Linear Equation (线性方程)

For $n \geq 1$, a linear equation in the variables $x_1, ... , x_n$ is an equation that can be written in the form

$$a_1x_1 + a_2x_2 + ... + a_nx_n = b$$

where $b$ and the coefficients $a_1, ... , a_n$ are real or complex numbers.

Examples

• $\sqrt{2}x_1 + \pi x_2 = 0$ is a linear equation.

• $x_1x_2 + x_3 = 0$ is NOT a linear equation because of the term $x_1x_2$.

• $x_1 + \sqrt{x_2} = 0$ is NOT a linear equation because of the term $\sqrt{x_2}$.

• $0x_1 = 0$ is a linear equation.

• $0x_1 = 1$ is a linear equation, although it has no solution.

Linear System (线性方程组)

Definition

A linear system is a collection of one or more linear equations involving the same variables (e.g., $x_1, ... , x_n$).

Solution of Linear Systems (线性方程组的解)

• A solution of a linear system is a (ordered) list $(s_1, s_2, ... ,s_n)$ of numbers that makes each equation a true statement when the values $s_1, s_2, ... ,s_n$ are substituted for $x_1, ... , x_n$, respectively.

• The set of all possible solutions is called the solution set of the linear system (线性方程组的解集).

• A linear system may have:

  • no solution
  • exactly one solution
  • infinitely many solutions

• A linear system that has at least one solution is called consistent (相容的); otherwise, it is called inconsistent (不相容的).

Equivalent Linear Systems (等价线性方程组)

Two linear systems are called equivalent if they have the same solution set.

Matrix Notation of Linear Systems (线性方程组的矩阵表示法)

Take the following linear system as an example:

$$\begin{cases} \begin{aligned} x_1 - 2x_2 + x_3 = 0 \\ 2x_2 - 8x_3 = 8 \\ 5x_1 - 5x_3 = 10 \end{aligned} \end{cases}$$

the matrix

$$\begin{bmatrix} 1 & -2 & 1 \\ 0 & 2 & -8 \\ 5 & 0 & -5 \end{bmatrix}$$

is called the coefficient matrix (系数矩阵), and the matrix

$$\begin{bmatrix} 1 & -2 & 1 & 0\\ 0 & 2 & -8 & 8\\ 5 & 0 & -5 & 10 \end{bmatrix}$$

is called the augmented matrix (增广矩阵).

Soving a Linear System with Augmented Matrix (用增广矩阵解线性方程组)

Take the following linear system as an example:

$$\begin{cases} \begin{aligned} x_1 &- 2x_2 &+ x_3 &= 0 \\ 3x_1 & &- 8x_3 &= 8 \\ 5x_1 &+ 2x_2 &- 5x_3 &= 10 \end{aligned} \end{cases}$$

The corresponding augmented matrix is

$$\begin{bmatrix} 1 & -2 & 1 & 0\\ 3 & 0 & -8 & 8\\ 5 & 2 & -5 & 10 \end{bmatrix}$$

We can apply a sequence of elementary row operations to transform it into its RREF.

Firstly, we can eliminate the first entry in the rows below the first row:

$$\begin{bmatrix} 1 & -2 & 1 & 0\\ 3 & 0 & -8 & 8\\ 5 & 2 & -5 & 10 \end{bmatrix} \overset{R_2 - 3R_1}{\underset{R_3 - 5R_1}{\longrightarrow}} \begin{bmatrix} 1 & -2 & 1 & 0\\ 0 & 6 & -11 & 8\\ 0 & 12 & -10 & 10 \end{bmatrix}$$

Next, we can eliminate the second entry in the third row:

$$\begin{bmatrix} 1 & -2 & 1 & 0\\ 0 & 6 & -11 & 8\\ 0 & 12 & -10 & 10 \end{bmatrix} \overset{R_3 - 2R_2}{\longrightarrow} \begin{bmatrix} 1 & -2 & 1 & 0\\ 0 & 6 & -11 & 8\\ 0 & 0 & 14 & -6 \end{bmatrix}$$

Finally, we can scale each row to make the leading entries equal to $1$, and then eliminate the entries above each leading $1$:

$$\begin{aligned} \begin{bmatrix} 1 & -2 & 1 & 0\\ 0 & 6 & -11 & 8\\ 0 & 0 & 14 & -6 \end{bmatrix} & \overset{\frac{R_2}{6}}{\underset{\frac{R_3}{14}}{\longrightarrow}} \begin{bmatrix} 1 & -2 & 1 & 0\\[3pt] 0 & 1 & -\frac{11}{6} & \frac{4}{3}\\[3pt] 0 & 0 & 1 & -\frac{3}{7} \end{bmatrix} \\ & \overset{R_2 + \frac{11}{6}R_3}{\underset{R_1 - R_3}{\longrightarrow}} \begin{bmatrix} 1 & 2 & 0 & \frac{3}{7}\\[3pt] 0 & 1 & 0 & \frac{23}{42}\\[3pt] 0 & 0 & 1 & -\frac{3}{7} \end{bmatrix} \\ & \overset{R_1 - 2R_2}{\longrightarrow} \begin{bmatrix} 1 & 0 & 0 & -\frac{14}{21}\\[3pt] 0 & 1 & 0 & \frac{23}{42}\\[3pt] 0 & 0 & 1 & -\frac{3}{7} \end{bmatrix} \end{aligned}$$

Now, according to the final RREF, we can directly read out the solution of the linear system:

$$(x_1, x_2, x_3) = (-\frac{14}{21}, \frac{23}{42}, -\frac{3}{7})$$

I'm sorry that I gave a complicated example here. There may be some calculation mistakes.

