What Determines the Dimension of Subspace S in R^4?

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Homework Statement



Find a basis for the subspace

S = {(a+2b,b-a+b,a+3b) | a,b \in R } \subseteq R^4

What is the dimension of S?

Homework Equations





The Attempt at a Solution



a(1,0,-1,1) + b(2,1,1,3) , a,b \in R

span { (1,0,-1,1) , (2,1,1,3) }

So I put (1,0,-1,1) as V1 and (2,1,1,3) as V2 and then formed a matrix with V1 and V2 in columns. Then i reduced it to row echelon form. The column corresponding to the leading entry form a basis which is essentially just (1,0,-1,1) and (2,1,1,3) . Thus the dimension is just 2. Am I doing the right thing because I don't have an answer to refer to.
 
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There has to be an error in your definition for this set.
S = {(a+2b,b-a+b,a+3b) | a,b R } R^4

Your vector has only three coordinates, which is too few for it to be in R^4. Are you missing a comma in there?
 
Hi Jeffrey,
jeffreylze said:

Homework Statement



Find a basis for the subspace

S = {(a+2b,b,[/color]-a+b,a+3b) | a,b \in R } \subseteq R^4

What is the dimension of S?

The Attempt at a Solution



a(1,0,-1,1) + b(2,1,1,3) , a,b \in R

span { (1,0,-1,1) , (2,1,1,3) }

So I put (1,0,-1,1) as V1 and (2,1,1,3) as V2 and then formed a matrix with V1 and V2 in columns. Then i reduced it to row echelon form. The column corresponding to the leading entry form a basis which is essentially just (1,0,-1,1) and (2,1,1,3) . Thus the dimension is just 2. Am I doing the right thing because I don't have an answer to refer to.
The first step you have done is shown that S = span {(1,0,-1,1), (2,1,1,3)}. Then if (1,0,-1,1) and (2,1,1,3) are linearly independent, they form a basis (recall the definition) for S.

It is correct that n vectors are linearly independent if and only if the matrix formed with them as its columns has rank n, but in this case it is probably simpler just to recognise that two non-zero vectors are linearly independent iff they're not scalar multiples of each other.
 
Mark44 said:
There has to be an error in your definition for this set.


Your vector has only three coordinates, which is too few for it to be in R^4. Are you missing a comma in there?

Oops, my bad. Yes, there is a comma in there. Should be, S = {(a+2b,b,-a+b,a+3b) | a,b R } R^4
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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