## Permutations & Combinations

I have now been blogging for 16 years, and my very first post (long gone) was on combinations and permutations. So, it’s fun to come back to the idea now. In 2004, my experience with the two concepts was limited to how textbooks often used the awkward “care about order” (permutations) or “don’t care about order” (combinations) language to introduce the ideas. So, that’s what I wrote about then. Now I want to talk about how the two concepts are related.

What They Are

When you count permutations, you count how many different ways you can sequentially arrange some things. When you count combinations, you count how many ways you can have some things. So, given 2 cards, there are 2 different ways you can sequentially arrange 2 cards, but given 2 cards, there’s just one way to have 2 cards.

Right off the bat, the language is weird, and it’s hard to see why combinations should ever be a thing (there’s always just 1 way to have a set of things). But combinations make better sense when you are not choosing from all the elements you are given.

So, for example, how many permutations and combinations can I make of 2 cards, chosen from a total of 3 cards?

Now having the two categories of permutation and combination makes a little more sense. There are 6 permutations of 2 cards chosen from 3 cards and there are 3 combinations of 2 cards chosen from 3 cards. That is, there are 6 different ways to sequentially arrange 2 cards chosen from 3 and just 3 different ways to have 2 cards chosen from 3. And you can see, by the way, that the combinations are a subset of the permutations.

In fact, let’s do an example with 4 cards to show the actual relationship between permutations and combinations. Here we’ll just use letters to save space. The permutations of JQKA if we choose 3 cards are:

JQK, JKQ, QJK, QKJ, KJQ, KQJ
JQA, JAQ, QJA, QAJ, AJQ, AQJ
QKA, QAK, KQA, KAQ, AQK, AKQ
KAJ, KJA, JKA, JAK, AJK, AKJ

That’s 24 permutations. For combinations, we get JQK, JQA, QKA, KAJ. That’s 4 combinations. What’s the relationship? We’ll come back on that next time.

## Rotating Coordinate Systems

I‘ve just started with Six Not-So-Easy Pieces, based on Feynman’s famous lectures, and already there’s some decently juicy stuff. In the beginning, Feynman discusses the symmetry of physical laws—that is, the invariance of physical laws under certain transformations (like rotations):

If we build a piece of equipment in some place and watch it operate, and nearby we build the same kind of apparatus but put it up on an angle, will it operate in the same way?

He goes on to explain that, of course, a grandfather clock will not operate in the same way under specific rotations. Assuming the invariance of physical laws under rotations, this change in operation tells us something interesting: that the operation of the clock is dependent on something outside of the “system” that is the clock itself.

The theorem is then false in the case of the pendulum clock, unless we include the earth, which is pulling on the pendulum. Therefore we can make a prediction about pendulum clocks if we believe in the symmetry of physical law for rotation: something else is involved in the operation of a pendulum clock besides the machinery of the clock, something outside it that we should look for.

Rotation Coordinates

Feynman then proceeds with a brief mathematical analysis of forces under rotations. A somewhat confusing prelude to this is a presentation that involves expressing the coordinates of a rotated system in terms of the original system. He uses the diagram below to derive those coordinates (except for the blue highlighting, which I use to show what (x’, y’) looks like in Moe’s system). What we want is to express (x’, y’) in terms of x and y—to describe Moe’s point P in terms of Joe’s point P.

“We first drop perpendiculars from P to all four axes and draw AB perpendicular to PQ.”

The first confusion that is not dealt with (because Feynman makes the assumption that his audience is advanced students) is what angles in the diagram are congruent to θ shown. And here again we see the value of the easy-to-forget art of eyeballing and common sense in geometric reasoning.

The y’ axis is displaced just as much as the x’ axis by rotation, and “displaced just as much by rotation” is a perfectly good definition of angle congruence that we tend to forget after hundreds of hours of deriving work. The same reasoning applies to the rotational displacement from AP to AB. If we imagine rotating AP to AB, we see that we are starting perpendicular to the x-axis and ending perpendicular to the x’-axis. The y-to-y’ rotation does the same thing, so the displacement angle must be the same. So let’s put in those new thetas, only one of which we’ll need.

Inspection of the figure shows that x’ can be written as the sum of two lengths along the x’-axis, and y’ as the difference of two lengths along AB.

Here is x’ as the sum of two lengths (red and orange): $\mathtt{x’=OA\cdot\color{red}{\frac{OC}{OA}}+AP\cdot\color{orange}{\frac{BP}{AP}}\quad\rightarrow\quad x\cdot\color{red}{cos\,θ}+y\cdot\color{orange}{sin\,θ}}$

And here is y’ as the difference of two lengths (green – purple): $\mathtt{y’=AP\cdot\color{green}{\frac{AB}{AP}}-OA\cdot\color{purple}{\frac{AC}{OA}}\quad\rightarrow\quad y\cdot\color{green}{cos\,θ}-x\cdot\color{purple}{sin\,θ}}$

So, if Joe describes the location of point P to Moe, and the rotational displacement between Moe and Joe’s systems is known (and it is known that the two systems share an origin), Moe can use the manipulations above to determine the location of point P in his system.

