What are some mathematically interesting computations involving matrices?

Solution 1:

If $(a,b,c)$ is a Pythagorean triple (i.e. positive integers such that $a^2+b^2=c^2$), then $$\underset{:=A}{\underbrace{\begin{pmatrix} 1 & -2 & 2\\ 2 & -1 & 2\\ 2 & -2 & 3 \end{pmatrix}}}\begin{pmatrix} a\\ b\\ c \end{pmatrix}$$ is also a Pythagorean triple. In addition, if the initial triple is primitive (i.e. $a$, $b$ and $c$ share no common divisor), then so is the result of the multiplication.

The same is true if we replace $A$ by one of the following matrices:

$$B:=\begin{pmatrix} 1 & 2 & 2\\ 2 & 1 & 2\\ 2 & 2 & 3 \end{pmatrix} \quad \text{or}\quad C:=\begin{pmatrix} -1 & 2 & 2\\ -2 & 1 & 2\\ -2 & 2 & 3 \end{pmatrix}. $$

Taking $x=(3,4,5)$ as initial triple, we can use the matrices $A$, $B$ and $C$ to construct a tree with all primitive Pythagorean triples (without repetition) as follows:

$$x\left\{\begin{matrix} Ax\left\{\begin{matrix} AAx\cdots\\ BAx\cdots\\ CAx\cdots \end{matrix}\right.\\ \\ Bx\left\{\begin{matrix} ABx\cdots\\ BBx\cdots\\ CBx\cdots \end{matrix}\right.\\ \\ Cx\left\{\begin{matrix} ACx\cdots\\ BCx\cdots\\ CCx\cdots \end{matrix}\right. \end{matrix}\right.$$

Source: Wikipedia's page Tree of primitive Pythagorean triples.

Solution 2:

(Just my two cents.) While this has not much to do with numerical computations, IMHO, a very important example is the modelling of complex numbers by $2\times2$ matrices, i.e. the identification of $\mathbb C$ with a sub-algebra of $M_2(\mathbb R)$.

Students who are first exposed to complex numbers often ask "$-1$ has two square roots. Which one is $i$ and which one is $-i$?" In some popular models of $-i$, such as $(0,-1)$ on the Argand plane or $-x+\langle x^2+1\rangle$ in $\mathbb R[x]/(x^2+1)$, a student may get a false impression that there is a natural way to identify one square root of $-1$ with $i$ and the other one with $-i$. In other words, they may wrongly believe that the choice should be somehow related to the ordering of real numbers. In the matrix model, however, it is clear that one can perfectly identify $\pmatrix{0&-1\\ 1&0}$ with $i$ or $-i$. The choices are completely symmetric and arbitrary. Neither one is more natural than the other.

Solution 3:

The roots of any polynomial $$p(x) = \sum_{i=1}^{n} c_i x^i$$ are the eigenvalues of the companion matrix

$$\begin{bmatrix} 0 & 0 & \cdots & 0 & -c_0/c_n \\ 1 & 0 & \cdots & 0 & -c_1/c_n \\ 0 & 1 & \cdots & 0 & -c_2/c_n \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & -c_{n-1}/c_n \end{bmatrix}$$

which you can have them compute via power iteration.

...at least in the case of real roots. Different methods may be needed for complex roots.

Solution 4:

Iterated Function Systems are fun to draw and are computed with matrices. They can also illustrate what matrix multiplication does.

As an example, this draws Barnsley's fern:

import numpy as np
import matplotlib.pyplot as plt
from random import randint

fig = plt.figure()

mats = [np.array([[0.8,0.03],[-0.03,0.8]])]
mats.append(np.array([[0,0.3],[0.3,0]]))
mats.append(np.array([[0,-0.3],[0.3,0]]))
mats.append(np.array([[0,-0.006],[0,0.2]]))

offsets = np.array([[0,1],[0.4,0.2],[-0.4,0.2],[0,-1]])

vec = np.array([0,0])
xlist, ylist  = [], []
for count in range(100000):
    r = randint(0,100)
    i = 0
    if r > 70: i += 1
    if r > 80: i += 1
    if r > 90: i += 1
    vec = mats[i].dot(vec) + offsets[i]
    xlist.append(vec[0])
    ylist.append(vec[1])

plt.plot(xlist, ylist, color='w', marker='o', markeredgewidth=0.1, markersize=0.3, markeredgecolor='k')
plt.show()