Linear Equation

What is a linear equation?

A linear equation in the set of variables $𝒳$ , where without loss of generality $𝒳 = {x_{1}, x_{2}, \dots, x_{n}}$ is an expression

a_{1} x_{1} + a_{2} x_{2} + \dots + a_{n} x_{n} = b

The value $b_{i}$ is the constant of the linear equation and belongs to the field $𝕂$ that contains the constants.

The linear equation

a_{1} x_{1} + a_{2} x_{2} + \dots + a_{n} x_{n} = b

will sometimes be described using the summation symbol $\sum$ as

\sum_{i = 1}^{n} a_{i} x_{i} = b

For a set of variable $𝒳 = {x_{1}, x_{2}, x_{3}, x_{4}}$ we have

\begin{matrix} linear equations & non-linear equations \\ x_{1} + 4 x_{2} - 8 x_{3} - 7 x_{4} = -8 & \sqrt{x_{1}} + 4 x_{2} - 7 x_{3} = 3 \\ 4 x_{1} - 8 x_{3} - 7 x_{4} = 5 & x_{1} x_{2} + 8 x_{3} - 7 x_{4} = 8 \end{matrix}

Homogeneous Linear Equation

The following is an important class of equations

A linear equation

a_{1} x_{1} + a_{2} x_{2} + \dots + a_{n} x_{n} = b

is called homogeneous if $b = 0$ .

\begin{matrix} non-homogeneous & homogeneous \\ x_{1} + 4 x_{2} - 8 x_{3} - 7 x_{4} = -8 & x_{1} + 4 x_{2} - 8 x_{3} - 7 x_{4} = 0 \\ 4 x_{1} - 8 x_{3} - 7 x_{4} = 5 & 4 x_{1} - 8 x_{3} - 7 x_{4} = 0 \end{matrix}

Solution set of a linear equation

An $n$ -tuple $(s_{1}, s_{2}, \dots, s_{n}) \in 𝕂^{n}$ is a solution (order is important) to the linear equation

a_{1} x_{1} + a_{2} x_{2} + \dots + a_{n} x_{n} = b

if and only if

a_{1} s_{1} + a_{2} s_{2} + \dots + a_{n} s_{n} = b

Observe that the second equation is concerned with numbers only; there are no variables.

$(1, 0, -3)$ is solutions to $4 x_{1} + 2 x_{2} + x_{3} = 1$ , but $(3, 1, 0)$ is not a solution.

$(1, 0, -3)$ is solutions to $4 x_{1} + 2 x_{2} + x_{3} = 1$ , but is not solution to $4 x_{1} + 2 x_{2} + x_{3} + 0 x_{4} = 1$

The tuples $(-2, 5, 0)$ and $(0, 4, -1)$ are solution to $x_{1} + x_{2} + x_{3} = 3$ , the tuple $(1, 5, 0)$ is not a solution. The set of all solutions is $x_{1} = s_{1}, x_{2} = s_{2}$ and $x_{3} = 3 - s_{1} - s_{2}$ where $s_{1}, s_{2} \in 𝕂$ that is

𝒮 = {(s_{1}, s_{2}, 3 - s_{1} - s_{2}) | s_{1}, s_{2} \in 𝕂}

Note that $𝒮 \subseteq 𝕂^{3}$

The solution set of $0 x_{1} + 2 x_{2} + x_{3} = -3$ is

𝒮 = {(s_{1}, \frac{-3 - s_{3}}{2}, s_{3}) | s_{1}, s_{3} \in 𝕂}

The solution can also be described as

𝒮 = {(t_{1}, t_{2}, -3 - 2 t_{2}) | t_{1}, t_{2} \in 𝕂}

Typically in case there is a non-zero coefficient the solution set is described using the smallest index $i$ for which $a_{i}$ is non-zero. For the above example it means the solution set is described with the former rather than the latter description.

The solution set of $0 x_{1} + 0 x_{2} = 0$ is

𝒮 = {(s_{1}, s_{2}) | s_{1}, s_{2} \in 𝕂} = 𝕂^{2}

The solution set of $0 x_{1} + 0 x_{2} + 0 x_{3} + 0 x_{4} = 4$ is empty, meaning it has no solution or is inconsistent.

𝒮 = \emptyset \subset 𝕂^{4}

The solution set of $2 x_{1} = 6$ contains a single element (singleton):

𝒮 = {3} \subset 𝕂^{1}

Consider the linear equation

a_{1} x_{1} + a_{2} x_{2} + \dots + a_{n} x_{n} = b

If at least one $a_{i}$ is non zero

𝒮 = {(s_{1}, \dots, s_{i - 1}, \frac{b - (a_{1} s_{1} + \dots + a_{i - 1} s_{i - 1} + a_{i + 1} s_{i + 1} + \dots + a_{n} s_{n})}{a_{i}}, s_{i + 1}, \dots, s_{n}) | \forall i, s_{i} \in 𝕂}

All $a_{i}$ 's are zero

if $b = 0$ then $𝒮 = 𝕂^{n}$
if $b \neq 0$ then $𝒮 = \emptyset^{n}$

Proof:

Suppose one coefficient is non zero. Without loss of generality $a_{i} \neq 0$ .

We will show that every solution satisfies the constraints in the description of the set $𝒮$ and every such tuple is a solution. Let $(s_{1} \dots s_{n})$ be a solution to the linear equation, by definition

a_{1} s_{1} + a_{2} s_{2} + \dots + a_{i - 1} s_{i - 1} + a_{i} s_{i} + a_{i + 1} s_{i + 1} + \dots + a_{n} s_{n} = b

rearranging

s_{i} = \frac{b - (a_{1} s_{1} + \dots + a_{i - 1} s_{i - 1} + a_{i + 1} s_{i + 1} + \dots + a_{n} s_{n})}{a_{i}}

Thus $(s_{1} \dots s_{n})$ satisfies the constraint described in the set $𝒮$ .

Take a tuple

(s_{1}, \dots, s_{i - 1}, \frac{b - (a_{1} s_{1} + \dots + a_{i - 1} s_{i - 1} + a_{i + 1} s_{i + 1} + \dots + a_{n} s_{n})}{a_{i}}, s_{i + 1}, \dots, s_{n})

and consider

\begin{matrix} a_{1} s_{1} + a_{2} s_{2} + \dots + a_{i - 1} s_{i - 1} \\ + a_{i} \frac{b - (a_{1} s_{1} + \dots + a_{i - 1} s_{i - 1} + a_{i + 1} s_{i + 1} + \dots + a_{n} s_{n})}{a_{i}} \\ + a_{i + 1} s_{i + 1} + \dots + a_{n} s_{n} \\ = a_{1} s_{1} + \dots + a_{i - 1} s_{i - 1} \\ + b - (a_{1} s_{1} + \dots + a_{i - 1} s_{i - 1} + a_{i + 1} s_{i + 1} + \dots + a_{n} s_{n}) \\ + a_{i + 1} s_{i + 1} + \dots + a_{n} s_{n} \\ = b \end{matrix}

Thus every such tuple is a solution to the given linear equation concluding this part of the argument.

Suppose now all coefficients are zero that is for all $i$ we have $a_{i} = 0$ . Take any tuple $(s_{1} \dots s_{n})$ . We have

a_{1} s_{1} + a_{2} s_{2} + \dots + a_{n} s_{n} = 0 s_{1} + 0 s_{2} + \dots + 0 s_{n} = 0

Thus if $b = 0$ $n$ -tuple is a solution. If $b \neq 0$ the equation has no solution, concluding the argument.