Correct the following definition.

Definition (Spanning Set): The vectors in a subset S of a vector space V over a field F spans V if and only if each `beta` in V can be expressed uniquely as a linear combination of vectors in S.

Solution:
Correct Definition (Spanning Set): A subset S of a vector space V over a field F spans V (or is a spanning set for V) if and only if every vector in V is a linear combination of vectors in S.
Comment on Error: The expression is not required to be unique. Uniqueness is a property of a basis (a set that spans and is linearly independent). END

Exercise (on Spanning Set): Consider the vector space `RR^2` over the field `RR`. Does the set
S = {(1 , 0) , (0 , 1) , (1 , 1)} = (x , y) span `RR` ?

Reformulate the given Exercise: Determine if every vector `(x , y) in RR^2` can be written as a linear combination of the vectors in S: `a(1 , 0) + b(0 , 1) + c(1 , 1) = (x , y)` where `a , b , c in RR`.

Give a general Strategy for a Solution:
1. Set up the vector equation for an arbitrary vector (x , y).
2. Convert the vector equation into a system of linear equations in terms of the scalar coefficients `a , b , c in RR`.
3. Check if the system has at least one solution for any choice of x and y.
If it does , the set spans `RR^2`.

Write down the specific Theory used :
Corrected Definition (Spanning Set): A subset S of a vector space V spans V if and only if every vector in V is a linear combination of vectors in S.

Do the Exercise:
1. Vector Equation: `a(1 , 0) + b(0 , 1) + c(1 , 1) = (x , y)`.
2. System of Equations: `(a + c , b + c) = (x , y) implies a + c = x \ , \ b + c = y`.
3. Checking for Solutions: Since we have 2 equations and 3 unknowns (a , b , c) , this system is underdetermined , meaning it has either no solution or infinitely many solutions. We can easily find a solution for any (x , y). For example , let c = 0. Then a = x and b = y:
`x * (1 , 0) + y * (0 , 1) + 0 * (1 , 1) = (x , y)`.
Since we found a set of scalars (a = x , b = y , c = 0) that works for any (x , y) , every vector in `RR^2` can be expressed as a linear combination of vectors in S.

Final Result: Yes , the set `S = {(1 ,0) , (0 , 1) , (1 , 1)}` spans `RR^2`. END