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(1) '''Bad Notation''' |
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'''(1) Bad Notation''' |
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Consider these three matrices: |
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Let |
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::::::::<math> |
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::::::::<math> |
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M_1 = |
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M_1 = |
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be matrices. We want to equate <math>span(M_1, M_2, M_3)</math> to the set of all symmetric <math>2 \times 2</math> matrices. Here is the wrong way to write this:
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We want to equate <math>span(M_1, M_2, M_3)</math> to the set of all symmetric <math>2 \times 2</math> matrices. Here is the wrong way to write this: |
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Firstly, <math>span(M_1, M_2, M_2)</math> is the set of all linear combinations of <math>M_1, M_2, M_3</math>. To equate it to a single |
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Firstly, <math>span(M_1, M_2, M_2)</math> is the set of all linear combinations of <math>M_1, M_2, M_3</math>. To equate it to a single symmetric <math>2 \times 2</math> matrix makes no sense. Secondly, the elements <math>a, b, c, d</math> 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: |
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symmetric |
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<math>2 \times 2</math> matrix makes no sense. Secondly, the elements <math>a, b, c, d</math> are undefined. What are they suppose to |
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represent? Rational numbers? Real numbers? Members of the field of two elements? The following way of writing erases those issues: |
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span(M_1, M_2, M_3) = \{ |
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span(M_1, M_2, M_3) = \{ |
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\begin{pmatrix} |
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\begin{pmatrix} |
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\end{pmatrix} |
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\end{pmatrix} |
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:a, b, c \in F \} |
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:a, b, c \in F \} |
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</math> where <math>F</math> is an arbitrary field. |
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</math> |
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where <math>F</math> is an arbitrary field. |
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'''(2) Algorithm vs. Proof (Boris's Section Only)''' |
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When solving a problem that requires a solution to a linear equation, it is not always obvious which of the following you should show: |
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(2) '''Algorithm vs. Proof (Boris's Section Only)''' |
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When solving a problem that requires a solution to a linear equation, it is not always obvious which of the following you |
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should show: |
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:a) An algorithm for finding the solution |
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:a) An algorithm for finding the solution |
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:b) A proof that a solution is correct |
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:b) A proof that a solution is correct |
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If the problem asks to solve a linear equation, then just show (a). Otherwise, consider problems such as this: |
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If the problem asks to solve a linear equation, then just show (a). Otherwise, consider problems such as this: |
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Show both (a) and (b) to be on the safe side. |
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Show both (a) and (b) to be on the safe side. |
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====Problem 5h) in Homework 3 for all Fields==== |
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====Problem 5h) on Page 34 in Homework 3 for all Fields==== |
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\end{pmatrix} |
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\end{pmatrix} |
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\in span(S) \iff char(F)=2</math>. ''Q.E.D.'' |
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\in span(S) \iff char(F)=2</math>. ''Q.E.D.'' |
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====A Field Problem==== |
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====A Field Problem==== |
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''Proof'' |
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''Proof:'' |
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We show that <math>dim(span(W_1 \cup W_2)) = dim(W_1) + dim(W_2)</math>. |
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:Since <math>V</math> is finite dimensional, then <math>W_1, W_2</math> are finite dimensional. |
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:Then we can let <math>B_1 = \{u_1, u_2, u_3, ..., u_m\}</math> be a basis of <math>W_1</math> and <math>B_2 = \{v_1, v_2, v_3, ..., v_n\}</math> be a basis of <math>W_2</math>. |
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:We show that <math>B_1 \cup B_2</math> is a basis of <math>span(W_1 \cup W_2)</math>. |
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::We show that <math>span(B_1 \cup B_2) = span(W_1 \cup W_2)</math>. |
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:::We show that <math>span(B_1 \cup B_2) \subset span(W_1 \cup W_2)</math>. |
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::::Since <math>(B_1 \cup B_2) \subset (W_1 \cup W_2)</math>, then <math>span(B_1 \cup B_2) \subset span(W_1 \cup W_2)</math>. |
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:::We show that <math>span(W_1 \cup W_2) \subset span(B_1 \cup B_2)</math>. |
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::::Since <math>B_1 \subset (B_1 \cup B_2)</math>, then <math>span(B_1) = W_1 \subset span(B_1 \cup B_2)</math>. |
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::::Since <math>B_2 \subset (B_1 \cup B_2)</math>, then <math>span(B_2) = W_2 \subset span(B_1 \cup B_2)</math>. |
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::::Then <math>(W_1 \cup W_2) \subset span(B_1 \cup B_2)</math> and <math>span(W_1 \cup W_2) \subset span(B_1 \cup B_2)</math>. |
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:::Then <math>span(B_1 \cup B_2) = span(W_1 \cup W_2)</math>. |
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::We show that <math>B_1 \cup B_2</math> is linearly independent. |
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:::Let <math>\displaystyle\sum_{i=1}^{m} b_iu_i + \displaystyle\sum_{j=1}^{n} c_jv_j = 0</math> where <math>b_i, c_j \in F</math>. |
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:::Then <math>\displaystyle\sum_{i=1}^{m} b_iu_i = \displaystyle\sum_{j=1}^{n} (-c_j)v_j</math>. |
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:::Since <math>W_1 \cap W_2 = \{ 0 \}</math>, then <math>\displaystyle\sum_{i=1}^{m} b_iu_i = \displaystyle\sum_{j=1}^{n} (-c_j)v_j = 0</math>. |
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:::Since <math>B_1, B_2</math> are linearly independent, then <math>b_i = (-c_j) = 0</math>. |
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:::For <math>\displaystyle\sum_{i=1}^{m} b_iu_i + \displaystyle\sum_{j=1}^{n} c_jv_j = 0</math>, then <math>b_i = c_j = 0</math>. |
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:::Then <math>B_1 \cup B_2</math> is linearly independent. |
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::Then <math>B_1 \cup B_2</math> is a basis of <math>span(W_1 \cup W_2)</math>. |
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:Since <math>\left| B_1 \cup B_2 \right| = \left| B_1 \right| + \left| B_2 \right|</math>, then <math>dim(span(W_1 \cup W_2)) = dim(W_1) + dim(W_2)</math>. ''Q.E.D.'' |
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==Nikita== |
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==Nikita== |
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[[File:1014.240.pdf]] |
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==Scanned Tutorial Notes by [[User Boyang.wu|Boyang.wu]]== |
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[[File:Tut6.pdf]] |
Welcome to Math 240! (additions to this web site no longer count towards good deed points)
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#
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Week of...
