How to find a basis of a subspace V = {(x1; x2;….; xn) | Σni=1 xi=0}

In summary, to find a basis for the given subspace V in Rn such that the sum of squares of each vector in the basis is equal to 1, we can solve for one variable in terms of the others in the single equation given. This reduces the dimension of the subspace by 1, so the basis will consist of n-1 vectors. Each vector in the basis can then be divided by its length to create a unit vector. This pattern can be applied for any value of n to find a basis for V with the desired properties.
  • #1
ahadmir
2
0
Given V = {(x1; x2;….; xn) | Σni=1 xi=0}
(sum of vectors is equal to zero) be a subspace of Rn. How can we find a basis of V such that for each vector {(x1; x2;….; xn) in the basis Σi=1n x2i=1 ( i.e. sum of squares is equal to 1).
 
Physics news on Phys.org
  • #2
For problems like this start by looking at the easy cases. If n= 2, we are looking at vectors <x1, x2> such that x1+ x2= 0 and are looking for a basis. That is the single equation x2= -x1 so <x1, -x1>= x1<1, -1>. That is, a basis consists of simply the single vector <1, -1>. (Generally, one equation reduces the dimension of the space by 1. Since the over all space, R2, has dimension 2, one equation reduces the dimension of the the subspace to 2- 1= 1.)
Since the length of a vector is "square root of sum of squares", the condition that the sum of the squares be 1 is exactly the same as requiring that the length of the vectors be 1. <1, -1> has length [itex]\sqrt{2}[/itex] so dividing that vector by [itex]\sqrt{2}[/itex] gives a unit vecotor: [itex]\{1/\sqrt{2}, -1/\sqrt{2}\}[/itex] is a basis.

Now try n= 3. Any vector in R3 can be written in the form <x1, x2, x3> and we are requiring that x1+ x2+ x3= 0. Again this is a single equation so our subspace has dimension 3- 1= 2. We can solve for x3, say, in terms of the other two: x3= -x1- x2 so any vector is of the form <x1, x2, -x1-x2>= <x1, 0, -x1)+ <0 x2, -x2>= x1<1, 0, -1>+ x2<0, 1, -1>. A basis is {<1, 0, -1>, <0, 1, -1>}. Again, those both have length [itex]\sqrt{2}[/itex] so you must divide by [itex]\sqrt{2}[/itex].

In R4, n= 4, we have x1+ x2+ x3+ x4= 0 or x4= -x1-x2-x3 so you should be able to see that a basis is {<1, 0, 0, -1>, <0, 1, 0, -1>, <0, 0, 1, -1>}. Again you will have to divide each by its length, [itex]\sqrt{2}[/itex].

Do you see the pattern now?
 

1. What is a basis of a subspace?

A basis of a subspace is a set of vectors that can be used to represent all other vectors within that subspace. It is a fundamental concept in linear algebra and is used to describe the structure of a vector space.

2. How do I find a basis of a subspace?

To find a basis of a subspace, you can use the following steps:

  1. Find a set of linearly independent vectors that span the subspace.
  2. If necessary, use the Gram-Schmidt process to orthogonalize the vectors.
  3. Normalize the vectors to obtain a set of orthonormal vectors.
  4. This set of orthonormal vectors is the basis of the subspace.

3. What does the notation V = {(x1; x2;….; xn) | Σni=1 xi=0} mean?

This notation represents a subspace V in which the vectors are of the form (x1, x2, ..., xn) and their components sum up to zero. This means that the subspace is a hyperplane passing through the origin in n-dimensional space.

4. Can a subspace have more than one basis?

Yes, a subspace can have multiple bases. The number of bases for a subspace is not unique and can vary depending on the vectors chosen to span the subspace.

5. How is a basis of a subspace related to linear independence?

A basis of a subspace is a set of linearly independent vectors. This means that none of the vectors in the basis can be expressed as a linear combination of the other vectors in the basis. Additionally, any vector in the subspace can be written as a unique linear combination of the basis vectors.

Similar threads

Replies
2
Views
739
  • Linear and Abstract Algebra
Replies
3
Views
3K
  • Linear and Abstract Algebra
Replies
8
Views
2K
  • Linear and Abstract Algebra
Replies
9
Views
940
  • Linear and Abstract Algebra
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
0
Views
418
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Replies
4
Views
3K
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
952
Back
Top