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

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?

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mfb
Mentor
Bases are defined for vector spaces. If you do not have a vector space, the concept of a basis is meaningless.

Stephen Tashi
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.

Kubilay Yazoglu and 1MileCrash
WWGD
Gold Member
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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.

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.

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?

Stephen Tashi
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 $R^3$ 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 $(x,y,z) = (x_0,y_0,z_0) + c_1( x_1,y_1,z_1) + c_2(x_2,y_2,z_2)$. The form of this equation prohibits the coefficient of the vector $(x_0,y_0,z_0)$ from being zero. So the equation does not define the "span" of the set of vectors $(x_i, y_i, z_i)$.

Kubilay Yazoglu
You can't solve the problem that you stated. The word "span" has a technical definition. Any basis for $R^3$ 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 $(x,y,z) = (x_0,y_0,z_0) + c_1( x_1,y_1,z_1) + c_2(x_2,y_2,z_2)$. The form of this equation prohibits the coefficient of the vector $(x_0,y_0,z_0)$ from being zero. So the equation does not define the "span" of the set of vectors $(x_i, y_i, z_i)$.
Yes! That is exactly what I've been thinking! But my friends almost convinced me :) Thank you!

HallsofIvy