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--- lecture_17 [2015/04/01 11:06] – rupert
+++ lecture_17 [2017/03/30 09:21] (current) – rupert
@@ Line 1: / Line 1: @@
-===== A more efficient way to find $A^{-1}$ =====
+====== Chapter 3: Vectors and geometry ======
-Given an $n\times n$ matrix $A$, form the $n\times 2n$ matrix
+Recall that a $2\times 1$ column vector such as $\def\m#1{\begin{bmatrix}#1\end{bmatrix}}\m{4\\3}$ is a pair of numbers written in a column. We are also used to writing points in the plane $\mathbb R^2$ as a pair of numbersl; for example $(4,3)$ is the point obtained by starting from the origin, and moving $4$ units to the right and $3$ units up.
-\[ \def\m#1{\left[
-\begin{array}{@{} c|c {}@} % it does autodetection
-#1
-\end{array}
-\right]}\m{A&I_n}\]
-and use [[EROs]] to put this matrix into [[RREF]]. One of two things can happen:
-  * Either you get a row of the form $[0~0~\dots~0~|~*~*~\dots~*]$ which starts with $n$ zeros. You can then conclude that $A$ is not invertible.
+We think of a (column) vector like $\vec v=\m{4\\3}$ as an instruction to move $4$ units to the right and $3$ units up. This movement is called "translation by $\vec v$".
-  * Or you end up with a matrix of the form $\m{I_n&B}$ for some $n\times n$ matrix $B$. You can then conclude that $A$ is invertible, and $A^{-1}=B$.
+=== Examples ===
+The vector $\vec v=\m{4\\3}$ moves:
+  * $(0,0)$ to $(4,3)$
+  * $(-2,6)$ to $(2,9)$
+  * $(x,y)$ to $(x+4,y+3)$.
+It is convenient to not be too fussy about the difference between a point like $(4,3)$ and the vector $\m{4\\3}$. If we agree to write points as column vectors, then we can perform algebra (addition, subtraction, scalar multiplication) as discussed in Chapter 2, 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 moves the point $A=(-1,3)$ to $B=(5,-4)$?
+Answer: we need a vector $\vec v$ with $A+\vec v=B$, so $\vec v=B-A = \def\m#1{\begin{bmatrix}#1\end{bmatrix}}\m{5\\-4}-\m{-1\\3}=\m{6\\-7}$. 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
+\[ \vec{AB}=B-A\]
+(where on the right hand side, we interpret the points as column vectors so we can subtract them to get a column vector).
+Thus $\vec{AB}$ is the vector which moves the point $A$ to the point $B$.
+=== Example ===
+In $\mathbb{R}^3$, the points $A=(3,-4,5)$ and $B=(11,6,-2)$ have $\vec{AB}=\m{11\\6\\-2}-\m{3\\-4\\5}=\m{8\\10\\-7}$.
+==== 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 ====
-  * Consider $A=\def\mat#1{\begin{matrix}#1\end{matrix}}\left[\mat{1&3\\2&6}\right]$. \begin{align*}\m{A&I_2}&=\m{\mat{1&3\\2&6}&\mat{1&0\\0&1}}
+  * $\left\|\m{4\\3}\right\|=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$
-\def\go#1#2{\m{\mat{#1}&\mat{#2}}}
+  * $\left\|\m{1\\0\\-2\\3}\right\|=\sqrt{1^2+0^2+(-2)^2+3^2}=\sqrt{1+0+4+9}=\sqrt{14}$
-\def\ar#1{\\[6pt]\xrightarrow{#1}&}
-\ar{R2\to R2-2R1}\go{1&3\\0&0}{1&0\\-2&1}
+==== Exercise ====
-\end{align*} Conclusion: $A$ is not invertible.
-  * Consider $A=\left[\mat{1&3\\2&7}\right]$.\begin{align*}\m{A&I_2}&=\m{\mat{1&3\\2&7}&\mat{1&0\\0&1}}
+Prove that if $c\in \mathbb{R}$ is a scalar and $\vec v$ is a vector in $\mathbb{R}^n$, then
-\ar{R2\to R2-2R1}\go{1&3\\0&1}{1&0\\-2&1}
+\[ \|c\vec v\|=|c|\,\|\vec v\|.\]
-\ar{R1\to R1-3R1}\go{1&0\\0&1}{7&-3\\-2&1}
+That is, multiplying a vector by a scalar $c$ scales its length by $|c|$, the absolute value of $c$.
-\end{align*} Conclusion: $A$ is invertible and $A^{-1}=\left[\mat{7&-3\\-2&1}\right]$.
