Correct the following definition.

Definition (Dimension): The dimension of a finite-dimensional vector space V over a field F is the minimum number of vectors required to span V.

Solution:
Correct Definition (Dimension): The dimension of a finite dimensional vector space V over a field F , denoted dim(V) , is the number of vectors in any basis for V.
Comment on Error: While the correct definition is equivalent to the minimum number of vectors required to span V , the standard , most fundamental definition is based on the size of a basis. END

Exercise (Dimension): What is the dimension of the subspace W of `RR^3` defined by the set of all vectors `(x , y , z)` such that `x - 2y + z = 0` ?

Reformulate the Given Exercise: Find a basis for the subspace `W = {(x , y , z) in RR^3 | x - 2y + z = 0}` and count the number of vectors in that basis. This count is the dimension dim(W).

Give a general Strategy for a Solution:
1. Express one variable in the defining equation in terms of the others (e.g. solve for x).
2. Substitute this expression back into the vector `(x , y , z)` to express it as a linear combination of the basis vectors, where the free variables act as the scalar coefficients. Or in general terms: A linear combination derived by factoring the free variables (y and z) out of the general vector, where the resulting bracketed vectors form the basis.
3. The number of resulting vectors is the dimension.

Write down the specific Theory used : The dimension of a finite dimensional vector space V is the number of vectors in any basis for V.

Do the Exercise:
1. Solve for a variable: `x - 2y + z =0 implies x = 2y - z`.
2. Express the general vector: Substitute this back into `(x , y , z)`, so we have `(x , y , z) = (2y - z , y , z) = (2y , y , 0) + (-z , 0 , z) = y(2 , 1 , 0) + z(-1 , 0 , 1)`.
3. Identify the basis and dimension: The set `B = {(2 , 1 , 0) , (-1 , 0 , 1)}` spans W. We check for linear independence : Are the vectors scalar multiples of each other ? No , because the y-component of `(2 , 1 , 0)` is non-zero while the y-component of `(-1 , 0 , 1)` is zero. Thus B is linearly independent. Therefore , B is a basis for W. The number of vectors in the basis B is 2.

Final Result: The dimension of the subspace W is 2. END