Consistence (相容性)

A linear system is consistent if and only if the rightmost column of the augmented matrix is NOT a pivot column.

Simply put, a linear system is inconsistent if and only if the augmented matrix (in RREF) has a row of the form $\begin{bmatrix}0 & 0 & \cdots & 0 & b\end{bmatrix}$, where $b$ is a nonzero number.

Basic and Free Variables (主变量和自由变量)

Take the following linear system as an example:

$$\begin{cases} \begin{aligned} x_1 & &+ x_3 &= 2 \\ x_1 &+ 2x_2 &+ x_3 &= 4 \end{aligned} \end{cases}$$

The corresponding augmented matrix is

$$\begin{bmatrix} 1 & 0 & 1 & 2\\ 1 & 2 & 1 & 4 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & 1 & 2\\ 0 & 1 & 0 & 1 \end{bmatrix}$$

The solution of the linear system can be written as

$$\begin{cases} \begin{aligned} x_1 &= 2 - x_3 \\ x_2 &= 1 \\ x_3 &\text{ is free} \end{aligned} \end{cases}$$

• The variables $x_1$ and $x_2$ corresponding to the pivot columns in the matrix are called basic variables.
• The other variable, $x_3$, is called a free variable.

The number of basic variables is equal to the number of pivots.

Parametric Vector Form (参数向量形式)

Following the above example, we can express the solution set of the linear system as

$$\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} = \begin{bmatrix}2 - x_3 \\ 1 \\ x_3\end{bmatrix} = \begin{bmatrix}2 \\ 1 \\ 0\end{bmatrix} + \begin{bmatrix}-x_3 \\ 0 \\ x_3\end{bmatrix} = \begin{bmatrix}2 \\ 1 \\ 0\end{bmatrix} + x_3\begin{bmatrix}-1 \\ 0 \\ 1\end{bmatrix}$$

$A\vec{x} = \vec{b}$ Notation of Linear Systems

A linear system can be written in the matrix equation form $A\vec{x} = \vec{b}$, where

• $A$ is the coefficient matrix,
• $\vec{x}$ is the column vector of the variables,
• $\vec{b}$ is the column vector of the constants on the right side of the equations.

For example, the linear system

$$\begin{cases} \begin{aligned} x_1 &- 2x_2 &+ x_3 &= 0 \\ 3x_1 & &- 8x_3 &= 8 \\ 5x_1 &+ 2x_2 &- 5x_3 &= 10 \end{aligned} \end{cases}$$

can be written as

$$\begin{bmatrix} 1 & -2 & 1 \\ 3 & 0 & -8 \\ 5 & 2 & -5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 8 \\ 10 \end{bmatrix}$$

Info

If $A$ is an invertible $n \times n$ matrix, then the linear system $A\vec{x} = \vec{b}$ has a unique solution $\vec{x} = A^{-1}\vec{b}$.

Homogeneity

Homogeneous Linear Systems

A linear system is called a homogeneous linear system (齐次线性方程组) if it can be written in the form $A\vec{x} = \vec{0}$.

Obviously, a homogeneous linear system always has at least one solution $\vec{x} = \vec{0}$, which is called the trivial solution (平凡解). If there exists a solution $\vec{x} \neq \vec{0}$, we call it a nontrivial solution (非平凡解).

If a homogeneous linear system has a nontrivial solution, then it must have infinitely many solutions. This is also obvious because if $\vec{x}$ is a nontrivial solution, then for any scalar $c$, $c\vec{x}$ is also a solution.

A homogeneous linear system has a nontrivial solution if and only if there is at least one free variable. Otherwise, if every variable is basic, then there's only one solution, which is the trivial solution.

The set of solutions of the homogeneous equation is equal to the set

Nonhomogeneous Linear Systems

A linear system is called a nonhomogeneous linear system (非齐次线性方程组) if it can be written in the form $A\vec{x} = \vec{b}$ with $\vec{b} \neq \vec{0}$.

According to the definition of matrix–vector multiplication, we can conclude that $A\vec{x}$ has the same solution set as the vector equation $x_1\vec{v_1} + x_2\vec{v_2} + \cdots + x_n\vec{v_n} = \vec{b}$, where $\vec{v_1}, \vec{v_2}, \ldots, \vec{v_n}$ are the column vectors of $A$. The augmented matrix of the vector equation can be written as $\begin{bmatrix} \vec{v_1} & \vec{v_2} & \cdots & \vec{v_n} & \vec{b} \end{bmatrix}$.

The following statements are equivalent (either they are all true or all false):

• For each $\vec{b}$ in $\mathbb{R}^m$, the equation $A\vec{x} = \vec{b}$ has a solution.
• Each $\vec{b}$ in $\mathbb{R}^m$ is a linear combination of the columns of $A$.
• The columns of $A$ span $\mathbb{R}^m$.
• $A$ (NOT the augmented matrix) has a pivot position in every row.

Suppose $A\vec{x} = \vec{b}$ is consistent for some $\vec{b}$. Let $\vec{p}$ be a solution to $A\vec{x} = \vec{b}$. Then the solution set of $A\vec{x} = \vec{b}$ can be expressed as $\vec{x} = \vec{p} + \vec{v}$, where $\vec{v}$ is any solution of the corresponding homogeneous equation $A\vec{x} = \vec{0}$.

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