Is it possible to determine a basis for non-subspace spaces?

In summary: Since those two vectors form a basis for the subspace that is the plane x- y+ z= 0, the set (not subspace) of vectors <x, y, z>= <x, x, 0>+ <0, z, z>+ <0, 0, 2> is a basis for the set of vectors representing the plane x- y+ z= 2.In summary, the concept of a basis can only be applied to vector spaces, and a plane not through the origin is an example of a "coset" in a vector space. It is possible to define a set of operations on a coset that makes it a vector space with a basis, but it will not be
  • #1
Kubilay Yazoglu
8
0
The title may seem a little confusing and possibly stupid :D

What I mean is like a plane doesn't go through the origin? Can we describe a basis for this? If so, how?
 
Physics news on Phys.org
  • #2
Bases are defined for vector spaces. If you do not have a vector space, the concept of a basis is meaningless.
 
  • #3
A plane not through the origin is an example of a "coset" in a vector space. A coset of a subspace is the subspace translated by a constant vector. If the coset C of a vector space V doesn't contain the zero vector of V then C can't be a vector space with respect to the operations used in V.

A set of things may not be a vector space under one set of operations and yet can become a vector space under a different set of operations. You can define a set of operations on a coset C in a vector space V that make the coset a vector space. Assume the coset is fomed by adding the vector b to each vector in the subspace S. You can define a new set of operations on C. The basic idea is to un-translate vectors involved in the operations by subtracting b from them. Do the operations on the un-translated vectors with the operations used in V. Then translate the result by adding b to it. A coset with this new set of operations would be a vector space and it could have a basis. However the coset would not be a subspace of the vector space V because subspace of a vector space V must satisfy properties that specify using exactly same operations that are used in V.
 
  • Like
Likes Kubilay Yazoglu and 1MileCrash
  • #4
Maybe you want to translate so that the vector space goes to the origin, find a base and translate. Then the space is generated by a linear combination plus translation.
 
  • #5
Stephen Tashi said:
et C of a vector space V doesn't contain the zero vector of V then C can't be a vector space with respect to the operations used in V.
 
  • #6
Can you determine a basis for the plane x − y + z = 2 that spans only the plane?

This is the original question. I forgot the "only" part. Now, is this possible? Can anyone solve please?
 
  • #7
Kubilay Yazoglu said:
Can you determine a basis for the plane x − y + z = 2 that spans only the plane?

This is the original question. I forgot the "only" part. Now, is this possible? Can anyone solve please?

You can't solve the problem that you stated. The word "span" has a technical definition. Any basis for [itex] R^3 [/itex] includes the zero vector in it's span. The span of a set of vectors includes the linear combination of those vectors whose coefficients are each the zero scalar. Such a linear combination is equal to the zero vector. So you can't exclude the zero vector from the "span" of a set of vectors.

Perhaps what you want is a representation of the plane in the form [itex] (x,y,z) = (x_0,y_0,z_0) + c_1( x_1,y_1,z_1) + c_2(x_2,y_2,z_2) [/itex]. The form of this equation prohibits the coefficient of the vector [itex] (x_0,y_0,z_0) [/itex] from being zero. So the equation does not define the "span" of the set of vectors [itex] (x_i, y_i, z_i) [/itex].
 
  • Like
Likes Kubilay Yazoglu
  • #8
Stephen Tashi said:
You can't solve the problem that you stated. The word "span" has a technical definition. Any basis for [itex] R^3 [/itex] includes the zero vector in it's span. The span of a set of vectors includes the linear combination of those vectors whose coefficients are each the zero scalar. Such a linear combination is equal to the zero vector. So you can't exclude the zero vector from the "span" of a set of vectors.

Perhaps what you want is a representation of the plane in the form [itex] (x,y,z) = (x_0,y_0,z_0) + c_1( x_1,y_1,z_1) + c_2(x_2,y_2,z_2) [/itex]. The form of this equation prohibits the coefficient of the vector [itex] (x_0,y_0,z_0) [/itex] from being zero. So the equation does not define the "span" of the set of vectors [itex] (x_i, y_i, z_i) [/itex].
Yes! That is exactly what I've been thinking! But my friends almost convinced me :) Thank you!
 
