# linear algebra- basis and dimensions

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#1
V is the set of all vectors (a,b,c) in R^3, with a+2b-2c=0; K=R.
Find the dimension and a basis of the vector space V over the given field K.

I said that the dimension is 3, and a basis could be (a,2b,-2c) but i'm not sure if this is correct. I'm quite confused as to how to work out the dimension/ basis.
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6 years ago
#2
A basis is a set of vectors, so (a,2b,-2c) is not a basis (it is just a vector).
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6 years ago
#3
(Original post by james22)
A basis is a set of vectors, so (a,2b,-2c) is not a basis (it is just a vector).
Well, almost. One vector can in some cases constitute a basis. OP, sounds like you need to go back and look at your definitions:

Basis: A subset of the vector space which is both (i) spanning and (ii) linearly independent.
(i) spanning: for any v in V, v can be written as a linear combination of elements of the basis
(ii) linearly independent: if there exist a linear combination of elements of the basis resulting in the zero vector, the scalars must ALL be zero.

Dimension: The number of vectors in any basis (the fact that any basis of a finite-dim vector space does have the same number of elements is not an obvious fact).

The dimension of R^3 IS 3. However, your set is clearly some subspace of R^3 which is not R^3 itself, and so MUST have a smaller dimension. Note that the dimension of your vector space follows immediately from the basis (you just count up the vectors), so try working out the basis first.

You have a+2b-2c = 0, hence a = 2c-2b, and so any vector in V can be written (2c-2b, b, c) for real numbers b and c. If I rewrite this as b(-2, 1, 0) + c(2, 0, 1), can you see what a basis for V might be? Then just count up the vectors
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