Finding a basis for vector spaces

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This discussion focuses on finding a basis for vector spaces defined by multiple linear conditions. The user successfully identifies a basis for the vector space in R^3 defined by the equation a + 2b + 3c = 0, resulting in the basis set {(-2,1,0),(-3,0,1)}. However, they struggle with the more complex case of two equations: a - b - c = 0 and 2a + 3b + 8c = 0. The solution involves eliminating two variables to express the remaining variable in terms of a free variable, ultimately leading to a basis of the form {(k1, k2, k3)}.

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I'm having trouble finding a basis for algebraically defined vector spaces where there is more than one condition. For instance, I can easily find a basis for the vector space in R^3 defined by a+2b+3c=0 (where a,b,c are the elements of the vector), but I have no idea what to do when the vector space is defined by something like a-b-c=0 & 2a+3b+8c=0.

For the first example I would write that every vector in the space has to be of the form (-2b-3c, b, c) = b(-2,1,0) + c(-3,0,1), so a basis for the vector space would be {(-2,1,0),(-3,0,1)}. I have no idea what to do for the second example.
 
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You got to (-2b-3c, b, c) by just algebraically eliminating a. Now b and c are free variables. For the second case do the same thing, but eliminate two variables. So the result is, say, (k1*c,k2*c,k3*c). Then the basis is {(k1,k2,k3)}.
 
Dang! I wrote out a brilliant explanation, then went back and read Dick's. His was better!
 

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