Linear Systems

Now that we’re familiar with linear functions and linear combinations, let’s dive into the process of using these concepts to define real-life systems as mathematical systems.

Linear modeling

We often come across data consisting of several attributes of interest to us. For example, the records of COVID-19 patients contain many attributes worth recording, including confirmation data, fever level, blood cell counts, medicine in use, and so on. One important attribute is the survival rate, which may depend on the other attributes. If we think that there are $d$ attributes in numeric form that are represented as vectors or arrays, then a typical $i^{th}$ record looks like the following: $\bold{x_i} = \begin{bmatrix} x_{1i}\\x_{2i}\\x_{3i} \\ \vdots \\x_{di}\end{bmatrix}$. The corresponding attribute, survival, can be denoted with $y_i$ having either the values, $0$ or $1$ , corresponding to “did not survive” and “survived,” respectively. If there’s a linear relationship between $\bold{x_i}$ and $y_i$, then there exists a linear function $f$ with parameters $\bold{w} = \begin{bmatrix} w_1\\w_2\\w_3 \\ \vdots \\w_d\end{bmatrix}$ that maps $\bold{x_i}$ to $y_i$ : $f_\bold{w}(\bold{x_i})=y_i, \forall i$. Although we don’t know if the linear relationship holds, we know that if it does, then: $f_\bold{w}(\bold{x_i})=w_1x_{1i}+w_2x_{2i}+...+w_dx_{di} = y_i, \forall i$.

Linear equation

An equation of the form $w_1x_{1i}+w_2x_{2i}+...+w_dx_{di} = y_i$ is called linear in $\bold{w}$ if $\bold{x_i}$ and $y_i$ are given. For example, with $d=4$, if we think of the $i^{th}$ record to be represented as $\bold{x_i} = \begin{bmatrix} -2.3\\-4.2\\9.6 \\76.2\end{bmatrix}$, and we assume that the patient survived, say, $y_i=1$, then the above equation becomes $-2.3w_1-4.2w_2+9.6w_3+76.2w_4 = 1$

Example 1: One linear equation

For what values of $w_1$ and $w_2$, $2w_1-4w_2=9$?

Solution: There are several solutions to this problem. For example, if we set $w_2=0$, then $w_1=\frac{9}{2}$. Also, if we set $w_2=-2$, then $w_1=\frac{1}{2}$. So, $(\frac{9}{2},0)$ and $(\frac{1}{2},-2)$ are two solutions. In general, if we set $w_2=\alpha$, then $w_1=\frac{9+4\alpha}{2}$. So, for every choice of $\alpha$, we have a solution $(\frac{9+4\alpha}{2},\alpha)$.

The following code prints k solutions to every equation in variables w1 and w2, provided that x1 isn’t zero.