Tag: linear algebra

When you add two fractions, you of course remember that you should never just add the numerators and add the denominators across. The resulting fraction will not be the sum of the two addends. But if you do add across (under certain conditions which I’ll show below), the result will be a fraction between the two “addend” fractions. So, you can use the add-across method to find a fraction between two other fractions.

For example, \(\mathtt{\frac{1}{2}+\frac{4}{3}\rightarrow\frac{5}{5}}\). The first “addend” is definitely less than 1, and the second definitely greater than 1. The Farey mean here is exactly 1 (or \(\mathtt{\frac{5}{5}}\)), which is between the two “addend” fractions.

Since this site is fast becoming all linear algebra all the time, let’s throw some linear algebra at this. What we want to show is that, given \(\mathtt{\frac{a}{b}<\frac{c}{d}}\) (we'll go with this assumption for now), \[\mathtt{\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}}\]

for certain positive integer values of \(\mathtt{a,b,c,}\) and \(\mathtt{d}\). I would probably do better to make those inequality signs less-than-or-equal-tos, but let’s stick with this for the present. We’ll start by representing the fraction \(\mathtt{\frac{a}{b}}\) as the vector \(\scriptsize\begin{bmatrix}\mathtt{b}\\\mathtt{a}\end{bmatrix}\) along with the fraction \(\mathtt{\frac{c}{d}}\) as the vector \(\scriptsize\begin{bmatrix}\mathtt{d}\\\mathtt{c}\end{bmatrix}\).

We’re looking specifically at the slopes or angles here (which is why we can represent a fraction as a vector in the first place), so we’ve made \(\scriptsize\begin{bmatrix}\mathtt{d}\\\mathtt{c}\end{bmatrix}\) have a greater slope to keep in line with our assumption above that \(\mathtt{\frac{a}{b}<\frac{c}{d}}\).

The fraction \(\mathtt{\frac{a+c}{b+d}}\) is the same as the vector \(\scriptsize\begin{bmatrix}\mathtt{b+d}\\\mathtt{a+c}\end{bmatrix}\). And since this vector is the diagonal of the vector parallelogram, it will of course have a greater slope than \(\mathtt{\frac{a}{b}}\) but less than \(\mathtt{\frac{c}{d}}\). You can keep going forever—just take one of the side vectors and use the diagonal vector as the other side. So long as you’re making parallelograms, you’ll get a new diagonal shallower than the two side vectors, and the result will be a fraction between the other two.

Incidentally, our assumption at the beginning that \(\mathtt{\frac{a}{b}<\frac{c}{d}}\) doesn't really matter to this picture. If we make \(\mathtt{\frac{c}{d}}\) less than \(\mathtt{\frac{a}{b}}\), our picture simply flips. The diagonal vector still has to be located between the two side vectors.

The linear algebra picture of this concept also tells us where this method fails to find a fraction between the two addend fractions. When the two “addend” fractions are equivalent, \(\mathtt{c}\) and \(\mathtt{d}\) are multiples of \(\mathtt{a}\) and \(\mathtt{b}\), respectively, or vice versa. In that case, the resulting fraction looks like this.

The slopes or angles for both addends and for the result are the same, producing a Farey mean that is equal to both fractions.

At any rate, here are the questions—edited a bit. Feel free to rate your answers on the scales provided. Careful! Once you click, you lock in your answer.

Would you rather . . .

Finally, one last question that is not a would-you-rather. Once you’ve answered this and the rest of the questions, you can press the I'm Finished! button to submit your responses.

I’m Finished!

Check out the results so far.

Next in the lesson, I move on to asking whether you think some of the survey responses are correlated. For example, if you scored “low” on the veterinarian-or-musician scale—meaning you would strongly prefer to be a veterinarian over a musician—would that indicate that you probably also scored “low” on Question (c) about talking to animals or speaking all the languages? In other words, are those two scores correlated? What about choosing the ability to fly and your fear of heights? Are those correlated? How could we measure this using a lot of responses from a lot of different people?

An ingenious way of looking at this question is by using cosine similarity from linear algebra. (We looked at the cosine of the angle between two vectors here and here.)

For example, suppose you really would rather have the ability to fly and you have almost no fear of heights. So, you answered Question (a) with a 1 and Question (f) with, say, a 2. Another person has no desire to fly and a terrible fear of heights, so they answer Question (a) with an 8 and Question (f) with a (10). From this description, we would probably guess that the two quantities wish-for-flight and fear-of-heights are strongly correlated. But we’ve also now got the vectors (1, 2) and (8, 10) to show us this correlation.

See that tiny angle between the vectors on the left? The cosine of tiny angles (as we saw) is close to 1, which indicates a strong correlation. On the right, you see the opposite idea. One person really wants to fly but is totally afraid of heights (1, 10) and another almost couldn’t care less about flying (or at least would really rather read minds) but has a low fear of heights (8, 2). The cosine of the close-to-90°-angle between these vectors will be close to 0, indicating a weak correlation between responses to our flight and heights questions.

That’s pretty cool, but it is not, in fact, how we measure correlation. The first difficulty we encounter happens after adding more people to the survey, giving us several angles to deal with—not impossible, but pretty messy for a hundred or a thousand responses. The second, more important, difficulty is that the graph on the right above doesn’t show a weak correlation; it shows a strong negative correlation. Given just the two response pairs to work from in that graph, we would have to conclude that a strong fear of heights would make you more likely to want the ability to fly (or vice versa) rather than less likely. But the “weakest” the cosine can measure in this kind of setup is 0.

The solution to the first difficulty is to take all the x-components of the responses and make one giant vector out of them. Then do the same to the y-components. Now we’ve got just two vectors to compare! For our data on the left, the vectors (1, 2) and (8, 10) become (1, 8) and (2, 10). The vectors on the right—(1, 10) and (8, 2)—become (1, 8) and (10, 2).

The solution to the second difficulty—no negative correlations—we can achieve by centering the data. Let’s take our new vectors for the right, uncorrelated, graph: (1, 8) and (10, 2). Add the components in each vector and divide by the number of components (2) to get an average. Then subtract the average from each component. So, our new centered vectors are

(1 – ((1 + 8) ÷ 2), 8 – ((1 + 8) ÷ 2)) and (10 – ((10 + 2) ÷ 2), 2 – ((10 + 2) ÷ 2))

Or (–3.5, 3.5) and (4, –4). It’s probably not too tough to see that a vector in the 2nd quadrant and a vector in the 4th quadrant are heading in opposite directions. And these vectors now form a close-to-180° angle, and the cosine of 180° is –1 which is the actual lowest correlation we can get, indicating a strong negative correlation.

To summarize, the way to determine correlation linear-algebra style is to determine the cosine of the centered x- and y-vectors of the data. That formula is \[\mathtt{\frac{(x-\overline{x}) \cdot (y-\overline{y})}{|x-\overline{x}||y-\overline{y}|} = cos(θ)}\]

Which is just another way of writing the more common version of the r-value correlation.

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.

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.

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.

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 \[\mathtt{c=\frac{u\cdot v}{v\cdot v}\quad\rightarrow\quad p=\left(\frac{u\cdot v}{v\cdot v}\right)v}\]

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).

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.

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).

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.

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.

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

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).

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.

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.

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.

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.)

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 \(\mathtt{b}\).

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.