Another, exactly equal, way of saying this—the way we said it when we talked about rotation matrices—is that, if we represent point P in Joe’s system as a position vector (x, y), then Moe’s point P vector is $\small{\begin{bmatrix}\mathtt{x’}\\\mathtt{y’}\end{bmatrix}=\begin{bmatrix}\mathtt{\,\,\,\,\,cos\,θ} & \mathtt{sin\,θ}\\\mathtt{-sin\,θ} & \mathtt{cos\,θ}\end{bmatrix}\begin{bmatrix}\mathtt{x}\\\mathtt{y}\end{bmatrix}=\mathtt{x}\begin{bmatrix}\mathtt{\,\,\,\,\,cos\,θ}\\\mathtt{-sin\,θ}\end{bmatrix}+\mathtt{y}\begin{bmatrix}\mathtt{sin\,θ}\\\mathtt{cos\,θ}\end{bmatrix}=\begin{bmatrix}\mathtt{\,\,\,\,\,\,x\cdot \color{red}{cos\,θ}\,\,\,+y\cdot \color{orange}{sin\,θ}}\\\mathtt{-(x\cdot \color{purple}{sin\,θ})+y\cdot \color{green}{cos\,θ}}\end{bmatrix}}$

The first rotation matrix above actually describes a clockwise rotation, which is both different from what we discussed at the link above (our final matrix there was for counterclockwise rotations) and unexpected, since we know that Moe’s system is a counterclockwise rotation of Joe’s system.

The resolution to that unexpectedness can again be found after a little eyeballing. The position vector for point P in Joe’s system is at a steep angle, whereas in Moe’s system, it is at a shallow angle. Only a clockwise rotation will change the coordinates in the appropriate way.

## Projections

A projection is like the shadow of a vector, say $$\mathtt{u}$$, on another vector, say $$\mathtt{v}$$, if light rays were coming in across $$\mathtt{u}$$ and perpendicular to $$\mathtt{v}$$. For the vectors at the right, imagine a light source above and to the left of the illustration, perpendicular to the vector $$\mathtt{v}$$. The projection, which we’ll call $$\mathtt{p}$$, will be a vector that will extend from the point shown to where the end of the shadow of $$\mathtt{u}$$ touches $$\mathtt{v}$$.

If you take a moment maybe to read that description twice (because it’s kind of dense), you may be able to tell that the vector $$\mathtt{p}$$ that we’re looking for will be some scalar multiple of vector $$\mathtt{v}$$, since it will lie exactly on top of $$\mathtt{v}$$. In fact, given the picture, the projection vector $$\mathtt{p}$$ will have the opposite sign as $$\mathtt{v}$$ and will have a scale factor pretty close to 0, since the projection vector looks like it will be much smaller than $$\mathtt{v}$$.

That information is shown at the right. Helpfully, the vector $$\mathtt{u-p}$$ is perpendicular to $$\mathtt{v}$$, so we know that $$\mathtt{\left(u-p\right)\cdot v=0}$$. And, using the Distributive Property, we get $$\mathtt{u\cdot v-p\cdot v=0}$$.

Since we know that our sought-after vector $$\mathtt{p}$$ will be some scalar multiple of $$\mathtt{v}$$, we can substitute, say, $$\mathtt{cv}$$ for $$\mathtt{p}$$ in the above to get $$\mathtt{u\cdot v-cv\cdot v=0}$$. And a property of dot product multiplication allows us to rewrite that as $$\mathtt{u\cdot v-c\left(v\cdot v\right)=0}$$.

This means that $$\mathtt{u\cdot v=c\left(v\cdot v\right)}$$, which means that the scale factor $$\mathtt{c}$$ that we’re after—the factor we can multiply by $$\mathtt{v}$$ to produce $$\mathtt{p}$$—is $\mathtt{c=\frac{u\cdot v}{v\cdot v}\quad\rightarrow\quad p=\left(\frac{u\cdot v}{v\cdot v}\right)v}$

## The Law of Total Probability

Where has this been all my life? The Law of Total Probability is really cool, and it seems accessible enough to be presented in high school, where it would be very useful as well, I think, although I’ve never seen it there. For example, from the book Causal Inference in Statistics we get this nice problem (in addition to the quote below): “Suppose we roll two dice, and we want to know the probability that the second roll (R2) is higher than the first (R1).” The Law of Total Probability can make answering this much more straightforward.

To understand this ‘law,’ we should start by understanding two simple things about probability. First, for any two mutually exclusive events (the events can’t happen together), the probability of $$\mathtt{A}$$ or $$\mathtt{B}$$ is the sum of the probability of $$\mathtt{A}$$ and the probability of $$\mathtt{B}$$:

$$\mathtt{P(A\text{ or }B)\,\,\,\,=\,\,\,\,\,\,P(A)\,\,\,\,+\,\,\,\,P(B)}$$

Second, thinking now of two mutually exclusive events “A and B” and “A and not-B”, we can write the probability of $$\mathtt{A}$$ this way, since if $$\mathtt{A}$$ is true, then either “A and B” or “A and not-B” must be true:

$$\mathtt{P(A)=P(A,B)\,\,+\,\,P(A,\text{not-}B)}$$

In different situations, however, $$\mathtt{B}$$ could take on many different values—for example, the six possible values of one die roll, $$\mathtt{B_1}$$–$$\mathtt{B_6}$$—even while we’re considering just one value for the mututally exclusive event $$\mathtt{A}$$—for example, rolling a 4. The Law of Total Probability tells us that

$$\mathtt{P(A)=P(A,B_1)+\cdots+P(A,B_n)}$$.

