Linear Algebra : equations, variables and solutions

September 23, 2023
Setup
Consider a simple equation \( x = 0 \). I can write this equation as:
$$x + 0x_{2} + 0x_{3} + ... + 0x_{n} = 0 $$
Graphical representation
The simplest way to represent this would be as a point at 0 on the number line. If I were to represent this on a 2-Dimensional plane, its representation would depend on the plane I am choosing.

For example, on the x-y plane, this would be the y-axis. Similarly, on the x-z plane, this would be the z-axis. In the y-z plane, it would be the entire plane. However, if it was the case that \(x \neq 0\), it would be irrelevant to talk about y-z plane. If I were to represent this in a 3-dimensional space, it would be the y-z plane. When I say y-z plane, I mean the plane formed by the y-axis and the z-axis

I cannot visualize what it would be in higher dimensions. But so far here is how I see the system to 'develop' its representation as the dimensions (starting at 1) are incremented by 1: $$\textrm{point}$$ $$\downarrow $$ $$\textrm{line}$$ $$\downarrow $$ $$\textrm{plane}$$ $$\downarrow $$ $$\textbf{magic}$$
Rank and restrictions
My observation here is that all these representations arise because the system that I started with is a 1x3 system with 1 leading variable and thus \(n-r = 3-1 = 2\) parameters that are y and z. The existence of these parameters allows the solution/visualization to extend infinitely in y and z. On the other hand, if the system was 3x3, with a rank of 3, it would imply that all the variables that I have are independent of each other and thus, the solution in this case would just be a point in 3D space. A system of 0 equations in 3 variables is then, all of 3D space. The introduction of m equations restricts the solution. Strictly considering our actual 3D space, we cannot restrict our system to more than 3 dimensions. It makes no sense to restrict it to dimensions more than the actual dimensions we have. Simply stated: $$r \leq n \iff \textrm{ consistent system} $$ $$r > n \iff \textrm{inconsistent system}$$ I think this helps with questions about rank of the augmented matrix. You CAN have an augmented matrix (in RREF) such that the leading one is in last column, giving a rank of n + 1. However, this would also mean that r > n, which implies that the system is inconsistent and no solutions exist.
An alternate perspective
Instead of my original attempt at visualization, this time I start with a 0x3 system. The solution for this is all of space. Increase m and r by 1, we have a 1x3 system and the solution is a plane. Increase m and r by 2, we have a 2x3 system and the solution is a line. Increase m and r by 3, we have a 3x3 system and the solution is a point. We get the following reductions as m (starting at m = 0) increments by 1. $$\textrm{Space = all planes}$$ $$\textrm{m + 1} \downarrow \textrm{r + 1}$$ $$\textrm{A specific plane = all lines in that plane}$$ $$\textrm{m + 1} \downarrow \textrm{r + 1}$$ $$\textrm{A specific line = all points in that line }$$ $$\textrm{m + 1} \downarrow \textrm{r + 1}$$ $$\textrm{No solution}$$
A meme