Welcome to Math 240! (additions to this web site no longer count towards good deed points)

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Week of...

Notes and Links

1

Sep 8

About This Class, What is this class about? (PDF, HTML), Monday, Wednesday

2

Sep 15

HW1, Monday, Wednesday, TheComplexField.pdf,HW1_solutions.pdf

3

Sep 22

HW2, Class Photo, Monday, Wednesday, HW2_solutions.pdf

4

Sep 29

HW3, Wednesday, Tutorial, HW3_solutions.pdf

5

Oct 6

HW4, Monday, Wednesday, Tutorial, HW4_solutions.pdf

6

Oct 13

No Monday class (Thanksgiving), Wednesday, Tutorial

7

Oct 20

HW5, Term Test at tutorials on Tuesday, Wednesday

8

Oct 27

HW6, Monday, Why LinAlg?, Wednesday, Tutorial

9

Nov 3

Monday is the last day to drop this class, HW7, Monday, Wednesday, Tutorial

10

Nov 10

HW8, Monday, Tutorial

11

Nov 17

MondayTuesday is UofT November break

12

Nov 24

HW9

13

Dec 1

Wednesday is a "makeup Monday"! EndofCourse Schedule, Tutorial

F

Dec 8

The Final Exam

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Boris
Elementary and (Not So Elementary) Errors in Homework
(1) Bad Notation
Consider these three matrices:
 $M_{1}={\begin{pmatrix}1&0\\0&0\end{pmatrix}},M_{2}={\begin{pmatrix}0&0\\0&1\\\end{pmatrix}},M_{3}={\begin{pmatrix}0&1\\1&0\\\end{pmatrix}}$
We want to equate $span(M_{1},M_{2},M_{3})$ to the set of all symmetric $2\times 2$ matrices. Here is the wrong way to write this:
 $span(M_{1},M_{2},M_{3})={\begin{pmatrix}a&b\\b&c\\\end{pmatrix}}$.
Firstly, $span(M_{1},M_{2},M_{2})$ is the set of all linear combinations of $M_{1},M_{2},M_{3}$. To equate it to a single symmetric $2\times 2$ matrix makes no sense. Secondly, the elements $a,b,c,d$ are undefined. What are they suppose to represent? Rational numbers? Real numbers? Members of the field of two elements? The following way of writing erases those issues:
 $span(M_{1},M_{2},M_{3})=\{{\begin{pmatrix}a&b\\b&c\\\end{pmatrix}}:a,b,c\in F\}$ where $F$ is an arbitrary field.
(2) Algorithm vs. Proof (Boris's Section Only)
When solving a problem that requires a solution to a linear equation, it is not always obvious which of the following you should show:
 a) An algorithm for finding the solution
 b) A proof that a solution is correct
If the problem asks to solve a linear equation, then just show (a). Otherwise, consider problems such as this:
 Determine if the vector $(2,2,2)$ is a linear combination of the vectors $((1,2,1),(3,3,3)$ in $R^{3}$.
Show both (a) and (b) to be on the safe side.
Problem 5h) on Page 34 in Homework 3 for all Fields
For an arbitrary field $F$, determine if the matrix
${\begin{pmatrix}1&0\\0&1\\\end{pmatrix}}$
is in span
$S=\{{\begin{pmatrix}1&0\\1&0\\\end{pmatrix}},{\begin{pmatrix}0&1\\0&1\\\end{pmatrix}},{\begin{pmatrix}1&1\\0&0\\\end{pmatrix}}\}$.
Proof:
We show that
${\begin{pmatrix}1&0\\0&1\\\end{pmatrix}}\in span(S)\iff char(F)=2$.
 We show that $char(F)=2\implies {\begin{pmatrix}1&0\\0&1\\\end{pmatrix}}\in span(S)$.
 Assume that $char(F)=2$.
 Let $c_{1}=0,c_{2}=1,c_{3}=1$.
 Then $c_{1},c_{2},c_{3}\in F$.
 Then $c_{1}{\begin{pmatrix}1&0\\1&0\\\end{pmatrix}}+c_{2}{\begin{pmatrix}0&1\\0&1\\\end{pmatrix}}+c_{3}{\begin{pmatrix}1&1\\0&0\\\end{pmatrix}}={\begin{pmatrix}1&2\\0&1\\\end{pmatrix}}$.
 Since $char(F)=2$ and the entries of the matrix are from $F$, then $0=2$.
 Then ${\begin{pmatrix}1&2\\0&1\\\end{pmatrix}}={\begin{pmatrix}1&0\\0&1\\\end{pmatrix}}\in span(S)$.
