# Disjoint sets/basis

• Jul 8th 2010, 05:35 AM
Disjoint sets/basis
In a past paper im stuck on this question

http://i32.tinypic.com/2u9izv9.jpg

Do i have to seperate them to linear dependency?
• Jul 9th 2010, 06:18 PM
Ackbeet
Are those definitions of vectors being a vector normal to the plane given? If so, the vectors $v_{j}$ are not unique, unless it's understood they go through the origin and are unit vectors, or something like that.

Part i) does not parse grammatically - I have no idea what's being asked there. "de" is not a word in the English language, that I'm aware of.

Part ii) makes sense only if you've correctly solved part i).

A little more clarification would be nice!
• Jul 10th 2010, 05:04 AM
Quote:

Originally Posted by Ackbeet
Are those definitions of vectors being a vector normal to the plane given? If so, the vectors $v_{j}$ are not unique, unless it's understood they go through the origin and are unit vectors, or something like that.

Part i) does not parse grammatically - I have no idea what's being asked there. "de" is not a word in the English language, that I'm aware of.

Part ii) makes sense only if you've correctly solved part i).

A little more clarification would be nice!

I have no idea sorry.

I think de is the.
• Jul 10th 2010, 06:11 AM
Also sprach Zarathustra
I think you should build a matrix:

3 2 1
1 2 3
5 -1 2
2 6 -8
• Jul 10th 2010, 07:07 AM
Quote:

Originally Posted by Also sprach Zarathustra
I think you should build a matrix:

3 2 1
1 2 3
5 -1 2
2 6 -8

Thanks thats what ive done but theres no linear dependency between any of them?

Is this the next step?

Or have i calculated it wrong?
• Jul 10th 2010, 07:25 AM
HallsofIvy
Quote:

In a past paper im stuck on this question

http://i32.tinypic.com/2u9izv9.jpg

Do i have to seperate them to linear dependency?

The problem says "being a basis". Yes, every basis is independent- if these are not already a basis for whatever set they span (I presume that these span some subset of $R^3$), then they must be dependent.

In each of those equations, you can solve for one of the variables, say z, in terms of the other two:
z= -3x- 2y, z= -(1/3)x- (2/3)y, z= -(5/2)x+ (1/2)y, and z= (1/4)x+ (3/4)y. That is, the subspaces spanned by each one separately are
(x, y, -3x- 2y)= x(1, 0, -3)+ y(0, 1, -2)
(x, y, -(1/3)x- (2/3)y)= x(1, 0, -1/3)+ y(0, 1, -2/3)
(x, y, -(5/2)x+ (1/2)y)= x(1, 0, -5/2)+ y(0, 1, 1/2)
(x, y, (1/4)x+ (3/4)y= x(1, 0, 1/4)+ y(0, 1, 3/4).

Of course, if it is true that you can separate these into sets that each span the same subspace, then that suspace must have dimension 2- you can separate them into two sets of two vectors each. I would suggest just trying pairs: do (v1, v2) and (v3, v4) span the same subspace? Do (v1, v3) and (v2, v4)? What about (v1, v4) and (v2, v3)?
• Jul 13th 2010, 05:20 AM
Quote:

Originally Posted by HallsofIvy

The problem says "being a basis". Yes, every basis is independent- if these are not already a basis for whatever set they span (I presume that these span some subset of $R^3$), then they must be dependent.

In each of those equations, you can solve for one of the variables, say z, in terms of the other two:
z= -3x- 2y, z= -(1/3)x- (2/3)y, z= -(5/2)x+ (1/2)y, and z= (1/4)x+ (3/4)y. That is, the subspaces spanned by each one separately are
(x, y, -3x- 2y)= x(1, 0, -3)+ y(0, 1, -2)
(x, y, -(1/3)x- (2/3)y)= x(1, 0, -1/3)+ y(0, 1, -2/3)
(x, y, -(5/2)x+ (1/2)y)= x(1, 0, -5/2)+ y(0, 1, 1/2)
(x, y, (1/4)x+ (3/4)y= x(1, 0, 1/4)+ y(0, 1, 3/4).

Of course, if it is true that you can separate these into sets that each span the same subspace, then that suspace must have dimension 2- you can separate them into two sets of two vectors each. I would suggest just trying pairs: do (v1, v2) and (v3, v4) span the same subspace? Do (v1, v3) and (v2, v4)? What about (v1, v4) and (v2, v3)?

Thanks forwriting tht out, i thought span the space was the same as linear independency?

They are all inearly independent from one another so i dont see how you can seperate them?
• Jul 14th 2010, 05:28 AM
Ackbeet
A set A of vectors spans a space if you can write any vector in the space as a linear combination of vectors in A. The vectors in A need not be linearly independent (i.e., you might have superfluous vectors in A). So, A spanning a space does not necessarily imply that the vectors in A are linearly independent.

A set B of vectors that are linearly independent might not span a particular vector space. In particular, if the number of vectors in B is less than the dimension of the vector space, you're guaranteed not to span it. So, linear independence does not necessarily imply that the vectors in B span the vector space.

However, if you have a set C of n linearly independent vectors in a vector space of dimension n, then C is a basis for the vector space, and spans it. A basis is kind of the "minimum required" set of vectors that spans a space.

Hopefully, this will clear up any confusion about the relationship between spanning and linear independence.
• Jul 18th 2010, 06:24 AM