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
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.
To show that a set forms a subspace, we need to demonstrate that it satisfies three conditions:
1. The set contains the zero vector.
2. The set is closed under vector addition.
3. The set is closed under scalar multiplication.
Let's examine each set individually:
A) The set of all (x, y) such that x = y:
1. Zero vector: The zero vector is (0, 0), which satisfies x = y. Therefore, the set contains the zero vector.
2. Vector addition: Let (a, a) and (b, b) be two vectors in the set. The sum of these vectors is (a + b, a + b). Since (a + b) = (a + b), the sum is also in the set.
3. Scalar multiplication: Let (a, a) be a vector in the set, and let c be a scalar. The scalar multiple of this vector is (ca, ca). Since (ca) = (ca), the scalar multiple is also in the set.
Therefore, the set of all (x, y) such that x = y forms a subspace.
B) The set of all (x, y) such that x – y = 0:
1. Zero vector: The zero vector is (0, 0), which satisfies x – y = 0. Therefore, the set contains the zero vector.
2. Vector addition: Let (a, a) and (b, b) be two vectors in the set. The sum of these vectors is (a + b, a + b). Since (a + b) – (a + b) = 0, the sum is also in the set.
3. Scalar multiplication: Let (a, a) be a vector in the set, and let c be a scalar. The scalar multiple of this vector is (ca, ca). Since (ca) – (ca) = 0, the scalar multiple is also in the set.
Therefore, the set of all (x, y) such that x – y = 0 forms a subspace.
C) The set of all (x, y, z) such that x + y = 3z:
1. Zero vector: The zero vector is (0, 0, 0), which satisfies x + y = 3(0). Therefore, the set contains the zero vector.
2. Vector addition: Let (a, b, c) and (d, e, f) be two vectors in the set. The sum of these vectors is (a + d, b + e, c + f). If (a + d) + (b + e) = 3(c + f), then (a + d) + (b + e) = 3c + 3f, which means the sum is also in the set.
3. Scalar multiplication: Let (a, b, c) be a vector in the set, and let d be a scalar. The scalar multiple of this vector is (da, db, dc). If da + db = 3dc, then d(a + b) = 3dc, which means the scalar multiple is also in the set.
Therefore, the set of all (x, y, z) such that x + y = 3z forms a subspace.
D) The set of all (x, y, z) such that x = y and z = 2y:
1. Zero vector: The zero vector is (0, 0, 0), which satisfies x = y and z = 2y. Therefore, the set contains the zero vector.
2. Vector addition: Let (a, a, b) and (c, c, d) be two vectors in the set. The sum of these vectors is (a + c, a + c, b + d). Since (a + c) = (a + d) and (b + d) = 2(a + c), the sum is also in the set.
3. Scalar multiplication: Let (a, a, b) be a vector in the set, and let c be a scalar. The scalar multiple of this vector is (ca, ca, cb). Since (ca) = (ca) and (cb) = 2(ca), the scalar multiple is also in the set.
Therefore, the set of all (x, y, z) such that x = y and z = 2y forms a subspace.