Topics#

  • Matrix multiplication
  • Linear transformation
  • Transposition

Notes#

Markov Process#

Let’s say we want to model how a certain species of bacteria can transition between different states of health. The bacteria can take on 3 states: 1) healthy, 2) sick, 3) dead. The following matrix represents the probability of transitioning the population of bacteria from one state to another state in 5 days.

\[ \left[\begin{array}{ccc|c} \text{from healthy} & \text{from sick} & \text{from dead} & \\ \hline 0.80 & 0.10 & 0.0 & \text{to healthy} \\ 0.15 & 0.60 & 0.0 & \text{to sick} \\ 0.05 & 0.30 & 1.0 & \text{to dead} \end{array}\right] \]\[ \begin{bmatrix} 0.80 & 0.10 & 0.0 \\ 0.15 & 0.60 & 0.0 \\ 0.05 & 0.30 & 1.0 \end{bmatrix} \]

This type of matrix is known as a stocastic matrix where all of the entries represent probabilities and the values in each column sum to 1. This is an example of a Markov process which is a process where the probability of entering one state depends only on the last state.

Let’s say we start with 1 billion healthy bacteria. What is the break down of bacteria expected to be healthy, sick and dead in 5 days?

\[ \begin{bmatrix} 0.80 & 0.10 & 0.0 \\ 0.15 & 0.60 & 0.0 \\ 0.05 & 0.30 & 1.0 \end{bmatrix} \begin{bmatrix} 1\mathrm{e}9 \\ 0 \\ 0 \end{bmatrix} \]\[\begin{aligned} (0,0) &= 0.80 \cdot 1\mathrm{e}9 + 0 + 0 = 8\mathrm{e}8 \\ (1,0) &= 0.15 \cdot 1\mathrm{e}9 + 0 + 0 = 15\mathrm{e}7 \\ (2,0) &= 0.05 \cdot 1\mathrm{e}9 + 0 + 0 = 5\mathrm{e}7 \end{aligned}\]
  • 800,000,000 are expected to be healthy.
  • 150,000,000 are expected to sick.
  • 50,000,000 are expected to be dead.

Exercises#

Problem 1#

Let

\[ \mathbf{A} = \begin{bmatrix} 4 & -2 & 3 \\ -2 & 1 & 6 \\ 1 & 2 & 2 \end{bmatrix}, \mathbf{B} = \begin{bmatrix} 1 & -3 & 0 \\ -3 & 1 & 0 \\ 0 & 0 & -2 \end{bmatrix} \]

Solve for

\[ \mathbf{A}\mathbf{B}, \mathbf{A}\mathbf{B}^{\top}, \mathbf{B}\mathbf{A}, \mathbf{B}^{\top}\mathbf{A} \]

\(\mathbf{A}\mathbf{B}\)#

\[\begin{aligned} (0,0) &= \begin{bmatrix} 4 & -2 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ -3 \\ 0 \end{bmatrix} = 4 \cdot 1 + -2 \cdot -3 + 3 \cdot 0 = 4 + 6 + 0 = 10 \\ (0,1) &= \begin{bmatrix} 4 & -2 & 3 \end{bmatrix} \begin{bmatrix} -3 \\ 1 \\ 0 \end{bmatrix} = 4 \cdot -3 + -2 \cdot 1 + 3 \cdot 0 = -12 + -2 + 0 = -14 \\ (0,2) &= \begin{bmatrix} 4 & -2 & 3 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ -2 \end{bmatrix} = -6 \\ (1,0) &= \begin{bmatrix} 2 & 1 & 6 \end{bmatrix} \begin{bmatrix} 1 \\ -3 \\ 0 \end{bmatrix} = 2 \cdot 1 + 1 \cdot -3 + 6 \cdot 0 = 2 + -3 + 0 = -1 \\ (1,1) &= \begin{bmatrix} 2 & 1 & 6 \end{bmatrix} \begin{bmatrix} -3 \\ 1 \\ 0 \end{bmatrix} = 2 \cdot -3 + 1 \cdot 1 + 6 \cdot 0 = -6 + 1 + 0 = -5 \\ (1,2) &= \begin{bmatrix} 2 & 1 & 6 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ -2 \end{bmatrix} = -6 \\ (2,0) &= \begin{bmatrix} 1 & 2 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ -3 \\ 0 \end{bmatrix} = 1 \cdot -3 + 2 \cdot 1 + 2 \cdot 0 = -3 + 2 + 0 = -1 \\ (2,1) &= \begin{bmatrix} 1 & 2 & 2 \end{bmatrix} \begin{bmatrix} -3 \\ 1 \\ 0 \end{bmatrix} = 1 \cdot -3 + 2 \cdot 1 + 2 \cdot 0 = -3 + 2 + 0 = -1 \\ (2,2) &= \begin{bmatrix} 1 & 2 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ -2 \end{bmatrix} = -4 \\ \end{aligned}\]\[ \mathbf{A}\mathbf{B} = \begin{bmatrix} 10 & -14 & -6 \\ -1 & -5 & -6 \\ -1 & -1 & -4 \end{bmatrix} \]