If we pull a random card from a standard deck, the probability that the card is a Jack [$$\mathtt{P(J)}$$] will be equal to the probability that it’s a Jack and a spade [$$\mathtt{P(J,C_S)}$$], plus the probability that it’s a Jack and a heart [$$\mathtt{P(J,C_H)}$$], plus the probability that it’s a Jack and a club [$$\mathtt{P(J,C_C)}$$], plus the probability that it’s a Jack and a diamond [$$\mathtt{P(J,C_D)}$$].

Now with Conditional Probabilities

Where this gets good is when we throw conditional probabilities into the mix. We can make use of the fact that $$\mathtt{P(A,B)=P(A|B)P(B)}$$, where $$\mathtt{P(A|B)}$$ means “the probability of A given B.” For example, the probability of randomly pulling a Jack, given that you pulled spades, is $$\mathtt{\frac{1}{13}}$$, and the probability of randomly pulling a spade is $$\mathtt{\frac{1}{4}}$$. Thus, the probability of pulling the Jack of spades is $$\mathtt{\frac{1}{13}\cdot \frac{1}{4}=\frac{1}{52}}$$. We can, therefore, rewrite the Law of Total Probability this way:

$$\mathtt{P(A)=P(A|B_1)P(B_1)+\cdots+P(A|B_n)P(B_n)}$$

And now we’re ready to determine the probability given in the opening paragraph, $$\mathtt{P(R2>R1)}$$, the probability that a second die roll is greater than the first die roll: $\mathtt{P(R2>R1)=P(R2>R1|R1=1)P(R1=1)+\cdots P(R2>R1|R1=6)P(R1=6)}$

The final result is $$\mathtt{\frac{5}{6}\cdot \frac{1}{6}+\frac{4}{6}\cdot \frac{1}{6}+\frac{3}{6}\cdot \frac{1}{6}+\frac{2}{6}\cdot \frac{1}{6}+\frac{1}{6}\cdot \frac{1}{6}+\frac{0}{6}\cdot \frac{1}{6}=\frac{5}{12}}$$.

## Monty Hall and the Colliders

Reading through Judea Pearl’s Book of Why, along with Causal Inference in Statistics, co-authored by Pearl, I came across a nifty, new-to-me explanation of the famous Monty Hall Problem.

To make it clearer, we can start with a toy mathematical problem, $$\mathtt{a+b=c}$$, which I model with the causal diagram below. A causal diagram is, of course, a little inappropriate for modeling a mathematical equation, but it’s also one that students may first use implicitly to think about operations and equations (and, without proper instruction, it’s one adults may use all their lives).

Here, we will think of the variables $$\mathtt{a}$$ and $$\mathtt{b}$$ as independent. Changing the number we substitute for $$\mathtt{a}$$ does not affect our choice for $$\mathtt{b}$$ and vice versa. However, $$\mathtt{a}$$ and $$\mathtt{c}$$ are dependent, and $$\mathtt{b}$$ and $$\mathtt{c}$$ are dependent as well. Increasing or decreasing $$\mathtt{a}$$ or $$\mathtt{b}$$ alone (or together) will have an effect on $$\mathtt{c}$$.

But once we fix $$\mathtt{c}$$, or “condition on $$\mathtt{c}$$” as Pearl would write, then $$\mathtt{a}$$ and $$\mathtt{b}$$ become dependent variables. That’s at once clear as a bell and pretty wacky. If we fix $$\mathtt{c}$$ at 10, then changing $$\mathtt{a}$$ will change $$\mathtt{b}$$ and vice versa. But $$\mathtt{a}$$ and $$\mathtt{b}$$ were entirely independent prior to knowing what $$\mathtt{c}$$ was. Afterwards, they’re dependent on each other. The diagram that represents this situation (above) is what Pearl calls a “collider.”

Pearl et. al also use a more everyday example:

Suppose a certain college gives scholarships to two types of students: those with unusual musical talents and those with extraordinary grade point averages. Ordinarily, musical talent and scholastic achievement are independent traits, so, in the population at large, finding a person with musical talent tells us nothing about that person’s grades. However, discovering that a person is on a scholarship changes things; knowing that the person lacks musical talent then tells us immediately that he is likely to have high grade point average. Thus, two variables that are marginally independent become dependent upon learning the value of a third variable (scholarship) that is a common effect of the first two.

For the Monty Hall problem, the collider model looks identical. And the correct model helps us see two things (forgiving some [I hope minor] mathematical sloppiness).

First, the model helps us see that Monty opening the door does not change the probability that you have made the correct initial choice (which is $$\mathtt{\frac{1}{3}}$$). It’s even accurate to say that the probability that the car is behind any of the three doors hasn’t changed from $$\mathtt{\frac{1}{3}}$$. The arrows are pointing the wrong way for those things to be possibilities. In and of itself, the correct model should prevent us from upgrading the probability from $$\mathtt{\frac{1}{3}}$$ to $$\mathtt{\frac{1}{2}}$$ after the freebie goat door is opened.

