1. Introduction and polynomials

How can an economist make use of linear algebra? I am not an economist, but I have managed to find out some answers to this question. I think I should start with those answers, because unsurprisingly students who know why they are studying the subject are better. 🙂

I have chosen two examples which show that many economic processes can be studied by means of linear algebra. The first one is so called input-output model. Assume that there are some producers, and each of them produces something. Each of them to produce needs some quantities of items produced by the others. So they depend on each other. Moreover, on the external market there is a need for their products. One can also consider this situation in a more macroeconomic set-up. For example, consider the classic branches in economic structure: agriculture, manufacturing and services. A farmer to do his job needs also some manufacture products and some services. Similarly a manufacturer and also a service provider. Let us assume that to produce agricultural products worth 100USD on average one needs to spend 20USD on other agricultural products, 30USD on manufacture products, and 10USD on services (the data is not actual, I have just made it up). Let us say that on average to make a product worth 100USD manufacturer needs to spend 10USD on agricultural products, 40USD on other manufacture products and 20USD on services. Meanwhile, providing services worth 100USD one needs to spend on average 5USD on agricultural products, 10USD on manufacture products and 30USD on other services. Moreover, the (external) market demands agricultural products, manufacture products and services worth 100 million USD, 150 million USD, and 3000 million USD respectively.

Therefore, if we denote by a, m, s the worth of agricultural products, manufacture products and services which will be produced in this setting, then the total worth of agricultural products given in millions of USD is a=0.2 a+0.1 m +0.05 s +100 (because e.g. manufacturers will buy 100000 USD worth of agricultural products for every 1 million USD worth of produced products). Similarly: m=0.3a+0.4m+0.1s+150 and s=0.1a+0.2m+0.3s+300. So we have the following set of equations:

    \[\begin{cases}a=0.2 a+0.1 m +0.05 s +100\\ m=0.3a+0.4m+0.1s+150\\ s=0.1a+0.2m+0.3s+300\end{cases}\]

This is what we call a system of linear equations (there are no squares or any other inconveniences in those equations) — and with such systems of equations we can deal using methods of linear algebra. Obviously, you may even now, without studying further, calculate the solution of this set of equations. It will be about: a=219, m=458, s=590 — those is the worth of agricultural products, manufacture products and services respectively given in millions of USD, which will be produced in our model of economy. But we can ask some further questions. How would those values change if e.g. the demand on services grows by 100 million dollars? And so on.

The second example is a problem of maximization of gains. Every company can choose some parameters of their business model. Often it can be assumed that some of those parameters depend linearly on each other. E.g. the amount of products which can be made during one day depends on the number of employees. The quality of of their work depends on their salaries. Production costs depend again on the number of employees. There are also some constraints, e.g. minimal wage or minimal volume of production given by technical parameters of the production line. The problem is how to choose those values (how many workers we should employ, what should be the volume of production, ad so on) to maximize our profits. Such problems are called problems of linear programming, and we will study them in this course of linear algebra.

But those specific application possibilities are not the only reason for studying linear algebra. This is simply a course which stimulates logical thinking and reasoning. And I suppose that all of you agree that those skills are crucial for an economist. 😉

Surprisingly, when studying linear algebra the knowledge of polynomials turns out to be useful. A polynomial is a function of the following form: a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots+a_{2}x^{2}+a_{1}x^+a_{0}, where a_{0}, a_{1}, \ldots are coefficients, i.e. given real numbers. So constant, linear or quadratic functions are examples of polynomials. E.g.: f(x)=4x-2, g(z)=-z^{2}+4z-\frac{1}{2}.

The highest exponent of the indeterminate is called its degree.

How to solve an equation f(x)=0, where f is a polynomial? If f is quadratic, then everyone knows how to solve it. Just calculate \Delta and so on.

If the polynomial is of higher degree, e.g. x^3-4x^2+x+6=0, then we first need to laboriously guess a root. In our case we guess that it is -1. Indeed, (-1)^3-4(-1)^2+(-1)-6=-1-4-1+6=0. Having any root a, one needs to know that the polynomial is dividable by (x-a) (see Bezout Theorem). The divided polynomial is of degree lower by one and has the same roots as the initial polynomial except from a.

But how to divide one polynomial by another one. One of a few possible methods resembles the usual number division:

    \[\begin{array}{l} x^2-5x+6\\ \overline{(x^3-4x^2+x+6) : (x+1)}\\ \underline{x^3+x^2}\\ \qquad -5x^2+x\\ \qquad \underline{-5x^2-5x}\\ \qquad \qquad 6x+6\\ \qquad \qquad \underline{6x+6}\\ \qquad \qquad \qquad 0 \end{array} \]

So: x^3-4x^2+x+6=(x+1)\cdot (x^2-5x+6).

And now it suffices to use \Delta-technique to solve equation x^2-5x+6=0. \Delta=25-24=1, so the roots are: 2 and 3. So the solutions of the equation x^3-4x^2+x+6=0 are: -1, 2 and 3.