Differences

This shows you the differences between two versions of the page.

--- lecture_16 [2016/03/30 12:02] – rupert
+++ lecture_16 [2017/03/30 09:20] (current) – [Chapter 3: Vectors and geometry] rupert
@@ Line 1: / Line 1: @@
+=== Example: $n=2$, general case ===
+If $A=\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\mat{a&b\\c&d}$, then $C=\mat{d&-c\\-b&a}$, so the adjoint of $A$ is $J=C^T=\mat{d&-b\\-c&a}$.
+Recall that $AJ=(\det A)I_2=JA$; we calculated this earlier when we looked at the inverse of a $2\times 2$ matrix. Hence for a $2\times 2$ matrix $A$, if $\det A\ne0$, then $A^{-1}=\frac1{\det A}J$.
 === Example: $n=3$ ===
@@ Line 30: / Line 36: @@
 If again we take $A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$, then $J=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}$ and $\det(A)=-1$, so $A$ is invertible and $A^{-1}=\frac1{-1}J=-J=\mat{4&-2&-3\\-11&6&9\\-12&7&10}$.
+==== Example ($n=4$) ====
+Let $A=\mat{1&0&0&0\\1&2&0&0\\1&2&3&0\\1&2&3&4}$.
+Recall that a matrix with a repeated row or a zero row has determinant zero.
+We have \[C=\mat{+\vm{2&0&0\\2&3&0\\2&3&4}&-\vm{1&0&0\\1&3&0\\1&3&4}&+0&-0\\-0&+\vm{1&0&0\\1&3&0\\1&3&4}&-\vm{1&0&0\\1&2&0\\1&2&4}&+0\\+0&-0&+\vm{1&0&0\\1&2&0\\1&2&4}&-\vm{1&0&0\\1&2&0\\1&2&3}\\-0&+0&-0&+\vm{1&0&0\\1&2&0\\1&2&3}}=\mat{24&-12&0&0\\0&12&-8&0\\0&0&8&-6\\0&0&0&6}\]
+so \[J=C^T=\mat{24&0&0&0\\-12&12&0&0\\0&-8&8&0\\0&0&-6&6}.\]
+Since $A$ is lower triangular, its determinant is given by multiplying together its diagonal entries: $\det(A)=1\times 2\times 3\times 4=24$. (Note that even if $A$ was not triangular, $\det A$ can be easily found from the matrix of cofactors $C$ by summing the entries of $A$ multiplied by the entries of $C$ (i.e., the minors) along any row or column.)
+So \[A^{-1}=\frac1{\det A}J = \frac1{24}\mat{24&0&0&0\\-12&12&0&0\\0&-8&8&0\\0&0&-6&6}=\mat{1&0&0&0\\-1/2&1/2&0&0\\0&-1/3&1/3&0\\0&0&-1/4&1/4}.\]
+You should check that this really is the inverse, by checking that $AA^{-1}=I_4=A^{-1}A$.
 ===== A more efficient way to find $A^{-1}$ =====
@@ Line 72: / Line 89: @@
-====== Chapter 3: Vectors and geometry ======
-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.
-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 $\vec v$".
-=== 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 $\vec v=B-A = \m{5\\-4}-\m{-1\\3}=\m{6\\-7}$. 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 ====
-  * $\left\|\m{4\\3}\right\|=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$
-  * $\left\|\m{1\\0\\-2\\3}\right\|=\sqrt{1^2+0^2+(-2)^2+3^2}=\sqrt{1+0+4+9}=\sqrt{14}$
-==== 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 $c$.