Second, when Monty opens a door to reveal a goat (thus fixing the value of “door Monty opens”), now changing your choice of door changes the probability that the car is behind that door, since these two variables are now dependent.

Thus, since the probability of being correct must change when I change my door selection, it must change from $$\mathtt{\frac{1}{3}}$$ to something else. And since the other door is the only other option, and all the probabilities in the situation must add to $$\mathtt{1}$$, switching must change the probability to $$\mathtt{\frac{2}{3}}$$.

## K-Means Clustering

K-means clustering is one way of taking some data and allowing a computer to do what you do pretty naturally with your eyes and brain—separate the data into distinguishable clusters. For example, in the graph shown below, you can very easily see two clumps of points (points A and D in a clump and points B and C in a clump). A computer, to the extent it sees anything, sees just four points and their coordinates.

Why not just use our eyes and brain? Because once we teach a computer to approximate our ability to cluster 2D or 3D data, it can cluster data with many more than just 2 or 3 components. And then its “seeing” outpaces ours by quite a lot.

Let’s take a look at the instructions a computer could follow to do k-means clustering, and then we’ll dress it all up in linear algebra symbolism some other time. To start, I’ve just made 2 clusters of points (which we know about, but the computer doesn’t) with 2 components each (an x-component and y-component, i.e., 2D data).

Determining Least Distances

To start, we select a $$\mathtt{k}$$, a number of clusters that we want, remembering that we know right now how many clusters there are but in most situations we would not. We choose $$\mathtt{k=2}$$. Then we place the two cluster points at random locations—here I’ve put $$\mathtt{\color{blue}{k_1}}$$ at $$\mathtt{(2,7)}$$ and $$\mathtt{\color{red}{k_2}}$$ at $$\mathtt{(4,2)}$$.

Next, we calculate the distance from each point to each center. This is the good ol’ Pythagorean Theoremish Euclidean distance of $$\mathtt{\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}}$$. The cluster that we assign to each point is given by the closest center to that point. You can run the code below to print out the 8 distances.

Going by shorter distances, our first result, then, is to group point A in cluster $$\mathtt{\color{blue}{k_1}}$$, because the first number in that pair is smaller, and points B, C, and D in cluster $$\mathtt{\color{red}{k_2}}$$, because the second number in each of those pairs is the smaller one.

Moving the Centers Based on the Means

The next step—and last before repeating the process—is to move each center to the mean of the points in the cluster to which it is currently assigned. The mean of the points is determined by calculating the mean of the components separately. So, for our current cluster $$\mathtt{\color{red}{k_2}}$$, the points B, C, and D have a mean of ($$\mathtt{\frac{2 + 2 + 6}{3}}$$, $$\mathtt{\frac{4 + 3 + 5}{3}}$$), or ($$\mathtt{\frac{10}{3}}$$, $$\mathtt{4}$$). And since $$\mathtt{\color{blue}{k_1}}$$ has just one point, A, it will move to smack dab on top of that point, at (5, 6).

Now we can do another round of distance comparisons, given the new center locations. These calculations give us what we can see automatically—that points A and D belong to one cluster and points B and C belong to another cluster. In this case, A and D belong to cluster $$\mathtt{\color{blue}{k_1}}$$ and B and C belong to cluster $$\mathtt{\color{red}{k_2}}$$.

The cluster centers now move to the means of each pair of points, placing them where we would likely place them to begin with (directly between the two points in the cluster). Further calculations won’t change these assignments, so the k-means algorithm is done when it stops changing things drastically (or at all).

## Linear Algebra Exercises I

We’ve done a fair amount of landscaping, as it were—running out a long way in the field of linear algebra to mark interesting points. Now it seems like a good time to turn back and start tending to each area a little more closely. An excellent way to do that—to make things more secure—is to practice.

The sets below can be completed after reading Lines the Linear Algebra Way. I would suggest completing this by working on a piece of paper and then checking your answers—one at a time at first and then go a stretch before checking. Try to complete each exercise before either checking back with the original post or looking at my answer. My answers can be uncovered by hovering under each exercise.

Convert each equation of a line to a vector equation in parametric form, $$\mathtt{l(k)=p+kv}$$, where $$\mathtt{p}$$ and $$\mathtt{v}$$ are 2D vectors and $$\mathtt{k}$$ is a scalar variable.

