Questions and Answers |
Question (Sep. 29):
I was wondering if there was an easy way to prove that vectors form
a spanning set. For example, one question from one of the previous
exams says something along the lines of: Is it true that {(1,0) ,
(1,1)} spans the two-dimensional real numbers? Is there some way to
prove this or is it simply a matter of being able to see that this
is true?
Answer: you must show that for any vector (x,y), there exist scalars a,b such that (x,y) = a(1,0) + b(1,1).This can be done by solving it for a and b. For instance, in this example, we get b=y, a=x-y. If the vectors do not form a spanning set, e.g. {(1,2), (2,4)}, then the above does not have a solution, e.g. (x,y) = a(1,2) + b(2,4)cannot be solved for a and b (try!). For now, this is the method you will have to use. Later, we will learn better methods. |
Question (Nov. 5):
As I was studying for the test I became quite confused. My question
is, are: row canonical form, Echelon form, triangular form, and row
reduced form all refering to the same thing? E.i. to the following
matrix?
1 1 5 4 7 0 4 8 6 4 0 0 1 2 -2 0 0 0 1 5 0 0 0 0 1 Answer: No, they don't refer to the same thing.
1 2 3 triangular: no 1 0 1 Echelon: no 0 0 1 row canonical: no 1 2 3 triangular: yes 0 0 1 Echelon: no 0 0 1 row canonical: no 1 2 3 triangular: yes 0 2 1 Echelon: yes 0 0 1 row canonical: no 1 0 0 triangular: yes 0 2 0 Echelon: yes 0 0 1 row canonical: no 1 0 0 triangular: yes 0 1 0 Echelon: yes 0 0 1 row canonical: yes 1 0 2 triangular: yes 0 1 2 Echelon: yes 0 0 0 row canonical: yes 1 2 3 4 triangular: n/a (no) 0 0 1 5 Echelon: no 0 0 1 6 row canonical: no 1 2 3 4 triangular: n/a (no) 0 0 1 5 Echelon: yes 0 0 0 0 row canonical: no 1 2 0 4 triangular: n/a (no) 0 0 2 5 Echelon: yes 0 0 0 0 row canonical: no 1 2 0 4 triangular: n/a (no) 0 0 1 5 Echelon: yes 0 0 0 0 row canonical: yesI hope these are enough examples to illustrate all 3 concepts. |