how is this not a subspace of R3?
The set of all vectors of the form
a
b
c
, where a > 0.
To determine if a set is a subspace of R^3, we have to check whether it satisfies three conditions:
1. The set is non-empty - It means that there must be at least one vector in the set.
2. Closure under addition - If we take any two vectors from the set and add them together, the result must still be in the set.
3. Closure under scalar multiplication - If we take any vector from the set and multiply it by any scalar, the result must still be in the set.
Now let's analyze the given set:
The set consists of all vectors of the form a, b, c, where a > 0.
1. Non-empty: The set is non-empty because we can choose any positive values for a, b, and c and we can have at least one vector that satisfies the condition.
2. Closure under addition: Let's take two arbitrary vectors from the set, (a1, b1, c1) and (a2, b2, c2), where a1 > 0 and a2 > 0. If we add these two vectors, we get (a1 + a2, b1 + b2, c1 + c2). Since a1 and a2 are both positive, their sum (a1 + a2) will also be positive. Therefore, the resulting vector is still in the set, and closure under addition is satisfied.
3. Closure under scalar multiplication: Let's take an arbitrary vector from the set, (a1, b1, c1), where a1 > 0, and multiply it by any scalar k. The resulting vector will be (k * a1, k * b1, k * c1). Since a1 is positive, the scalar multiple k * a1 will still be positive for any non-zero scalar k. Therefore, the resulting vector is still in the set, and closure under scalar multiplication is satisfied.
Since the set satisfies all three conditions, we can conclude that the set of all vectors of the form (a, b, c), where a > 0, is indeed a subspace of R^3.