1. $$\mathtt{y=x+3}$$

$$\mathtt{l(k)=}\begin{bmatrix}\mathtt{0}\\\mathtt{3}\end{bmatrix}+\begin{bmatrix}\mathtt{1}\\\mathtt{1}\end{bmatrix}\mathtt{k}$$

1. $$\mathtt{y=2x+3}$$

$$\mathtt{l(k)=}\begin{bmatrix}\mathtt{0}\\\mathtt{3}\end{bmatrix}+\begin{bmatrix}\mathtt{1}\\\mathtt{2}\end{bmatrix}\mathtt{k}$$

1. $$\mathtt{y=-2x}$$

$$\mathtt{l(k)=}\begin{bmatrix}\mathtt{0}\\\mathtt{0}\end{bmatrix}+\begin{bmatrix}\mathtt{\,\,\,\,1}\\\mathtt{-2}\end{bmatrix}\mathtt{k}$$,    or just $$\begin{bmatrix}\mathtt{\,\,\,\,1}\\\mathtt{-2}\end{bmatrix}\mathtt{k}$$

1. $$\mathtt{y=x}$$

$$\mathtt{l(k)=}\begin{bmatrix}\mathtt{0}\\\mathtt{0}\end{bmatrix}+\begin{bmatrix}\mathtt{1}\\\mathtt{1}\end{bmatrix}\mathtt{k}$$,    or just $$\begin{bmatrix}\mathtt{1}\\\mathtt{1}\end{bmatrix}\mathtt{k}$$

1. $$\mathtt{y=10}$$

$$\mathtt{l(k)=}\begin{bmatrix}\mathtt{0}\\\mathtt{10}\end{bmatrix}+\begin{bmatrix}\mathtt{1}\\\mathtt{0}\end{bmatrix}\mathtt{k}$$

1. $$\mathtt{x+y=1}$$

$$\mathtt{l(k)=}\begin{bmatrix}\mathtt{0}\\\mathtt{1}\end{bmatrix}+\begin{bmatrix}\mathtt{\,\,\,\,1}\\\mathtt{-1}\end{bmatrix}\mathtt{k}$$

1. $$\mathtt{2x+3y=9}$$

$$\mathtt{l(k)=}\begin{bmatrix}\mathtt{0}\\\mathtt{3}\end{bmatrix}+\begin{bmatrix}\mathtt{\,\,\,\,3}\\\mathtt{-2}\end{bmatrix}\mathtt{k}$$

1. $$\mathtt{x=5}$$

$$\mathtt{l(k)=}\begin{bmatrix}\mathtt{5}\\\mathtt{0}\end{bmatrix}+\begin{bmatrix}\mathtt{0}\\\mathtt{1}\end{bmatrix}\mathtt{k}$$

1. $$\mathtt{-3x-\frac{1}{4}y=15}$$

$$\mathtt{l(k)=}\begin{bmatrix}\mathtt{\,\,\,\,0}\\\mathtt{-60}\end{bmatrix}+\begin{bmatrix}\mathtt{\,\,\,\,1}\\\mathtt{-12}\end{bmatrix}\mathtt{k}$$

1. Devise an algorithm you could use to convert an equation for a line in slope-intercept form to a vector equation for the line in parametric form.

For slope-intercept form, $$\mathtt{y=mx+b}$$, the conversion $$\begin{bmatrix}\mathtt{0}\\\mathtt{b}\end{bmatrix}+\begin{bmatrix}\mathtt{1}\\\mathtt{m}\end{bmatrix}\mathtt{k}\,$$ seems to work for most lines.

There are many ways to write the above vector equations. In particular, the intercept vector does not have to have a first component of $$\mathtt{0}$$, and the slope vector does not have to be in simplest form. All that is required is for the intercept vector to get you to some point on the line and for the slope vector to correctly represent the slope of the line.

Now let’s go the other way. Identify the slope, y-intercept, and x-intercept of each line.

1. $$\mathtt{l_{1}(k)=}\begin{bmatrix}\mathtt{1}\\\mathtt{2}\end{bmatrix}+\begin{bmatrix}\mathtt{\,\,\,\,2}\\\mathtt{-1}\end{bmatrix}\mathtt{k}$$

slope → $$\mathtt{-\frac{1}{2}}$$, y-intercept → $$\mathtt{\frac{5}{2}}$$, x-intercept → $$\mathtt{5}$$

1. $$\mathtt{l_{2}(k)=}\begin{bmatrix}\mathtt{-3}\\\mathtt{-5}\end{bmatrix}+\begin{bmatrix}\mathtt{0}\\\mathtt{4}\end{bmatrix}\mathtt{k}$$

slope → undefined, y-intercept → none, x-intercept → $$\mathtt{-3}$$

1. $$\mathtt{l_{3}(k)=}\begin{bmatrix}\mathtt{\,\,\,\,0}\\\mathtt{-1}\end{bmatrix}+\begin{bmatrix}\mathtt{2}\\\mathtt{2}\end{bmatrix}\mathtt{k}$$

slope → $$\mathtt{1}$$, y-intercept → $$\mathtt{-1}$$, x-intercept → $$\mathtt{1}$$

1. $$\mathtt{l_{4}(k)=}\begin{bmatrix}\mathtt{\,\,\,\,3}\\\mathtt{-9}\end{bmatrix}+\begin{bmatrix}\mathtt{7}\\\mathtt{0}\end{bmatrix}\mathtt{k}$$

slope → $$\mathtt{0}$$, y-intercept → $$\mathtt{-9}$$, x-intercept → none

1. $$\mathtt{l_{5}(k)=}\begin{bmatrix}\mathtt{-5}\\\mathtt{\,\,\,\,1}\end{bmatrix}+\begin{bmatrix}\mathtt{-3}\\\mathtt{\,\,\,\,4}\end{bmatrix}\mathtt{k}$$

slope → $$\mathtt{-\frac{4}{3}}$$, y-intercept → $$\mathtt{-\frac{17}{3}}$$, x-intercept → $$\mathtt{-\frac{17}{4}}$$

And just two more. Determine if each point is on the line mentioned from above.

