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

In summary, the conversation discusses finding a basis and determining the dimension of a subspace S, which is defined as {(a+2b,b,-a+b,a+3b) | a,b R } R^4. The method used involves finding linearly independent vectors by forming a matrix with two given vectors and reducing it to row echelon form, ultimately resulting in a basis of S with dimension 2.
  • #1
jeffreylze
44
0

Homework Statement



Find a basis for the subspace

S = {(a+2b,b-a+b,a+3b) | a,b [tex]\in[/tex] R } [tex]\subseteq[/tex] 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 [tex]\in[/tex] 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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  • #2
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?
 
  • #3
Hi Jeffrey,
jeffreylze said:

Homework Statement



Find a basis for the subspace

S = {(a+2b,b,-a+b,a+3b) | a,b [tex]\in[/tex] R } [tex]\subseteq[/tex] 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 [tex]\in[/tex] 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.
 
  • #4
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
 

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

What is a subspace?

A subspace is a subset of a vector space that satisfies the properties of being closed under addition and scalar multiplication. In other words, if you add two vectors or multiply a vector by a scalar within a subspace, the result will still be within that subspace.

Why is it important to find a basis for a subspace?

Finding a basis for a subspace allows us to represent all vectors within that subspace using a linear combination of the basis vectors. This can help simplify calculations and provide a better understanding of the structure of the subspace.

How do you find a basis for a subspace?

To find a basis for a subspace, we can use the method of Gaussian elimination to reduce the vectors in the subspace to their simplest form. The resulting vectors will form the basis for the subspace.

Can a subspace have more than one basis?

Yes, a subspace can have multiple bases. This is because there can be multiple ways to represent the same subspace using different combinations of basis vectors.

Is it possible for a basis of a subspace to contain linearly dependent vectors?

No, a basis for a subspace must contain linearly independent vectors. This ensures that the basis set is unique and there is no redundancy in the representation of the subspace.

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