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

  1. Dec 9, 2014 #1
    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?
  2. jcsd
  3. Dec 9, 2014 #2


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    Bases are defined for vector spaces. If you do not have a vector space, the concept of a basis is meaningless.
  4. Dec 10, 2014 #3

    Stephen Tashi

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    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.
  5. Dec 11, 2014 #4


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    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.
  6. Dec 11, 2014 #5
  7. Dec 11, 2014 #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?
  8. Dec 11, 2014 #7

    Stephen Tashi

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    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].
  9. Dec 11, 2014 #8
    Yes! That is exactly what I've been thinking! But my friends almost convinced me :) Thank you!
  10. Dec 11, 2014 #9


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    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>.
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