\(\mathbf{A}\mathbf{B}^{\top}\)#

Because \(\mathbf{B}\) is symmetrical across the diagnol \(\mathbf{B}=\mathbf{B}^{\top}\) so

\[ \mathbf{A}\mathbf{B} = \mathbf{A}\mathbf{B}^{\top} = \begin{bmatrix} 10 & -14 & -6 \\ -1 & -5 & -6 \\ -1 & -1 & -4 \end{bmatrix} \]

\(\mathbf{B}\mathbf{A}\), \(\mathbf{B}^{\top}\mathbf{A}\)#

I’m skipping this calculation out of time but we can see that \(\mathbf{B}\mathbf{A}=\mathbf{B}^{\top}\mathbf{A}\) because \(\mathbf{B}=\mathbf{B}^{\top}\).

Problem 2#

\[ \mathbf{A} = \begin{bmatrix} 4 & -2 & 3 \\ -2 & 1 & 6 \\ 1 & 2 & 2 \end{bmatrix}, \mathbf{a} = \begin{bmatrix} 1 & -2 & 0 \end{bmatrix}, \mathbf{b} = \begin{bmatrix} 3 \\ 1 \\ -1 \end{bmatrix} \]\[ \mathbf{A}\mathbf{a}, \mathbf{A}\mathbf{a}^{\top}, (\mathbf{A}\mathbf{a})^{\top}, \mathbf{b}^{\top}\mathbf{A}^{\top} \]

\(\mathbf{A}\mathbf{a}\)#

\[ \mathbf{A}\mathbf{a} = \begin{bmatrix} 4 & -2 & 3 \\ -2 & 1 & 6 \\ 1 & 2 & 2 \end{bmatrix} \begin{bmatrix} 1 & -2 & 0 \end{bmatrix}, \]

\(\mathbf{A}\mathbf{a}\) is undefined because the number of columns in \(\mathbf{A}\) don’t match the number of rows in \(\mathbf{a}\).

\(\mathbf{A}\mathbf{a}^{\top}\)#

\[ \mathbf{A}\mathbf{a}^{\top} = \begin{bmatrix} 4 & -2 & 3 \\ -2 & 1 & 6 \\ 1 & 2 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix}, \]\[\begin{aligned} (0,0) &= \begin{bmatrix} 4 & -2 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix} = 4 \cdot 1 + -2 \cdot -2 + 3 \cdot 0 = 4 + -4 + 0 = 0 \\ (1,0) &= \begin{bmatrix} -2 & 1 & 6 \end{bmatrix} \begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix} = -2 \cdot 1 + 1 \cdot -2 + 6 \cdot 0 = -2 + -2 + 0 = -4 \\ (2,0) &= \begin{bmatrix} 1 & 2 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix} = 1 \cdot 1 + 2 \cdot -2 + 2 \cdot 0 = 1 + -4 + 0 = -3 \end{aligned}\]\[ \mathbf{A}\mathbf{a}^{\top} = \begin{bmatrix} 0 \\ -4 \\ 3 \end{bmatrix} \]

\((\mathbf{A}\mathbf{a})^{\top}\)#

\((\mathbf{A}\mathbf{a})^{\top}\) is undefined because the number of columns in \(\mathbf{A}\) don’t match the number of rows in \(\mathbf{a}\).

