Subspaces, etc.

[hotcommodity]

I need to know if I understand subspaces correctly so if anyone can point out any flaws in my thinking it would be well appreciated...

 

If you have 2 vectors, each with 3 real numbers, the vectors exist in [math]R^3[/math]. Since there can only be 2 pivots at the most, the subspace that the vectors generate is in [math]R^2[/math], but not any higher.

 

A question here: Is there any way that the subspace could exist in [math]R^3[/math]?

 

If you have a set of vectors and you're looking to generate a basis for a subspace, the vectors must be linearly independent, that is, having the trivial solution [math] 0 [/math] as its only solution when the vectors are augmented into A[math]x[/math]=[math]0[/math] form. Additionally, the vectors must span [math]R^m[/math] for an mXn matrix ( if the vectors were linearly dependent, they would not span [math]R^m[/math]).

 

Given these conditions we can generate the basis for a nullspace and/or column space. The vectors that form a basis for the column space of A are simply the columns in the original matrix A that have a pivot in their columns when row reduced. The vectors that form a basis for the nullspace of A is the solution to the homogeneous equation A[math]x[/math]=[math]0[/math].

 

A question here: If the conditions for the basis of a nullspace are the vector solutions given in parametric form, then how can the original vectors augmented to form the matrix A have been linearly independent? When I get the solution, it looks as tho' my text ignores the parameters in the solution and gives the vectors as the basis for a nullspace.

 

Finally, if we want to find a vector in the column space we can simply choose any column of A as a vector and it will be in the span generated by the columns of A. If we want to find a vector in the nullspace then we solve for the equation A[math]x[/math]=[math]0[/math] and choose any real number for the parameter.

 

Last question: Can the zero vector count as a subspace or a nullspace?

[Dave]

If one takes two vectors in [math]v_1, v_2 \in \mathbb{R}^3[/math], then indeed the subspace generated by the spanning set V of v₁ and v₂ has dimension of at most 2. If the dimension is indeed 2 (i.e. v₁ and v₂ are linearly independent), then there exists a bijection between [math]\mathbb{R}^2[/math] and V. So V is "equal" to [math]\mathbb{R}^2[/math] up to a bijection. This doesn't mean the spanning space actually is [math]\mathbb{R}^2[/math], so be careful as to how you phrase your wording in this regard.

 

Secondly, is it possible for this subspace to be equal to [math]\mathbb{R}^3[/math]? No. Clearly [math]\mathbb{R}^3[/math] has dimension 3. As discussed previously, the maximal dimension of V is 2, and so the complement of V in [math]\mathbb{R}^3[/math] must be non-empty.

 

I don't quite have the time to answer your other questions right now, but I will hopefully be able to do so a little later. Does this help at all?

[hotcommodity]

It does help, and I appreciate you taking the time to answer. I finally get that if you have 2 vectors in [math]\mathbb{R}^3[/math] that they are in [math]\mathbb{R}^3[/math] but do not span [math]\mathbb{R}^3[/math]. That was one of things I was confused about, and it makes sense now that if you only have 2 vectors the most they can span is [math]\mathbb{R}^2[/math]. I'm not sure what you mean by "bijection" however. Beyond that I think I have a decent grasp on the rest, and again I appreciate your help