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Notes and Links
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1
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Sep 8
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About This Class, What is this class about? (PDF, HTML), Monday, Wednesday
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2
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Sep 15
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HW1, Monday, Wednesday, TheComplexField.pdf,HW1_solutions.pdf
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3
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Sep 22
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HW2, Class Photo, Monday, Wednesday, HW2_solutions.pdf
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4
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Sep 29
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HW3, Wednesday, Tutorial, HW3_solutions.pdf
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5
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Oct 6
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HW4, Monday, Wednesday, Tutorial, HW4_solutions.pdf
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6
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Oct 13
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No Monday class (Thanksgiving), Wednesday, Tutorial
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7
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Oct 20
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HW5, Term Test at tutorials on Tuesday, Wednesday
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8
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Oct 27
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HW6, Monday, Why LinAlg?, Wednesday, Tutorial
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9
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Nov 3
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Monday is the last day to drop this class, HW7, Monday, Wednesday, Tutorial
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10
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Nov 10
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HW8, Monday, Tutorial
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11
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Nov 17
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Monday-Tuesday is UofT November break
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12
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Nov 24
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HW9
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13
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Dec 1
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Wednesday is a "makeup Monday"! End-of-Course Schedule, Tutorial
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F
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Dec 8
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The Final Exam
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Register of Good Deeds
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![Class Photo](/images/thumb/6/6b/14-240-ClassPhoto.jpg/310px-14-240-ClassPhoto.jpg) Add your name / see who's in!
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Boris
Elementary and (Not So Elementary) Errors in Homework
(1) Bad Notation
Consider these three matrices:
![{\displaystyle 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/43940ba5803b8280e842a0a8fbf117050282035b)
We want to equate
to the set of all symmetric
matrices. Here is the wrong way to write this:
.
Firstly,
is the set of all linear combinations of
. To equate it to a single symmetric
matrix makes no sense. Secondly, the elements
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:
where
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
is a linear combination of the vectors
in
.
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
, determine if the matrix
is in span
.
Proof:
We show that
.
- We show that
.
- Assume that
.
- Let
.
- Then
.
- Then
.
- Since
and the entries of the matrix are from
, then
.
- Then
.
- Then
.
- We show that
.
- Assume to the contrary that
.
- Then
.
- Then this system of linear equations has a solution:
![{\displaystyle (11)c_{1}+c_{3}=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/78273a3f0fe933fdfe674f76ad77f08e2409a534)
![{\displaystyle (21)-c_{1}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d42cca7e74c6f2324e15dbf0045c2d54762a2378)
![{\displaystyle (12)c_{2}+c_{3}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e66edef2ea6aefb0c67742bffc52afcb6e25710f)
.
- When solving this system, we see that it has no solution.
- This contradicts the assumption that it has a solution.
- Then
.
- Then
. Q.E.D.
A Field Problem
Find the solution to
in
.
Note that a polynomial of
degree has at most
solutions.
Algorithm:
We find the solution to
in
.
- Since in
, then
.
- Since
is additive inverse of
, then
.
- By the result that we proved in Question 2 of Homework 1, then
.
- Then
are the solutions.
A Dimension Problem
Let
be subspaces of a finite dimensional vector space
over a field
where
. Then
.
Proof:
We show that
.
- Since
is finite dimensional, then
are finite dimensional.
- Then we can let
be a basis of
and
be a basis of
.
- We show that
is a basis of
.
- We show that
.
- We show that
.
- Since
, then
.
- We show that
.
- Since
, then
.
- Since
, then
.
- Then
and
.
- Then
.
- We show that
is linearly independent.
- Let
where
.
- Then
.
- Since
, then
.
- Since
are linearly independent, then
.
- For
, then
.
- Then
is linearly independent.
- Then
is a basis of
.
- Since
, then
. Q.E.D.
Nikita
File:1014.240.pdf
Scanned Tutorial Notes by Boyang.wu
File:Tut6.pdf