Section03 Matrices

Intro

• $A = \left[ \begin{matrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \\ \end{matrix} \right]$ is a 3 by 2 matrix

Combination of the columns

$$\begin{array}{ll} \boldsymbol{Ax} = \left[ \begin{matrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn}\\ \end{matrix} \right] \left[ \begin{matrix} x_{1}\\ x_{2}\\ \vdots\\ x_{n} \end{matrix} \right] \\ = \left[ \begin{matrix} \sum_{j=1}^{n}a_{1j}x_{j}\\ \sum_{j=1}^{n}a_{2j}x_{j}\\ \vdots \\ \sum_{j=1}^{n}a_{mj}x_{j}\\ \end{matrix} \right] = \left[ \begin{matrix} \boldsymbol{a}_{1 \boldsymbol{\cdot}} \cdot \boldsymbol{x} \\ \boldsymbol{a}_{2 \boldsymbol{\cdot}} \cdot \boldsymbol{x} \\ \vdots \\ \boldsymbol{a}_{m \boldsymbol{\cdot}} \cdot \boldsymbol{x} \\ \end{matrix} \right] \\ = \color{#D0104C}{ x_{1} \cdot \left[ \begin{matrix} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \end{matrix} \right] + x_{2} \cdot \left[ \begin{matrix} a_{12} \\ a_{22} \\ \vdots \\ a_{m2} \end{matrix} \right] + \cdots + x_{n} \cdot \left[ \begin{matrix} a_{1n} \\ a_{2n} \\ \vdots \\ a_{mn} \end{matrix} \right]} \\ = \color{#D0104C}{ \sum_{j=1}^{n}{x_{j} \cdot \boldsymbol{a}_{\boldsymbol{\cdot}j}}} \end{array}$$

• When a Matrix mutiple a vector is euqal to the rows of Dot Product with the vector to form a new vector.
• $\boldsymbol{Ax}$ is a Linear combination of the columns of $\boldsymbol{A}$

Linear Equations

$$\begin{array}{c} \boldsymbol{Ax} = \boldsymbol{b} \\\\ \left\{ \begin{array}{ll} \boldsymbol{A} \text{ and } \boldsymbol{b} \text{ has known.}\\ \boldsymbol{x} \text{ is unknown.} \end{array} \right. \end{array}$$

• When equations have solutions, $\boldsymbol{A}$ is invertible.
• $\boldsymbol{x} = \boldsymbol{A}^{-1}\boldsymbol{b}$

The Inverse Matrix

• example $$\begin{array}{c} \boldsymbol{Ax} = \boldsymbol{b} \\\\ \left\{ \begin{array}{ll} \boldsymbol{A} = \left[ \begin{matrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 0 \end{matrix} \right] \\ \boldsymbol{b} = \left[ \begin{matrix} b_{1} \\ b_{2} \\ b_{3} \end{matrix} \right] \\ \boldsymbol{x} = \left[ \begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix} \right] \end{array} \right. \\ \boldsymbol{x} = \boldsymbol{A}^{-1}\boldsymbol{x} \\\\ \left\{ \begin{array}{ll} \boldsymbol{A}^{-1} = \left[ \begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{matrix} \right] \end{array} \right. \end{array}$$

  • $\boldsymbol{A}^{-1}$ is the Inverse Matrix of $\boldsymbol{A}$

• $\boldsymbol{A}$ is the difference matrix which convert $\boldsymbol{x}$ to $\boldsymbol{b}$; $\boldsymbol{A}^{-1}$ is the difference matrix which convert $\boldsymbol{b}$ to $\boldsymbol{x}$

• function perspective The vector $\boldsymbol{x}$ changes to a function $x(t)$. The differences $\boldsymbol{Ax}$ become the derivative $\frac{x(t)}{t} = b(t)$. In the inverse direction, the $\boldsymbol{A}^{-1}\boldsymbol{x}$ become the integral of $b(t)$ Sums of differences are like integrals of derivatives $$\boldsymbol{Ax} = \boldsymbol{b} \text{ and } \boldsymbol{x} = \boldsymbol{A}^{-1}\boldsymbol{b}\ \textbf{ is similar to }\ \frac{dx}{dt} = b \text{ and } x(t) = \int_{0}^{t}{b} \cdot dt$$

Difference

$x(t) = t^{2}$

Backward Difference

$$x(t) - x(t-1) = t^{2} - (t-1)^{2} = 2t - 1$$

Centered Difference

$$\frac{x(t+1)-x(t-1)}{(t+1) - (t-1)} = \frac{(t+1)^{2} - (t-1)^{2}}{2} = 2t$$

Cyclic Differences

• example $$\left\{ \begin{array}{ll} \boldsymbol{C} = \left[ \begin{matrix} \boldsymbol{u} & \boldsymbol{v} & \boldsymbol{w} \end{matrix} \right]\\ \boldsymbol{u} = \left[ \begin{matrix} u_{1} \\ u_{2} \\ u_{3} \end{matrix} \right] \text{ and } \boldsymbol{v} = \left[ \begin{matrix} v_{1} \\ v_{2} \\ v_{3} \end{matrix} \right] \text{ and } \boldsymbol{w} = \left[ \begin{matrix} w_{1}\\ w_{2}\\ w_{3} \end{matrix} \right] \end{array} \right.\\\\ \boldsymbol{Cx} = \left[ \begin{matrix} 1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{matrix} \right] \left[ \begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix} \right] = \left[ \begin{matrix} x_{1} - x_{3} \\ x_{2} - x_{1} \\ x_{3} - x_{2} \end{matrix} \right] = \boldsymbol{b}$$

  • sums of $\boldsymbol{b}$ must is equal to 0

Independence and Dependence

• Independence
  $$\boldsymbol{w} \text{ is not in the plane of } \boldsymbol{u} \text{ and } \boldsymbol{v}.$$
  • 当$\boldsymbol{u}, \boldsymbol{v}, \boldsymbol{w}$ 相互独立时。仅有$0 \boldsymbol{u} + 0 \boldsymbol{v} + 0 \boldsymbol{w} = \boldsymbol{0}$ 时，$\boldsymbol{b} = \boldsymbol{0}$
  • more general defination $$\boldsymbol{Ax} = \boldsymbol{0} \text{ has only one solution. } \boldsymbol{A} \text{ is an } \textbf{invertible matrix}.$$
• Dependence $$\boldsymbol{w} \text{ is in the plane of } \boldsymbol{u} \text{ and } \boldsymbol{v}. \text{ (}\boldsymbol{w} \text{ is the linear combination of } \boldsymbol{v} \text{ and } \boldsymbol{w}\text{)}$$
  • 当$\boldsymbol{u},\boldsymbol{v}, \boldsymbol{w}$不独立时。存在其他某种线性组合$a \boldsymbol{u} + b \boldsymbol{v} + c \boldsymbol{w} (a,b,c \text{不全为0})$
  • more general defination $$\boldsymbol{Cx} = \boldsymbol{0} \text{ has many solution. } \boldsymbol{C} \text{ is an } \textbf{singular matrix}.$$

if $\left[ \begin{matrix} a & b \\ c & d \end{matrix} \right]$ has dependent rows, then it also has dependent columns.