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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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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

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Stephen Tashi

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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.

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WWGD

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

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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.

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

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].

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Yes! That is exactly what I've been thinking! But my friends almost convinced me :) Thank you!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].

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HallsofIvy

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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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