Final answer:
To prove whether u and w are subspaces of IR³, we need to show that they satisfy three conditions: closure under addition, closure under scalar multiplication, and contain the zero vector. Using these conditions, we can prove that both u and w are subspaces of IR³. The intersection of u and w, denoted as u ∩ w, is described by a set of vectors in three-dimensional space that have a specific relationship between their components.
Step-by-step explanation:
(a) To prove whether u and w are subspaces of IR³, we need to show that they satisfy three conditions: closure under addition, closure under scalar multiplication, and contain the zero vector.
To show closure under addition, let's take two arbitrary vectors from u and w: (x₁, x₁, 3x₁) and (x₂, -x₂, -3x₂). Their sum will be (x₁ + x₂, x₁ - x₂, 3x₁ - 3x₂). This sum satisfies the condition for u, so u is closed under addition. Similarly, if we take two arbitrary vectors from w and add them, we get (x₁ + x₂, -x₁ - x₂, -3x₁ - 3x₂), which satisfies the condition for w, so w is closed under addition as well.
To show closure under scalar multiplication, let's take an arbitrary vector (x, x, 3x) from u and multiply it by a scalar c. The result will be (cx, cx, 3cx), which satisfies the condition for u. Similarly, if we take an arbitrary vector (x, -x, -3x) from w and multiply it by a scalar c, we get (cx, -cx, -3cx), which satisfies the condition for w.
Lastly, both u and w contain the zero vector (0, 0, 0), so they satisfy the third condition.
Therefore, u and w are indeed subspaces of IR³.
(b) and (c) u ∩ w = {(x, x, 3x) : x ∈ IR, x ∈ {-(30), -(20), -(15), -(12), -(10), -9, -8, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 20, 30}}
(c) u ∩ w can be described as the set of all vectors in three-dimensional space whose components have the property that the sum of the first and second components is equal to three times the third component.