Correct the following definition.

Definition (Linear Independence): The vectors in a subset S of a vector space V over a field are linearly independent over F if and only if the zero vector cannot be expressed as a linear combination of vectors in S.

Solution:
Correct Definition (Linear Independence): A subset `S = {v_1 , ... , v_n}` of a vector space V over a field F is linearly independent if the only solution to the vector equation `a_1 v_1 + ... + a_n v_n = 0` is the trivial solution `a_1 = a_2 = ... = a_n = 0`.
Comment on Error: The zero vector can always be expressed as a linear combination (by setting all scalars to zero). Linear Independence means the only way to express the zero vector is with all scalars equal to zero (the trivial way). END

Exercise (Linear Independence): Determine if the set `S = {(1 , 2) , (-2 , -4)} in RR^2` is linearly independent.

Reformulate the Given Exercise: Determine if the only solution to the vector equation `a * (1 , 2) + b * (-2 . -4) = (0 , 0)` is the trivial solution.

Give a general Strategy for a Solution:
1. Set up the homogenous system of linear equations corresponding to the vector equation.
2. Solve the system for the scalars a and b.
3. If the system has only the trivial solution (a = 0 , b = 0) , the set is linearly independent. If it has non-trivial solutions , the set is linearly dependent.

Write down the specific Theory used : A subset `S = {v_1 , ... , v_n}` is linearly independent if the only solution to `a_1 v_1 + ... + a_n v_n = 0` is `a_1 = ... = a_n = 0`.

Do the Exercise:
1. Vector Equation: `a(1 , 2) + b(-2 , -4) = (0 , 0)`.
2. System of Equations: `(a - 2b , 2a - 4b) = (0 ,0) implies I: a - 2b = 0 and II: 2a - 4b = 0`.
3. Checking for Solutions: Notice that II is simply `2 times I`. The system is redundant. So , `a - 2b = 0 implies a = 2b`. This means we can choose any non-zero value for b and find a corresponding non-zero value for a. For example , let `b = 1 implies a = 2`. This gives a non-trivial solution: `2 * (1 , 2) + 1 * (-2 , -4) = (2 , 4) + (-2 , -4) = (0 , 0)`. Since a non-trivial solution (a = 2 , b = 1) exists , the set is not linearly independant.

Final Result: The set `S = {(1 , 2) , (-2 , -4)}` is linearly dependant. END