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lecture_16_slides
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| * Conclusion: $A$ is invertible, and $A^{-1}=\left[\mat{4& | * Conclusion: $A$ is invertible, and $A^{-1}=\left[\mat{4& | ||
| - | |||
| - | ====== Chapter 3: Vectors and geometry ====== | ||
| - | |||
| - | ==== Vectors ==== | ||
| - | |||
| - | * $\def\m# | ||
| - | * 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 $\vec v$ ==== | ||
| - | |||
| - | The vector $\vec v=\m{4\\3}$ moves: | ||
| - | |||
| - | * $(0,0)$ to $(4,3)$ | ||
| - | * $(-2,6)$ to $(2,9)$ | ||
| - | * $(x,y)$ to $(x+4, | ||
| - | |||
| - | * 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, | ||
| - | |||
| - | ==== ==== | ||
| - | |||
| - | 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 $\vec{AB}=\m{6\\-7}$, | ||
| - | |||
| - | ==== Definition of $\vec{AB}$ ==== | ||
| - | |||
| - | If $A$ and $B$ are any points in $\mathbb{R}^n$, | ||
| - | \[ \vec{AB}=B-A\] | ||
| - | (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, | ||
| - | * and $B=(11, | ||
| - | * 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/ | ||
| - | |||
| - | * 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 $\mathbb{R}^n$, | ||
| - | \[ \|\vec v\|=\sqrt{v_1^2+v_2^2+\dots+v_n^2}.\] | ||
| - | |||
| - | ==== 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\|& | ||
| - | |||
| - | ==== Exercise ==== | ||
| - | |||
| - | Prove that if $c\in \mathbb{R}$ is a scalar and $\vec v$ is a vector in $\mathbb{R}^n$, | ||
| - | \[ \|c\vec v\|=|c|\, | ||
| - | That is, multiplying a vector by a scalar $c$ scales its length by $|c|$, the absolute value of $c$. | ||
| - | |||
| - | * Hints: $\sqrt{xy}=\sqrt{x}\sqrt{y}$, | ||
| - | |||
| - | ==== 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# | ||
| - | |||
| - | * 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
