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lecture_16_slides [2017/03/29 13:06] – [Example: $n=3$] rupertlecture_16_slides [2017/04/03 15:26] (current) rupert
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   * Conclusion: $A$ is invertible, and $A^{-1}=\left[\mat{4&-2&3\\-11&6&9\\-12&7&10}\right]$.   * Conclusion: $A$ is invertible, and $A^{-1}=\left[\mat{4&-2&3\\-11&6&9\\-12&7&10}\right]$.
  
- 
-====== Chapter 3: Vectors and geometry ====== 
- 
-==== Vectors ==== 
- 
-  * $\def\m#1{\begin{bmatrix}#1\end{bmatrix}}\m{4\\3}$ is a $2\times 1$ column vector 
-  * i.e., a pair of numbers written in a column 
-  * We also use pairs of numbers to write points in the plane $\mathbb R^2$ 
-  * e.g., $(4,3)$ is a point 
-    * you get there by starting from the origin, moving $4$ units to the right and $3$ units up. 
-  * We think $\vec v=\m{4\\3}$ as an instruction to move $4$ units to the right and $3$ units up.  
-  * This movement is called "translation by $\vec v$" 
- 
-==== Translation by $\vec v$ ==== 
- 
-The vector $\vec v=\m{4\\3}$ moves: 
- 
-  * $(0,0)$ to $(4,3)$ 
-  * $(-2,6)$ to $(2,9)$ 
-  * $(x,y)$ to $(x+4,y+3)$. 
- 
-  * We're not too fussy about the difference between points like $(4,3)$ and vectors like $\m{4\\3}$. 
-  * If we write points as column vectors, we can perform algebra (addition, subtraction, scalar multiplication) using points and column vectors. 
- 
-==== ==== 
- 
-For example, we could rewrite the examples above by saying that $\vec v=\m{4\\3}$ moves: 
- 
-  * $\m{0\\0}$ to $\m{0\\0}+\m{4\\3}=\m{4\\3}$ 
-  * $\m{-2\\6}$ to $\m{-2\\6}+\m{4\\3}=\m{2\\9}$ 
-  * $\m{x\\y}$ to $\m{x\\y}+\m{4\\3}=\m{x+4\\y+3}$. 
- 
-  * More generally: a column vector $\vec v$ moves a point $\vec x$ to $\vec x+\vec v$. 
- 
-==== Example ==== 
- 
-Which vector $\vec v$ moves the point $A=(-1,3)$ to $B=(5,-4)$? 
- 
-  * We need a vector $\vec v$ with $A+\vec v=B$ 
-  * So $\vec v=B-A = \m{5\\-4}-\m{-1\\3}=\m{6\\-7}$.  
- 
-  * We write $\vec{AB}=\m{6\\-7}$, since this is the vector which moves $A$ to $B$. 
- 
-==== Definition of $\vec{AB}$ ==== 
- 
-If $A$ and $B$ are any points in $\mathbb{R}^n$, then the vector $\vec{AB}$ is defined by  
-\[ \vec{AB}=B-A\] 
-(on the right, we interpret points as column vectors so we can subtract them to get a column vector). 
- 
-  * $\vec{AB}$ is the vector which moves $A$ to $B$. 
- 
-==== Example ==== 
- 
-In $\mathbb{R}^3$,  
-  * if $A=(3,-4,5)$  
-  * and $B=(11,6,-2)$  
-  * then $\vec{AB}=\m{11\\6\\-2}-\m{3\\-4\\5}=\m{8\\10\\-7}$. 
- 
-==== The uses of vectors ==== 
- 
-Vectors are used in geometry and science to represent quantities with both a **magnitude** (size/length) and a **direction**. For example: 
- 
-  * displacements (in geometry) 
-  * velocities 
-  * forces 
- 
-Recall that a column vector moves points. Its magnitude, or length, is how far it moves points. 
- 
-==== Definition: the length of a vector ==== 
- 
-If $\vec v=\m{v_1\\v_2\\\vdots\\v_n}$ is a column vector in $\mathbb{R}^n$, then its **magnitude**, or **length**, or **norm**, is the number 
-\[ \|\vec v\|=\sqrt{v_1^2+v_2^2+\dots+v_n^2}.\] 
- 
-==== Examples ==== 
- 
-  * $\left\|\m{4\\3}\right\|=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$ 
- 
-  * \begin{align*}\left\|\m{1\\0\\-2\\3}\right\|&=\sqrt{1^2+0^2+(-2)^2+3^2}\\&=\sqrt{1+0+4+9}=\sqrt{14}\end{align*} 
- 
-==== Exercise ==== 
- 
-Prove that if $c\in \mathbb{R}$ is a scalar and $\vec v$ is a vector in $\mathbb{R}^n$, then  
-\[ \|c\vec v\|=|c|\,\|\vec v\|.\] 
-That is, multiplying a vector by a scalar $c$ scales its length by $|c|$, the absolute value of $c$. 
- 
-  * Hints: $\sqrt{xy}=\sqrt{x}\sqrt{y}$, and $\sqrt{c^2}=|c|$. 
- 
-==== Distance between two points ==== 
- 
-$\|\vec{AB}\|$ is the distance from point $A$ to point $B$ 
-  * since this is the length of vector which takes point $A$ to point $B$. 
- 
-  * e.g. how far from from $A=(1,2)$ to $B=(-3,4)$? 
-  * $\small\|\def\m#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\vec{AB}\|=\left\|\m{-3\\4}-\m{1\\2}\right\|=\left\|\m{-4\\2}\right\|=\sqrt{(-4)^2+2^2}=\sqrt{20}$. 
- 
-  * e.g. what's the length of the main diagonal of the unit cube in $\mathbb{R}^3$? 
-  * = distance from $0=(0,0,0)$ to $A=(1,1,1)$ 
-  * $\|\vec{0A}\|=\left\|\m{1\\1\\1}\right\|=\sqrt{1^2+1^2+1^2}=\sqrt3$. 
lecture_16_slides.1490792804.txt.gz · Last modified: by rupert

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