14-240/Tutorial-October14

DBN: Classes: 2014-15: 14-240 / Navigation Welcome to Math 240! (additions to this web site no longer count towards good deed points) # 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 Monday-Tuesday is UofT November break 12 Nov 24 HW9 13 Dec 1 Wednesday is a "makeup Monday"! End-of-Course Schedule, Tutorial F Dec 8 The Final Exam Register of Good Deeds Add your name / see who's in!

Boris

Elementary and (Not So Elementary) Errors in Homework

(1) Bad Notation

Let [math]\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} }[/math] be matrices.

We want to equate [math]\displaystyle{ span(M_1, M_2, M_3) }[/math] to the set of all symmetric [math]\displaystyle{ 2 \times 2 }[/math] matrices. Here is the wrong way to write this:

[math]\displaystyle{ span(M_1, M_2, M_3) = \begin{pmatrix} a & b \\ b & c \\ \end{pmatrix} }[/math].

Firstly, [math]\displaystyle{ span(M_1, M_2, M_2) }[/math] is the set of all linear combinations of [math]\displaystyle{ M_1, M_2, M_3 }[/math]. To equate it to a single

symmetric [math]\displaystyle{ 2 \times 2 }[/math] matrix makes no sense. Secondly, the elements [math]\displaystyle{ 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]\displaystyle{ span(M_1, M_2, M_3) = \{ \begin{pmatrix} a & b \\ b & c \\ \end{pmatrix} :a, b, c \in F \} }[/math] where [math]\displaystyle{ F }[/math] is an arbitrary field.

(2) Algorithm vs. Proof

When solving a problem that requires a solution to a linear equation, it is not always obvious if 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, for problems such as this:

Determine if the vector [math]\displaystyle{ (-2, 2, 2) }[/math] is a linear combination of the vectors [math]\displaystyle{ (- (1, 2, -1), (-3, -3, 3) }[/math] in [math]\displaystyle{ R^3 }[/math].

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

Problem 5h) of Homework 3 for all Fields

For a field [math]\displaystyle{ F }[/math], determine if the matrix [math]\displaystyle{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} }[/math] is in span [math]\displaystyle{ S=\{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \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]\displaystyle{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \in span(S) \iff char(F)=2 }[/math].

We show that [math]\displaystyle{ char(F)=2 \implies \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \in span(S) }[/math].

Assume that [math]\displaystyle{ char(F)=2 }[/math].

Let [math]\displaystyle{ c_1=0, c_2=1, c_3=1 }[/math].

Then [math]\displaystyle{ c_1, c_2, c_3 \in F }[/math].

Then [math]\displaystyle{ c_1 \begin{pmatrix} 1 & 0 \\ -1 & 0 \\ \end{pmatrix} + c_2 \begin{pmatrix} 0 & 1 \\ 0 & 1 \\ \end{pmatrix} + c_3 = 0 \begin{pmatrix} 1 & 0 \\ -1 & 0 \\ \end{pmatrix} + 1 \begin{pmatrix} 0 & 1 \\ 0 & 1 \\ \end{pmatrix} + 1 = \begin{pmatrix} 1 & 0 \\ -1 & 0 \\ \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ 0 & 1 \\ \end{pmatrix} }[/math]

[math]\displaystyle{ = \begin{pmatrix} 1 & 2 \\ 0 & 1 \\ \end{pmatrix} }[/math].

Since [math]\displaystyle{ char(F)=2 }[/math] and the entries of the matrix are from [math]\displaystyle{ F }[/math], then [math]\displaystyle{ 0=2 }[/math].

Then [math]\displaystyle{ \begin{pmatrix} 1 & 2 \\ 0 & 1 \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \in span(S) }[/math].

Then [math]\displaystyle{ char(F)=2 \implies \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \in span(S) }[/math].

We show that [math]\displaystyle{ char(F) \neq 2 \implies \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \notin span(S) }[/math].

Assume to the contrary that [math]\displaystyle{ char(F) \neq 2 \and \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \in span(S) }[/math].

Then [math]\displaystyle{ \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]\displaystyle{ (11)c_1+c_3=1 }[/math]

[math]\displaystyle{ (21)-c_1=0 }[/math]

[math]\displaystyle{ (12)c_2+c_3=0 }[/math]

[math]\displaystyle{ (22)c_2=1 }[/math].

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

This contradicts the assumption that it has a solution.

Then [math]\displaystyle{ char(F) \neq2 \implies \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \notin span(S) }[/math].

Then [math]\displaystyle{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \in span(S) \iff char(F)=2 }[/math]. Q.E.D.