1. Is $$\mathtt{(-11,3)}$$ on the line $$\mathtt{l_1}$$?

No: $$\begin{bmatrix}\mathtt{1}\\\mathtt{2}\end{bmatrix}+\begin{bmatrix}\mathtt{\,\,\,\,2}\\\mathtt{-1}\end{bmatrix}\mathtt{k=}\begin{bmatrix}\mathtt{-11}\\\mathtt{\,\,\,\,3}\end{bmatrix}$$.
No value of $$\mathtt{k}$$ solves both $$\mathtt{2k+1=-11}$$ and $$\mathtt{-k+2=3}$$.

1. Is $$\mathtt{(5,5)}$$ on the line $$\mathtt{l_3}$$?

No. Since the line has a slope of $$\mathtt{1}$$, it would contain $$\mathtt{(5,5)}$$ if the line passed through the origin, but it does not.

## More with Least Squares

One very cool thing about our formula for the least squares regression line, $$\mathtt{\left(X^{T}X\right)^{-1}X^{T}y}$$, is that it is the same no matter whether we have one independent variable (univariate) or many independent variables (multivariate).

Consider these data, showing the selling prices of some grandfather clocks at auction. The first scatter plot shows the age of the clock in years on the x-axis (100–200), and the second shows the number of bidders on the x-axis (0–20). Price (in pounds or dollars) is on the y-axis on each plot (500–2500).

Age (years)BiddersPrice (\$)
127131235
115121080
1277845
15091522
15661047
182111979
156121822
132101253
13791297
1139946
137151713
117111024
13781147
15361092
117131152
126101336
170142131
18281550
162111884
184102041
1436854
15991483
108141055
17581545
1086729
17991792
111151175
18781593
1117785
1157744
19451356
16871262

You can see in the notebook below that the first regression line, for the price of a clock as a function of its age, is approximately $$\mathtt{10.5x-192}$$. The second regression line, for the price of a clock as a function of the number of bidders at auction, is approximately $$\mathtt{55x+806}$$. As mentioned above, each of these univariate least squares regression lines can be calculated with the formula $$\mathtt{\left(X^{T}X\right)^{-1}X^{T}y}$$.

Combining both age and number of bidders together, we can calculate, using the same formula, a multivariate least squares regression equation. This of course is no longer a line. In the case of two input variables as we have here, our line becomes a plane.

Our final regression equation becomes $$\mathtt{12.74x_{1}+85.82x_{2}-1336.72}$$, with $$\mathtt{x_1}$$ representing the age of a clock and $$\mathtt{x_2}$$ representing the number of bidders.

## Linear Algebra Connections

There are so many connections within and applications of linear algebra—I can only imagine that this will be a series of posts, to the extent that I continue writing about the subject. Here are a few connections that I’ve come across in my reading recently.

Compound Interest and Matrix Powers

We can multiply a matrix by itself $$\mathtt{n}$$ times. The result is the matrix to the power $$\mathtt{n}$$. We can use this when setting up a compound interest situation. For example, suppose we have three accounts, which each have a different interest rate compounded annually—say, 5%, 3%, and 2%. Without linear algebra, the amount in the first account can be modeled by the equation $\mathtt{A(t)=p \cdot 1.05^t}$ where $$\mathtt{p}$$ represents the starting amount in the account, and $$\mathtt{t}$$ represents the time in years. With linear algebra, we can group all of the account interest rates into a matrix. The first year for each account would look like this: $\mathtt{A(1)}=\begin{bmatrix}\mathtt{1.05}&\mathtt{0}&\mathtt{0}\\\mathtt{0}&\mathtt{1.03}&\mathtt{0}\\\mathtt{0}&\mathtt{0}&\mathtt{1.02}\end{bmatrix}^\mathtt{1}\begin{bmatrix}\mathtt{p_1}\\\mathtt{p_2}\\\mathtt{p_3}\end{bmatrix}=\begin{bmatrix}\mathtt{1.05p_1}\\\mathtt{1.03p_2}\\\mathtt{1.02p_3}\end{bmatrix}$

For years beyond the first year, all we have to do is raise the matrix to the appropriate power. Since it’s diagonal, squaring it, cubing it, etc., will square, cube, etc., each entry. This computation can be a little more organized—and more straightforward for programming. A matrix has to be square ($$\mathtt{m \times m})$$ in order to raise it to a power. Below we calculate the amount in each account after 100 years.

Centroids and Areas

We have seen that the determinant can be thought about as the area of the parallelogram formed by two vectors. We can use this fact to determine the area of a complex shape like the one shown below.