 Then $char(F)=2\implies {\begin{pmatrix}1&0\\0&1\\\end{pmatrix}}\in span(S)$.
 We show that $char(F)\neq 2\implies {\begin{pmatrix}1&0\\0&1\\\end{pmatrix}}\notin span(S)$.
 Assume to the contrary that $char(F)\neq 2\land {\begin{pmatrix}1&0\\0&1\\\end{pmatrix}}\in span(S)$.
 Then $\exists c_{1},c_{2},c_{3}\in F,{\begin{pmatrix}1&0\\0&1\\\end{pmatrix}}=c_{1}{\begin{pmatrix}1&0\\1&0\\\end{pmatrix}}+c_{2}{\begin{pmatrix}0&1\\0&1\\\end{pmatrix}}+c_{3}{\begin{pmatrix}1&1\\0&0\\\end{pmatrix}}$.
 Then this system of linear equations has a solution:
 $(11)c_{1}+c_{3}=1$
 $(21)c_{1}=0$
 $(12)c_{2}+c_{3}=0$
 $(22)c_{2}=1$.
 When solving this system, we see that it has no solution.
 This contradicts the assumption that it has a solution.
 Then $char(F)\neq 2\implies {\begin{pmatrix}1&0\\0&1\\\end{pmatrix}}\notin span(S)$.
 Then ${\begin{pmatrix}1&0\\0&1\\\end{pmatrix}}\in span(S)\iff char(F)=2$. Q.E.D.
A Field Problem
Find the solution to $x^{2}=2$ in $Z_{11}$.
Note that a polynomial of $n$ degree has at most $n$ solutions.
Algorithm:
We find the solution to $x^{2}=2$ in $Z_{11}$.
 Since in $Z_{11},2=9$, then $x^{2}=9$.
 Since $9$ is additive inverse of $9$, then $x^{2}9=99=0$.
 By the result that we proved in Question 2 of Homework 1, then $(x^{2}9)=(x3)(x+3)=0$.
 Then $x=\pm 3$ are the solutions.
A Dimension Problem
Let $W_{1},W_{2}$ be subspaces of a finite dimensional vector space $V$ over a field $F$ where $W_{1}\cap W_{2}=\{0\}$. Then
$dim(span(W_{1}\cup W_{2}))=dim(W_{1})+dim(W_{2})$.
Proof:
We show that $dim(span(W_{1}\cup W_{2}))=dim(W_{1})+dim(W_{2})$.
 Since $V$ is finite dimensional, then $W_{1},W_{2}$ are finite dimensional.
 Then we can let $B_{1}=\{u_{1},u_{2},u_{3},...,u_{m}\}$ be a basis of $W_{1}$ and $B_{2}=\{v_{1},v_{2},v_{3},...,v_{n}\}$ be a basis of $W_{2}$.
 We show that $B_{1}\cup B_{2}$ is a basis of $span(W_{1}\cup W_{2})$.
 We show that $span(B_{1}\cup B_{2})=span(W_{1}\cup W_{2})$.
 We show that $span(B_{1}\cup B_{2})\subset span(W_{1}\cup W_{2})$.
 Since $(B_{1}\cup B_{2})\subset (W_{1}\cup W_{2})$, then $span(B_{1}\cup B_{2})\subset span(W_{1}\cup W_{2})$.
 We show that $span(W_{1}\cup W_{2})\subset span(B_{1}\cup B_{2})$.
 Since $B_{1}\subset (B_{1}\cup B_{2})$, then $span(B_{1})=W_{1}\subset span(B_{1}\cup B_{2})$.
 Since $B_{2}\subset (B_{1}\cup B_{2})$, then $span(B_{2})=W_{2}\subset span(B_{1}\cup B_{2})$.
 Then $(W_{1}\cup W_{2})\subset span(B_{1}\cup B_{2})$ and $span(W_{1}\cup W_{2})\subset span(B_{1}\cup B_{2})$.
 Then $span(B_{1}\cup B_{2})=span(W_{1}\cup W_{2})$.
 We show that $B_{1}\cup B_{2}$ is linearly independent.
 Let $\displaystyle \sum _{i=1}^{m}b_{i}u_{i}+\displaystyle \sum _{j=1}^{n}c_{j}v_{j}=0$ where $b_{i},c_{j}\in F$.
 Then $\displaystyle \sum _{i=1}^{m}b_{i}u_{i}=\displaystyle \sum _{j=1}^{n}(c_{j})v_{j}$.
 Since $W_{1}\cap W_{2}=\{0\}$, then $\displaystyle \sum _{i=1}^{m}b_{i}u_{i}=\displaystyle \sum _{j=1}^{n}(c_{j})v_{j}=0$.
 Since $B_{1},B_{2}$ are linearly independent, then $b_{i}=(c_{j})=0$.
 For $\displaystyle \sum _{i=1}^{m}b_{i}u_{i}+\displaystyle \sum _{j=1}^{n}c_{j}v_{j}=0$, then $b_{i}=c_{j}=0$.
 Then $B_{1}\cup B_{2}$ is linearly independent.
 Then $B_{1}\cup B_{2}$ is a basis of $span(W_{1}\cup W_{2})$.
 Since $\leftB_{1}\cup B_{2}\right=\leftB_{1}\right+\leftB_{2}\right$, then $dim(span(W_{1}\cup W_{2}))=dim(W_{1})+dim(W_{2})$. Q.E.D.
Nikita
File:1014.240.pdf
Scanned Tutorial Notes by Boyang.wu
File:Tut6.pdf