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Homework 2 collaboration area

Here is the Homework 2 collaboration area. Since HWK 2 is due the Wednesday after Labor Day, I won't have a chance to answer questions on Monday like usual. I will answer any and all questions here on the Rhea on Tuesday with help from Eun Young Park. - Steve Bell



Back to MA527, Fall 2013

Questions from a student :

When finding a basis, does it always have to be fully reduced? For example, if you have a basis [4 -2 6] does it need to be reduced to [2 -1 3] or is either answer acceptable? Jones947

Answer from Eun Young :

No, it doesn't need to be reduced. If { [4 -2 6] } is a basis for some vector space $ V $, then { [2 -1 3] } is also a basis for $ V $ and vice versa. If v belongs to span{[4 -2 6 ]}, v = c[4 -2 6] = 2c [2 -1 3 ] for some c. Hence, v belongs to span{[2 -1 3]}. The opposite direction is same. So, span{[4 -2 6]}= span{[2 -1 3]}. A basis for a vector space is not unique but a dimension of a vector space is unique.

Question from a student:

From what I understand, a basis is a set of vectors that can be used to create any vector in the span. So for example, if the basis is [1 0] [0 1], then the span could be [1 0] [0 1] [2 2] [2 0]. Is that correct?

Answer from Steve Bell:

The span is ALL vectors you get by taking linear combinations. Hence, the span is x*[1,0] + y*[0,1] = [x,y] as x and y range over all possible values, i.e., the span is R^2.

Question from a student:

I also have a question about how to find the basis. Should I be reducing the matrix to RREF? Or leave it when I get it to REF? I do not think it should matter either way, as the basis can still make up any vector in the set even if I use RREF. My answer is different from the back of the book but I think that is typical for linear algebra problems.

Answer from Steve Bell:

I tried to explain in class why the RREF matrix is better. It is easier to read off information from it and it has the virtue of being unique. But the plain old RE form does give you the answers with a little extra work (like back substitution). You are right that it is not uniquely determined, so you might get the right answer, even if it looks different than the one given in the back of the book.

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