Since determinants are signed areas, as we move around the shape counterclockwise, calculating the determinant of each vector pair (and multiplying each determinant by one half so we just get each triangle), we get the total area of the shape.$\frac{1}{2}\left(\begin{vmatrix}\mathtt{6}&\mathtt{6}\\\mathtt{0}&\mathtt{4}\end{vmatrix}+\begin{vmatrix}\mathtt{6}&\mathtt{3}\\\mathtt{4}&\mathtt{4}\end{vmatrix}+\begin{vmatrix}\mathtt{3}&\mathtt{3}\\\mathtt{4}&\mathtt{6}\end{vmatrix}+\begin{vmatrix}\mathtt{3}&\mathtt{-2}\\\mathtt{6}&\mathtt{\,\,\,\,6}\end{vmatrix}+\begin{vmatrix}\mathtt{-2}&\mathtt{-2}\\\mathtt{\,\,\,\,6}&\mathtt{\,\,\,\,3}\end{vmatrix}+\begin{vmatrix}\mathtt{-2}&\mathtt{0}\\\mathtt{\,\,\,\,3}&\mathtt{3}\end{vmatrix}\right)=\mathtt{36}\text{ units}\mathtt{^2}$

That’s pretty hand-wavy, but it’s something that you can probably figure out with a little experimentation.

Another counterclockwise-moving calculation (though this one can be clockwise without changing the answer) is the calculation of the centroid of a closed shape. All that is required here is to calculate the sum of the position vectors of the vertices of the figure and then divide by the number of vertices.

$\mathtt{\frac{1}{5}}\left(\begin{bmatrix}\mathtt{3}\\\mathtt{4}\end{bmatrix}+\begin{bmatrix}\mathtt{0}\\\mathtt{6}\end{bmatrix}+\begin{bmatrix}\mathtt{-3}\\\mathtt{\,\,\,\,4}\end{bmatrix}+\begin{bmatrix}\mathtt{-1}\\\mathtt{\,\,\,\,3}\end{bmatrix}\right)=\begin{bmatrix}\mathtt{-\frac{1}{5}}\\\mathtt{\,\,\,\,\frac{17}{5}}\end{bmatrix}$

Here again, it’s just magic, but you can figure it out with a little play. In each case—for both complex areas and centroids—we assign a point to be the origin and go from there.

## Least Squares with Linear Algebra

In brief, linear regression is about finding the line of best fit to a data set. If you’re looking for the linear algebra way of doing this, you will most likely find it searching for the term least squares.

In the basic scenario, you’ve got some two-dimensional data, $$\mathtt{(x, y)}$$ coordinates, and you want to find the equation for a straight line that is as close as possible to each point. Such a scenario is shown below, though here the line has already been graphed and the equation for the line of best fit displayed. (But feel free to change the data in the table or move the points around to see how the line of best fit changes.)

xy
108.04
86.95
137.58
98.81
118.33
149.96
67.24
44.26
1210.84
74.82
55.68

So, let’s imagine that we haven’t found this line yet. We know what we are looking for is a line of the form $$\mathtt{y=mx+b}$$. In linear algebra terms, we want the vector $$\mathtt{y}$$ (11 rows, 1 column) to equal the vector $$\mathtt{x}$$ (11 rows, 1 column) times a slope vector $$\mathtt{m}$$ (1 row, 2 columns) plus an intercept vector $$\mathtt{b}$$ (11 rows, 1 column).

The first problem with this that we have to fix is that $$\mathtt{x}$$ the 11 × 1 vector needs to be $$\mathtt{X}$$ the 11 × 2 matrix so that we get our equations right. So we’ll pad $$\mathtt{x}$$ with some 1s and then we’ll be able to call it $$\mathtt{X}$$. (We can pad on either side, left or right, just remembering to interchange $$\mathtt{b}$$ and $$\mathtt{m}$$ to keep the equations straight.) The second problem we can fix is that we don’t need a separate intercept vector and a separate slope vector. We can combine things so that we form an equivalent matrix equation that means the same thing as $$\mathtt{y=mx+b}$$. We need the equation to look like this: $\begin{bmatrix}\mathtt{8.04}\\\mathtt{6.95}\\\mathtt{7.58}\\\mathtt{8.81}\\\mathtt{8.33}\\\mathtt{9.96}\\\mathtt{7.24}\\\mathtt{4.26}\\\mathtt{10.84}\\\mathtt{4.82}\\\mathtt{5.68}\end{bmatrix} = \begin{bmatrix}\mathtt{1}&\mathtt{10}\\\mathtt{1}&\mathtt{8}\\\mathtt{1}&\mathtt{13}\\\mathtt{1}&\mathtt{9}\\\mathtt{1}&\mathtt{11}\\\mathtt{1}&\mathtt{14}\\\mathtt{1}&\mathtt{6}\\\mathtt{1}&\mathtt{4}\\\mathtt{1}&\mathtt{12}\\\mathtt{1}&\mathtt{7}\\\mathtt{1}&\mathtt{5}\end{bmatrix}\begin{bmatrix}\mathtt{b}\\\mathtt{m}\end{bmatrix}$

Multiply the matrix and vector on the right side of that equation, and you get, for the first equation, $$\mathtt{8.04=b+(m)(10)}$$. That’s equivalent to $$\mathtt{y=mx+b}$$ for the first $$\mathtt{(x, y)}$$ data point $$\mathtt{(10, 8.04)}$$. The matrix and vector setup ensures that all the equations for all the points are of the correct form.

