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2014-10-29T15:33:40Z

H ZhuKL:

File:1029.240-0.pdf

2014-10-29T15:32:47Z

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File:1029.240-3.pdf

2014-10-29T15:30:30Z

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File:1029.240-2.pdf

2014-10-29T15:27:23Z

H ZhuKL: Wednesday Oct 29, 2014

Wednesday Oct 29, 2014

14-240/Tutorial-October14

2014-10-27T23:44:52Z

H ZhuKL:

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==Boris==

====Elementary and (Not So Elementary) Errors in Homework====

(1) '''Bad Notation'''

Let
::::::::<math>
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}
</math>

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:

::::::::::::<math>
span(M_1, M_2, M_3) =
\begin{pmatrix}
a & b \\
b & c \\
\end{pmatrix}
</math>.

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:

::::::::<math>
span(M_1, M_2, M_3) = \{
\begin{pmatrix}
a & b \\
b & c \\
\end{pmatrix}
:a, b, c \in F \}
</math>

where <math>F</math> 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 <math>(-2, 2, 2)</math> is a linear combination of the vectors <math>(- (1, 2, -1), (-3, -3, 3)</math> in <math>R^3</math>.

Show both (a) and (b) to be on the safe side.

====Problem 5h) in Homework 3 for all Fields====

For an arbitrary field <math>F</math>, determine if the matrix
<math>
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}
</math>
is in span
<math>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}
\}
</math>.

'''Proof:'''

We show that
<math>
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}
\in span(S) \iff char(F)=2</math>.

:We show that <math>char(F)=2 \implies
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}
\in span(S)
</math>.
::Assume that <math>char(F)=2</math>.

::Let <math>c_1=0, c_2=1, c_3=1</math>.

::Then <math>c_1, c_2, c_3 \in F</math>.

::Then <math>
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}
</math>.

::Since <math>char(F)=2</math> and the entries of the matrix are from <math>F</math>, then <math>0=2</math>.

::Then <math>
\begin{pmatrix}
1 & 2 \\
0 & 1 \\
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}
\in span(S)
</math>.

::Then <math>
char(F)=2 \implies
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}
\in span(S)
</math>.

:We show that <math>char(F) \neq 2 \implies
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}
\notin span(S)
</math>.
::Assume to the contrary that <math>
char(F) \neq 2 \and
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}
\in span(S)
</math>.

::Then <math>\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}
</math>.

::Then this system of linear equations has a solution:

:::<math>(11)c_1+c_3=1</math>

:::<math>(21)-c_1=0</math>

:::<math>(12)c_2+c_3=0</math>

:::<math>(22)c_2=1</math>.

::When solving this system, we see that it has no solution.

::This contradicts the assumption that it has a solution.

::Then <math>
char(F) \neq2 \implies
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}
\notin span(S)
</math>.

:Then <math>
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}
\in span(S) \iff char(F)=2</math>. ''Q.E.D.''

====A Field Problem====

Find the solution to <math>x^2 = -2</math> in <math>Z_{11}</math>.

Note that a polynomial of <math>n</math> degree has at most <math>n</math> solutions.

''Algorithm:''

We find the solution to <math>x^2 = -2</math> in <math>Z_{11}</math>.

:Since in <math>Z_{11}, -2 = 9</math>, then <math>x^2 = 9</math>.

:Since <math>-9</math> is additive inverse of <math>9</math>, then <math>x^2 - 9 = 9 - 9 = 0</math>.

:By the result that we proved in ''Question 2 of Homework 1'', then <math>(x^2 - 9) = (x - 3)(x + 3) = 0</math>.

:Then <math>x = \pm 3</math> are the solutions.

====A Dimension Problem====

Let <math>W_1, W_2</math> be subspaces of a finite dimensional vector space <math>V</math> over a field <math>F</math> where <math>W_1 \cap W_2 = \{0\}</math>. Then

<math>dim(span(W_1 \cup W_2)) = dim(W_1) + dim(W_2)</math>.

''Proof:''

We show that <math>dim(span(W_1 \cup W_2)) = dim(W_1) + dim(W_2)</math>.

:Since <math>V</math> is finite dimensional, then <math>W_1, W_2</math> are finite dimensional.

: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>.

:We show that <math>B_1 \cup B_2</math> is a basis of <math>span(W_1 \cup W_2)</math>.