  • #9
What you can do, and this is equivalent to what Stephen Tashi said, is look at the plane through the origin, parallel to the given plane. For your example, x- y+ z= 0. That is equivalent to y= x+ z so any vector in it is of the form <x, x+ z, z>= <x, x, 0>+ <0, z, z>= x<1, 1, 0>+ z<0, 1, 1>. That is, any vector can be written as a linear combination of the two vectors < 1, 1, 0> and <0, 1, 1> so they form a basis for the subspace.

Now, one point on the plane x- y+ z= 2 (a plane but NOT a subspace as others have said) is (0, 0, 2) so one vector in the set (not subspace) of vectors representing that plane can be written as the sum of (0, 0, 2) (or any other single vector in the plane) plus a linear combination of <1, 1, 0> and <0, 1, 1>.
 

1. Can a non-subspace space have a basis?

Yes, it is possible for a non-subspace space to have a basis. A basis is a set of linearly independent vectors that span the entire space. While a subspace must contain the zero vector and be closed under addition and scalar multiplication, a non-subspace space does not have these restrictions. Therefore, a non-subspace space can have a basis that does not follow these rules.

2. How do you determine a basis for a non-subspace space?

The process of determining a basis for a non-subspace space is similar to determining a basis for a subspace. First, you need to find a set of linearly independent vectors that span the entire space. This can be done using methods such as Gaussian elimination or finding the null space of a matrix. Then, you can check if the set of vectors satisfies the definition of a basis by making sure they are linearly independent and span the entire space.

3. Is the basis for a non-subspace space unique?

No, the basis for a non-subspace space is not necessarily unique. While a subspace always has a unique basis, a non-subspace space can have multiple sets of linearly independent vectors that span the entire space. However, any basis for a non-subspace space will have the same number of vectors, known as the dimension of the space.

4. Can a non-subspace space have a finite basis?

Yes, a non-subspace space can have a finite basis. The dimension of a non-subspace space can be any positive integer, including finite values. This means that a non-subspace space can have a basis with a finite number of vectors that span the entire space.

5. How do you know if a set of vectors is a basis for a non-subspace space?

To determine if a set of vectors is a basis for a non-subspace space, you can use the definition of a basis. A basis for a non-subspace space must satisfy two conditions: the vectors must be linearly independent and they must span the entire space. This means that the vectors cannot be scaled versions of each other and together they can represent any vector in the space. You can also use methods such as the rank-nullity theorem to check if the set of vectors spans the entire space.

Similar threads

I Vector subspace and basis vectors in the context of data science
    • Linear and Abstract Algebra
Replies
6
Views
878
I Standard basis and other bases...
    • Linear and Abstract Algebra
Replies
9
Views
575
I Non orthogonal basis and the lines of its coordinate grid
    • Linear and Abstract Algebra
Replies
9
Views
199
I Row space, Column space, Null space & Left null space
    • Linear and Abstract Algebra
Replies
8
Views
880
I Proof that if T is Hermitian, eigenvectors form an orthonormal basis
    • Linear and Abstract Algebra
Replies
3
Views
2K
B Understanding Bases of a Vector Space
    • Linear and Abstract Algebra
Replies
3
Views
298
I Basis of a Subspace of a Vector Space
    • Linear and Abstract Algebra
Replies
8
Views
2K
I Question about an eigenvalue problem: range space
    • Linear and Abstract Algebra
Replies
1
Views
1K
MHB What is the basis of the trivial vector space {0}
    • Linear and Abstract Algebra
Replies
17
Views
9K
I Representing vectors with respect to a basis
    • Linear and Abstract Algebra
Replies
8
Views
3K

Hot Threads

Recent Insights

Back
Top