least, we can zero out the tail end of $S$ as well as the corresponding columns of $U$ and $V$ to produce a lightweight set of matrices whose products approximate $A$.

Images are matrices, so this method is the basis of many widely used image-compression algorithms. Here, I give a practical demonstration of how SVD works and the level of quality one can expect when compressing a typical image.

First, we read in the image file. It’s a JPEG, so technically it’s already been compressed (you can even see some compression artifacts if you squint), but we won’t sweat it.

I cropped the image a bit at this stage to show its detail. The area we are working with is 1200 by 1300 pixels; these will be the eventual dimensions of $S$ in SVD. The array has an additional dimension of 3, corresponding to the image’s red, green, and blue channels.

In :
img = imread('linalg-img2.jpg')[500:1700,0:1300,:]
img.shape

Out:
(1200, 1300, 3)
In :
plt.figure(figsize=(12,13))
plt.axis('off')
plt.imshow(img)

Out:
<matplotlib.image.AxesImage at 0x26f70d3b550>