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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 |
<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 |
represent? Rational numbers? Real numbers? Members of the field of two elements? The following way of writing erases those issues: |
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issues: |
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Show both (a) and (b) to be on the safe side. |
Show both (a) and (b) to be on the safe side. |
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====Problem 5h) of Homework 3 for all Fields==== |
====Problem 5h) of Homework 3 for all Fields==== |
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Revision as of 00:30, 15 October 2014
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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.
