14240/TutorialOctober14

Contents 
Boris
Elementary and (Not So Elementary) Errors in Homework
(1) Bad Notation
Consider these three matrices:
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:
 .
 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.