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[sspeedy]

you guys helped me out with the probability questions last night, so i thought someone might be able to help me out here.

The book gives an example. It says V = R(sub3). S = {v1, v12, v3, v4, v5} (1,...,5 are all subs).

It specifies each of the vectors and then says "we find that S spans Rsub3 and we now wish to find a subset of S that is a basis for Rsub3.

But then, in the hw exercises, there's this question:

Which of the following sets of vectors are bases for Rsub3?
a) {a set of 2vectors}
b){set of 4 vectors}
c){set of 3 vectors}
d){set of 4 vectors}

(I purposely didn't give the exact vectors.)
So for the solution to this question, the TA seems automatically disregard a, b, and c since they don't have 3 vectors.

If we can do that, then how come the book's example gives FIVE vectors for Rsub3? since it's five set of vectors, shouldn't it be automatically be disregarded as not being a bases for Rsub3 just like the hw exercise?

I ask because if given a set of vectors, should I automatically disregard it if its # of vectors doesn't match n from Rsubn? Is this a correct way of going about?
Or am i just not seeing something here?!

Thanks in advance.

 

[Dilbert]

A basis for an n-dimensional space consists of n linearly independent vectors. To put it in another way: A basis is a minimal set of vectors which span a space.

In this example, since you are working in three-dimensional space (R sub 3), you know that any basis can only consist of three vectors. A set of four or five vectors may span the space, but they are not a minimal set. (Obviously two vectors cannot span a three-dimensional space, so that's easy to rule out.)

The problem gets harder if they give you more than one solution with three vectors. In that case, all of them may be valid bases, but the trick is usually that at least one of them will have two or more vectors which are not linearly independent, meaning that they do not span the space.

Hope this answers your question...