Section03 The Complete Solution to Ax = b

• From solutions of $\boldsymbol{Ax} = \boldsymbol{0}$ to solutions of $\boldsymbol{Ax} = \boldsymbol{b}$ $$\left[ \begin{matrix} \boldsymbol{A} & \boldsymbol{b} \end{matrix} \right] \underset{\text{Elimination}}{==\Rightarrow} \left[ \begin{matrix} \boldsymbol{R} & \boldsymbol{d} \end{matrix} \right]$$

One Particular Solution $\boldsymbol{Ax}_{p} = \boldsymbol{b}$

• Choose all free variables as zeros.

Particular Solution and Nullspace

$\boldsymbol{x}_{particular}$

The particular solution solves $\boldsymbol{Ax}_{p} = \boldsymbol{b}$

$\boldsymbol{x}_{nullspace}$

The $n-r$ special solutions solve $\boldsymbol{Ax}_{n} = \boldsymbol{0}$

Complete solution $\boldsymbol{x}$

$\color{#D0104C}{\boldsymbol{x} = \boldsymbol{x}_{particular} + \boldsymbol{x}_{nullspace}}$

$$\begin{array}{ll} \because & \boldsymbol{Ax}_{particular} = \boldsymbol{b} \text{ and } \boldsymbol{Ax}_{n} = \boldsymbol{0} \\ \therefore & \boldsymbol{A}(\boldsymbol{x}_{particular} + \boldsymbol{x}_{n}) = \boldsymbol{Ax}_{particular} + \boldsymbol{Ax}_{n} = \boldsymbol{b} + \boldsymbol{0} = \boldsymbol{b} \\ \end{array}$$

Computing in Julia

using LinearAlgebra
x = A \ b

Full Column Rank and Full Row Rank

$$\boldsymbol{A} \text{ is a } m \text{ by } n \text{ matrix}$$

Full column rank ($r = n$)

Number of Solutions ($m > n$)

• $\boldsymbol{Ax} = \boldsymbol{b}$ has one or no solution.
• $\boldsymbol{R}$ of $\boldsymbol{A}$ is $\left[ \begin{matrix} \boldsymbol{I} \\ \boldsymbol{0}_{(m-n)\times (n)} \end{matrix} \right]$
• therefore, $\boldsymbol{Ax} = \boldsymbol{b}$ is solvable if and only if the elimination of $\boldsymbol{b}$ like $\left[ \begin{matrix} \boldsymbol{d} \\ \boldsymbol{0}_{(m-n)\times 1} \end{matrix} \right]$, and there is only one solution.

Properties

• All columns of $\boldsymbol{A}$ are pivot columns. (This $\boldsymbol{A}$ has independent columns.)
• There are no free variables or special solutions.
• The nullspace $\boldsymbol{N}(\boldsymbol{A})$ contains only the zero vector $\boldsymbol{x} = \boldsymbol{0}$
• If $\boldsymbol{Ax} = \boldsymbol{b}$ has a solution (it might not) then it has only one solution.

Full row rank ($r = m$)

Number of Solutions ($n > m$)

• It is solvable always, and the solutions is $$\begin{array}{c} \boldsymbol{x} = \boldsymbol{x}_{particular} + \boldsymbol{x}_{nullspace} = \boldsymbol{x}_{particular} + \sum_{j}{\boldsymbol{x}_{j}\cdot (\text{the correspond special solution of } \boldsymbol{Ax} = \boldsymbol{0})} \\ \boldsymbol{x}_{j} \text{ is the free variables of } \boldsymbol{x} \end{array}$$

Properties

• All rows have pivots, and $\boldsymbol{R}$ has no zero rows. (This $\boldsymbol{A}$ has independent rows.)
• $\boldsymbol{Ax} = \boldsymbol{b}$ has a solution for every right side $\boldsymbol{b}$.
• The column space is the whole space $\mathbb{R}^{m}$.
• There are $n - r = n - m$ special solutions in the nullspace of $\boldsymbol{A}$.

Four possibilities for linear equations

Properties from Exercises

1. Column space of $\boldsymbol{A}$
   • All linear combination of pivot columns of $\boldsymbol{A}$.
   • All $\boldsymbol{b}$, which make $\boldsymbol{Ax} = \boldsymbol{b}$ is solvable.
2. If $\boldsymbol{Ax} = \boldsymbol{b}$ has two solutions $\boldsymbol{x}_{1}$ and $\boldsymbol{x}_{2}$, then $c(\boldsymbol{x}_{1} - \boldsymbol{x}_{2})$ is the solutions of $\boldsymbol{Ax} = \boldsymbol{0}$ $which$ how
3. If $\boldsymbol{Ax} = \boldsymbol{b}$ has infinitely many solutions, it's impossible for $\boldsymbol{Ax} = \boldsymbol{B} \text{(} \boldsymbol{B} \text{ is a new right side.)}$ to have only one solution.
4. proof $$\begin{array}{ll} \because & \boldsymbol{Ax} = \boldsymbol{b} \text{ has infinitely many solutions.} \\ \therefore & \text{Exsit solutions } \boldsymbol{x}_{1} \text{ and } \boldsymbol{x}_{2} \text{ make } \boldsymbol{A}(\boldsymbol{x}_{1} - \boldsymbol{x}_{2}) = \boldsymbol{0} \\ \therefore & \text{There should exsit solutions to } \boldsymbol{Ax} = \boldsymbol{B} \text{: } \boldsymbol{x}_{p} + c(\boldsymbol{x}_{1} - \boldsymbol{x}_{2})\\ \therefore & \text{It's impossible fot } \boldsymbol{Ax} = \boldsymbol{B} \text{ has only one solution, if } \boldsymbol{Ax} = \boldsymbol{b} \text{ exsits infinitely many solutions.}\\ \end{array}$$
5. If $\boldsymbol{Ax} = \boldsymbol{b}$ and $\boldsymbol{Cx} = \boldsymbol{b}$ have the same solutions, then $\boldsymbol{A} = \boldsymbol{C}$.
6. proof: $\boldsymbol{A}$ and $\boldsymbol{C}$ have the same shape and the samle nullspace. If $\boldsymbol{b} = \text{column 1 of } \boldsymbol{A}$, $\boldsymbol{x} = (1,0,\cdots,0)$ solves $\boldsymbol{Ax} = \boldsymbol{b}$ so it also solves $\boldsymbol{Cx} = \boldsymbol{b}$, therefore, $\boldsymbol{a}_{\boldsymbol{\cdot}1} = \boldsymbol{c}_{\boldsymbol{\cdot}1}$. Ohter columns too, so $\boldsymbol{A} = \boldsymbol{C}$.