In linear algebra terms, we have rewritten the equation to be $$\mathtt{y=Xv}$$, where $$\mathtt{y}$$ is an 11 × 1 vector, $$\mathtt{X}$$ is an 11 × 2 matrix (padded with some 1s), and $$\mathtt{v}$$ is a 2 × 1 vector which contains the unknown slope $$\mathtt{m}$$ of the best fit line and the unknown intercept $$\mathtt{b}$$.

It’s an Approximation

At this point in the explanation, it’s important to realize that we will make another shift. The first was from the real data to the matrix algebra setup. We shift again below, away from that setup, per se, and toward just finding out what that unknown $$\mathtt{v}$$ is.

As an analogy, the system shown below features a 3d vector (2, 0, –2) which does not live in the same plane as the one formed by the two other column vectors of the matrix. Thus, there is no solution $$\mathtt{(j, k)}$$. $\begin{bmatrix}\mathtt{1}&\mathtt{\,\,\,\,1}\\\mathtt{1}&\mathtt{-3}\\\mathtt{1}&\mathtt{\,\,\,\,1}\end{bmatrix}\begin{bmatrix}\mathtt{j}\\\mathtt{k}\end{bmatrix}=\begin{bmatrix}\mathtt{\,\,\,\,2}\\\mathtt{\,\,\,\,0}\\\mathtt{-2}\end{bmatrix}\longleftarrow\text{no solutions}$

Similarly, the columns of our $$\mathtt{X}$$ matrix form a plane in 11-dimensional space. In order for $$\mathtt{v}$$ to be a solution to our original matrix equation above, $$\mathtt{y}$$ has to live in this plane too. But we already know that it doesn’t. If it did, the points would lie along some line. It’s true that $$\mathtt{y}$$ is an 11-dimensional vector, just like each column of $$\mathtt{X}$$, but $$\mathtt{y}$$ doesn’t live in the same plane.

The closest approximation we can get to $$\mathtt{y}$$ in the plane of $$\mathtt{X}$$ is $$\mathtt{Xv=p}$$, where $$\mathtt{p}$$ is the projection of $$\mathtt{y}$$ onto the plane (another thing we’ll have to come back to). The vector $$\mathtt{q}$$ connecting $$\mathtt{y}$$ with its projection $$\mathtt{p}$$ is perpendicular to $$\mathtt{p}$$ (and, thus, to the plane).

Now we can write some more equations. For example, we know we want $$\mathtt{Xv=p}$$, but it’s also true that $$\mathtt{p+q=y}$$. And we can do some manipulation to show that all the column vectors of $$\mathtt{X}$$ are perpendicular to $$\mathtt{q}$$. Assuming for a second that our $$\mathtt{X}$$ is a 2 × 2 matrix, we write $$\mathtt{X}$$ as a 3D matrix, transpose it (so that multiplication is defined), and multiply it by $$\mathtt{q}$$. The same basic idea applies to our original $$\mathtt{X}$$ (11 × 2) matrix. $\begin{bmatrix}\mathtt{x_{11}}&\mathtt{x_{12}}\\\mathtt{x_{21}}&\mathtt{x_{22}}\end{bmatrix}\rightarrow\begin{bmatrix}\mathtt{x_{11}}&\mathtt{x_{12}}\\\mathtt{x_{21}}&\mathtt{x_{22}}\\\mathtt{0}&\mathtt{0}\end{bmatrix}\rightarrow\begin{bmatrix}\mathtt{x_{11}}&\mathtt{x_{21}}&\mathtt{0}\\\mathtt{x_{12}}&\mathtt{x_{22}}&\mathtt{0}\end{bmatrix}$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,\,\begin{bmatrix}\mathtt{x_{11}}&\mathtt{x_{21}}&\mathtt{0}\\\mathtt{x_{12}}&\mathtt{x_{22}}&\mathtt{0}\end{bmatrix}\begin{bmatrix}\mathtt{0}\\\mathtt{0}\\\mathtt{q_3}\end{bmatrix}=\begin{bmatrix}\mathtt{0}\\\mathtt{0}\end{bmatrix}$ This gives us a key equation: $$\mathtt{X^{T}q=0}$$.

Since $$\mathtt{q=y-p}$$, substituting for $$\mathtt{q}$$ gets us $$\mathtt{X^{T}(y-p)=0}$$. Then since $$\mathtt{Xv=p}$$, substituting for $$\mathtt{p}$$ brings us to $$\mathtt{X^{T}(y-Xv)=0}$$. Distribute to get $$\mathtt{X^{T}y-X^{T}Xv=0}$$, which means $$\mathtt{X^{T}y=X^{T}Xv}$$. Multiply both sides on the left by the inverse of $$\mathtt{X^{T}X}$$, and the vector $$\mathtt{v}$$ that we’re after, then, is $\mathtt{v=(X^{T}X)^{-1}X^{T}y}$

That formula gives us the slope and intercept of our best fit line. Below is one way the least squares can be calculated with a little Python. The calculation I used for the above interactive to get the best fit line is way more complicated. If I had known about the linear algebra way when I made it, I would have definitely gone with that instead.