-  * Consider $A=\left[\mat{3&1&0\\-2&-4&3\\5&4&-2}\right]$.\begin{align*}\m{A&I_3}&=\go{3&1&0\\-2&-4&3\\5&4&-2}{1&0&0\\0&1&0\\0&0&1}
+==== Remark ====
-\ar{R1\to R1+R2}
-\go{1&-3&3\\-2&-4&3\\5&4&-2}{1&1&0\\0&1&0\\0&0&1}
+$\|\vec{AB}\|$ is the distance from point $A$ to point $B$, since this is the length of vector which takes point $A$ to point $B$.
-\ar{R2\to R2+2R1,\ R3\to R3-5R1}
-\go{1&-3&3\\0&-10&9\\0&19&-17}{1&1&0\\2&3&0\\-5&-5&1}
+=== Examples ===
-\ar{R3\leftrightarrow R2}
-\go{1&-3&3\\0&19&-17\\0&-10&9}{1&1&0\\-5&-5&1\\2&3&0}
+  * The distance from $A=(1,2)$ to $B=(-3,4)$ is $\|\def\m#1{\begin{bmatrix}#1\end{bmatrix}}\vec{AB}\|=\left\|\m{-3\\4}-\m{1\\2}\right\|=\left\|\m{-4\\2}\right\|=\sqrt{(-4)^2+2^2}=\sqrt{20}=2\sqrt{5}$.
-\ar{R2\to R2+2R3}
+  * The length of the main diagonal of the unit cube in $\mathbb{R}^3$ is the distance between $0=(0,0,0)$ and $A=(1,1,1)$, which is $\|\vec{0A}\|=\left\|\m{1\\1\\1}\right\|=\sqrt{1^2+1^2+1^2}=\sqrt3$.
-\go{1&-3&3\\0&-1&1\\0&-10&9}{1&1&0\\-1&1&1\\2&3&0}
-\ar{R1\to R1+3R2,\ R3\to R3-10R2}
+==== Scalar multiplication and direction ====
-\go{1&0&0\\0&-1&1\\0&0&-1}{4&-2&3\\-1&1&1\\12&-7&-10}
-\ar{R2\to R2+R3}
+Multiplying a vector by a scalar changes its length, but doesn't change its direction.
-\go{1&0&0\\0&-1&0\\0&0&-1}{4&-2&3\\11&-6&-9\\12&-7&-10}
-\ar{R2\to -R2,\ R3\to -R3}
+==== Definition: unit vectors ====
-\go{1&0&0\\0&1&0\\0&0&1}{4&-2&3\\-11&6&9\\-12&7&10}
-\end{align*}
+{{page>unit vector}}
+==== Proposition: finding a unit vector in the same direction as a given vector ====
+If $\vec v$ is a non-zero vector, then $\vec w=\frac1{\|\vec v\|}\vec v$ is a unit vector (in the same direction as $\vec v$).
+=== Proof ===
+Using the formula $\|c\vec v\|=|c|\,\|\vec v\|$ and the fact that $\|\vec v\|>0$, we have
+\[ \|\vec w\|=\left\|\frac1{\|\vec v\|}\vec v\right\|=\left|\frac1{\|\vec v\|}\right|\,\|\vec v\|=\frac1{\|\vec v\|}\,\|\vec v\| = 1.\]
+So $\vec w$ is a unit vector, and since it's scalar multiple of $\vec v$, it's in the same direction as $\vec v$. ■
+==== Example ====
+What is unit vector in the same direction as $\vec v=\m{1\\2}$?
+We have $\|\vec v\|=\sqrt{1^2+2^2}=\sqrt5$, so the proposition tells us that is $\vec w=\frac1{\|\vec v\|}\vec v = \frac1{\sqrt 5}\vec v=\frac1{\sqrt5}\m{1\\2}=\m{1/\sqrt{5}\\2/\sqrt5}$ is a unit vector in the same direction as $\vec v$.
+===== Addition of vectors =====
+If $\vec v=\vec{AB}$, then $\vec v$ moves $A$ to $B$, so $A+\vec v=B$.
+If $\vec w=\vec {BC}$, then $\vec w$ moves $B$ to $C$, so $B+\vec w=C$.
+What about $\vec v+\vec w$? We have $A+\vec v+\vec w=B+\vec w=C$. So $\vec v+\vec w=\vec{AC}$.
+This gives us the **triangle law for vector addition**: $\vec v$, $\vec w$ and $\vec v+\vec w$ may be arranged to form a triangle:
+{{ :tri.png?nolink&300 |}}
+We get another triangle by starting at $A$ and translating first by $\vec w$ and then by $\vec v$; the other side of this triangle is $\vec w+\vec v$.
+But we know that $\vec v+\vec w=\vec w+\vec v$! So we can put these two triangles together to get the **parallelogram law for vector addition**:
+{{ :par.png?nolink&300 |}}