To show that a set is a subspace of a vector space, we need to show that it satisfies three conditions: closure under addition, closure under scalar multiplication, and contains the zero vector.
A) The set of all (x, y) such that x = y is a subspace because it satisfies all three conditions.
1. Closure under addition: Let (x₁, y₁) and (x₂, y₂) be two arbitrary vectors in the set. We need to show that their sum, (x₁ + x₂, y₁ + y₂), is also in the set. Since x₁ = y₁ and x₂ = y₂, we have x₁ + x₂ = y₁ + y₂, which means that (x₁ + x₂, y₁ + y₂) satisfies the condition x = y. Therefore, the set is closed under addition.
2. Closure under scalar multiplication: Let c be any scalar and (x, y) be an arbitrary vector in the set. We need to show that c(x, y) = (cx, cy) is also in the set. Since x = y, then cx = cy, which satisfies the condition x = y. Therefore, the set is closed under scalar multiplication.
3. Contains the zero vector: The zero vector is (0, 0), which satisfies the condition x = y. Therefore, the set contains the zero vector.
B) The set of all (x, y) such that x - y = 0 is a subspace because it satisfies all three conditions.
1. Closure under addition: Let (x₁, y₁) and (x₂, y₂) be two arbitrary vectors in the set. We need to show that their sum, (x₁ + x₂, y₁ + y₂), is also in the set. If we have x₁ - y₁ = 0 and x₂ - y₂ = 0, then (x₁ + x₂) - (y₁ + y₂) = (x₁ - y₁) + (x₂ - y₂) = 0 + 0 = 0. Therefore, the sum satisfies the condition x - y = 0. Therefore, the set is closed under addition.
2. Closure under scalar multiplication: Let c be any scalar and (x, y) be an arbitrary vector in the set. We need to show that c(x, y) = (cx, cy) is also in the set. If x - y = 0, then c(x - y) = cx - cy = 0 - 0 = 0. Therefore, the scalar multiple satisfies the condition x - y = 0. Therefore, the set is closed under scalar multiplication.
3. Contains the zero vector: The zero vector is (0, 0), which satisfies the condition x - y = 0. Therefore, the set contains the zero vector.
C) The set of all (x, y, z) such that x + y = 3z is a subspace because it satisfies all three conditions.
1. Closure under addition: Let (x₁, y₁, z₁) and (x₂, y₂, z₂) be two arbitrary vectors in the set. We need to show that their sum, (x₁ + x₂, y₁ + y₂, z₁ + z₂), is also in the set. If we have x₁ + y₁ = 3z₁ and x₂ + y₂ = 3z₂, then (x₁ + x₂) + (y₁ + y₂) = (x₁ + y₁) + (x₂ + y₂) = 3z₁ + 3z₂ = 3(z₁ + z₂). Therefore, the sum satisfies the condition x + y = 3z. Therefore, the set is closed under addition.
2. Closure under scalar multiplication: Let c be any scalar and (x, y, z) be an arbitrary vector in the set. We need to show that c(x, y, z) = (cx, cy, cz) is also in the set. If x + y = 3z, then c(x + y) = cx + cy = 3cz. Therefore, the scalar multiple satisfies the condition x + y = 3z. Therefore, the set is closed under scalar multiplication.
3. Contains the zero vector: The zero vector is (0, 0, 0), which satisfies the condition x + y = 3z. Therefore, the set contains the zero vector.
D) The set of all (x, y, z) such that x = y and z = 2y is a subspace because it satisfies all three conditions.
1. Closure under addition: Let (x₁, y₁, z₁) and (x₂, y₂, z₂) be two arbitrary vectors in the set. We need to show that their sum, (x₁ + x₂, y₁ + y₂, z₁ + z₂), is also in the set. Since x₁ = y₁ and x₂ = y₂, then (x₁ + x₂) = (y₁ + y₂). Also, since z₁ = 2y₁ and z₂ = 2y₂, (z₁ + z₂) = 2(y₁ + y₂). Therefore, the sum satisfies the conditions x = y and z = 2y. Therefore, the set is closed under addition.
2. Closure under scalar multiplication: Let c be any scalar and (x, y, z) be an arbitrary vector in the set. We need to show that c(x, y, z) = (cx, cy, cz) is also in the set. Since x = y, then cx = cy. Also, since z = 2y, cz = 2(cy). Therefore, the scalar multiple satisfies the conditions x = y and z = 2y. Therefore, the set is closed under scalar multiplication.
3. Contains the zero vector: The zero vector is (0, 0, 0), which satisfies the conditions x = y and z = 2y. Therefore, the set contains the zero vector.
In conclusion, all four sets (A, B, C, D) are subspaces of the given vector space.
Show that the following sets form subspaces
A)The set of all (x, y) in such that x = y
B)The set of all (x, y) in such that x – y = 0
C)The set of all (x, y, z) in such that x + y = 3z
D)The set of all (x, y, z) in such that x = y and z = 2y
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