::We show that <math>span(B_1 \cup B_2) = span(W_1 \cup W_2)</math>.

:::We show that <math>span(B_1 \cup B_2) \subset span(W_1 \cup W_2)</math>.

::::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>.

:::We show that <math>span(W_1 \cup W_2) \subset span(B_1 \cup B_2)</math>.

::::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>.

::::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>.

::::Then <math>(W_1 \cup W_2) \subset span(B_1 \cup B_2)</math>.

:::Then <math>span(B_1 \cup B_2) = span(W_1 \cup W_2)</math>.

::We show that <math>B_1 \cup B_2</math> is linearly independent.

:::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>.

:::Then <math>\displaystyle\sum_{i=1}^{m} b_iu_i = \displaystyle\sum_{j=1}^{n} (-c_j)v_j</math>.

:::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>.

:::Since <math>B_1, B_2</math> are linearly independent, then <math>b_i = (-c_j) = 0</math>.

:::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>.

:::Then <math>B_1 \cup B_2</math> is linearly independent.

::Then <math>B_1 \cup B_2</math> is a basis of <math>span(W_1 \cup W_2)</math>.

: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.''

==Nikita==
[[File:1014.240.pdf]]

File:1014.240.pdf

2014-10-27T23:44:29Z

H ZhuKL: Tutorial

Tutorial

14-240/Classnotes for Monday October 27

2014-10-27T15:01:56Z

H ZhuKL:

{{14-240/Navigation}}

A video link for today's riddle is [https://media.library.utoronto.ca/play.php?DJ6CPFxByy2J&id=8503&access=public here], and the transcript is [https://cmc.math.ca/home/videos/game-of-15-and-isomorphisms/ here]. Enjoy!

{{14-240:Dror/Students Divider}}
The pdf notes of Linear Tranformation class of today:[[File:1027.240-lt.pdf]]

14-240/Classnotes for Wednesday October 22

2014-10-27T15:01:18Z

H ZhuKL:

The pdf notes for Wednesday class is[[File:1021.240.pdf]]{{14-240/Navigation}}

By Yue Jiang--[[User:Yue.Jiang|Yue.Jiang]] ([[User talk:Yue.Jiang|talk]]) 11:29, 22 October 2014 (EDT)

----

<gallery>
File:Oct 22 note1.jpeg|Oct 22 note1

File:Oct 22 note2.jpg|Oct 22 note2

File:Oct 22 note3.jpg|Oct 22 note3

</gallery>

14-240/Classnotes for Monday October 27

2014-10-27T15:00:40Z

H ZhuKL:

14-240/Classnotes for Monday October 27

2014-10-27T14:59:31Z

H ZhuKL:

{{14-240/Navigation}}
[[File:1027.240-lt.pdf]]
A video link for today's riddle is [https://media.library.utoronto.ca/play.php?DJ6CPFxByy2J&id=8503&access=public here], and the transcript is [https://cmc.math.ca/home/videos/game-of-15-and-isomorphisms/ here]. Enjoy!

{{14-240:Dror/Students Divider}}

File:1027.240-lt.pdf

2014-10-27T14:58:50Z

H ZhuKL:

14-240/Classnotes for Wednesday October 22

2014-10-22T15:50:44Z

H ZhuKL:

[[File:1021.240.pdf]]{{14-240/Navigation}}

By Yue Jiang--[[User:Yue.Jiang|Yue.Jiang]] ([[User talk:Yue.Jiang|talk]]) 11:29, 22 October 2014 (EDT)

----

<gallery>
File:Oct 22 note1.jpeg|Oct 22 note1

File:Oct 22 note2.jpg|Oct 22 note2

File:Oct 22 note3.jpg|Oct 22 note3

</gallery>

File:1021.240.pdf

2014-10-22T15:49:47Z

H ZhuKL: October 22, 2014 Linear Transformation

October 22, 2014
Linear Transformation

Drorbn - User contributions [en]

File:1029.240-1(1).pdf

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File:1029.240-3.pdf

File:1029.240-2.pdf

14-240/Tutorial-October14

File:1014.240.pdf

14-240/Classnotes for Monday October 27

14-240/Classnotes for Wednesday October 22

14-240/Classnotes for Monday October 27

14-240/Classnotes for Monday October 27

File:1027.240-lt.pdf

14-240/Classnotes for Wednesday October 22

File:1021.240.pdf