\(\mathbf{b}^{\top}\mathbf{A}^{\top}\)#

\[ \mathbf{A}^{\top} = \begin{bmatrix} 4 & -2 & 1 \\ -2 & 1 & 2 \\ 3 & 6 & 2 \end{bmatrix}, \mathbf{b}^{\top} = \begin{bmatrix} 3 & 1 & -1 \end{bmatrix} \]\[ \mathbf{b}^{\top}\mathbf{A}^{\top} = \begin{bmatrix} 3 & 1 & -1 \end{bmatrix} \begin{bmatrix} 4 & -2 & 1 \\ -2 & 1 & 2 \\ 3 & 6 & 2 \end{bmatrix} \]\[\begin{aligned} (0,0) &= \begin{bmatrix} 3 & 1 & -1 \end{bmatrix} \begin{bmatrix} 4 \\ -2 \\ 3 \end{bmatrix} = 3 \cdot 4 + 1 \cdot -2 + -1 \cdot 3 = 12 + -2 + -3 = 7 \\ (0,1) &= \begin{bmatrix} 3 & 1 & -1 \end{bmatrix} \begin{bmatrix} -2 \\ 1 \\ 6 \end{bmatrix} = 3 \cdot -2 + 1 \cdot 1 + -1 \cdot 6 = -6 + 1 + -6 = -11 \\ (0,2) &= \begin{bmatrix} 3 & 1 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} = 3 \cdot 1 + 1 \cdot 2 + -1 \cdot 2 = 3 + 2 + -2 = 3 \end{aligned}\]\[ \mathbf{b}^{\top}\mathbf{A}^{\top} = \begin{bmatrix} 7 & -11 & 3 \end{bmatrix} \]

Problem 3#

Let

\[ \mathbf{A} = \begin{bmatrix} 4 & -2 & 3 \\ -2 & 1 & 6 \\ 1 & 2 & 2 \end{bmatrix}, \mathbf{B} = \begin{bmatrix} 1 & -3 & 0 \\ -3 & 1 & 0 \\ 0 & 0 & -2 \end{bmatrix} \mathbf{a} = \begin{bmatrix} 1 & -2 & 0 \end{bmatrix}, \mathbf{b} = \begin{bmatrix} 3 \\ 1 \\ -1 \end{bmatrix} \]

Solve for

\[ 1.5\mathbf{a} + 3.0\mathbf{b}, 1.5\mathbf{a}^{\top} + 3.0\mathbf{b}, (\mathbf{A}-\mathbf{B})\mathbf{b}, \mathbf{A}\mathbf{a} - \mathbf{B}\mathbf{b} \]

Skipping

Problem 4#

Write a program for a Markov process to calculate further steps for the bacteria example above.

import numpy as np
def bacteria_markov_process(starting_population, iterations):
  stochastic_matrix = np.array([
    [ 0.8, 0.1, 0.0 ],
    [ 0.15, 0.60, 0.0],
    [ 0.05, 0.30, 1.0],
  ])
  current_population = np.array([
    [starting_population],
    [0],
    [0],
  ])

  for i in range(iterations):
    current_population = stochastic_matrix @ current_population
  return current_population

print(bacteria_markov_process(1e9, 1))

Problem 5#

Two grocery stores \(S_1\) and \(S_2\) sell apples, oranges, and bananas with a profit of $3.50, $6.20, and $3.00, respectively. Let the sales in a certain week be given by the matrix

\[ \left[\begin{array}{ccc|c} \text{apples} & \text{oranges} & \text{bananas} & \\ \hline 40 & 6 & 24 & S_1 \\ 10 & 12 & 50 & S_2 \end{array}\right] \]

Find a vector that represents the total profits of \(S_1\) and \(S_2\).

\[ \begin{bmatrix} 40 & 6 & 24 \\ 10 & 12 & 50 \end{bmatrix} \begin{bmatrix} 3.5 \\ 6.2 \\ 3.0 \end{bmatrix} = \begin{bmatrix} 40 \cdot 3.5 + 6 \cdot 6.2 + 24 \cdot 3.0 \\ 10 \cdot 3.5 + 12 \cdot 6.2 \cdot 50 + 3.0 \end{bmatrix} = \begin{bmatrix} 140 + 37.2 + 72 \\ 35 + 74.4 + 150 \end{bmatrix} = \begin{bmatrix} 249.2 \\ 259.4 \end{bmatrix} \]