To show that a set is a subspace, we need to verify two conditions:
1. The set is closed under addition.
2. The set is closed under scalar multiplication.
Let's go through each set one by one:
1. The set of all (x, y) such that x = y:
- Addition: Suppose (x1, y1) and (x2, y2) are in the set. Then x1 = y1 and x2 = y2.
Taking the sum, we have x1 + x2 = y1 + y2. Thus, (x1 + x2, y1 + y2) is also in the set.
- Scalar multiplication: Suppose (x, y) is in the set, and c is a scalar. Then x = y.
Multiplying both sides by c, we have cx = cy. Thus, (cx, cy) is also in the set.
Therefore, the set of all (x, y) such that x = y forms a subspace.
2. The set of all (x, y) such that x – y = 0:
- Addition: Suppose (x1, y1) and (x2, y2) are in the set. Then x1 - y1 = 0 and x2 - y2 = 0.
Adding the equations, we have (x1 + x2) - (y1 + y2) = 0, which means (x1 + x2, y1 + y2) is also in the set.
- Scalar multiplication: Suppose (x, y) is in the set, and c is a scalar. Then x - y = 0.
Multiplying both sides by c, we have cx - cy = 0. Thus, (cx, cy) is also in the set.
Therefore, the set of all (x, y) such that x – y = 0 forms a subspace.
3. The set of all (x, y, z) such that x + y = 3z:
- Addition: Suppose (x1, y1, z1) and (x2, y2, z2) are in the set. Then x1 + y1 = 3z1 and x2 + y2 = 3z2.
Adding the equations, we have (x1 + x2) + (y1 + y2) = 3(z1 + z2), which means (x1 + x2, y1 + y2, z1 + z2) is also in the set.
- Scalar multiplication: Suppose (x, y, z) is in the set, and c is a scalar. Then x + y = 3z.
Multiplying both sides by c, we have cx + cy = 3cz. Thus, (cx, cy, cz) is also in the set.
Therefore, the set of all (x, y, z) such that x + y = 3z forms a subspace.
4. The set of all (x, y, z) such that x = y and z = 2y:
- Addition: Suppose (x1, y1, z1) and (x2, y2, z2) are in the set. Then x1 = y1 and z1 = 2y1, and x2 = y2 and z2 = 2y2.
Adding the equations, we have (x1 + x2) = (y1 + y2), and (z1 + z2) = 2(y1 + y2).
Thus, (x1 + x2, y1 + y2, z1 + z2) is also in the set.
- Scalar multiplication: Suppose (x, y, z) is in the set, and c is a scalar. Then x = y and z = 2y.
Multiplying both sides of each equation by c, we have cx = cy and cz = 2cy.
Thus, (cx, cy, cz) is also in the set.
Therefore, the set of all (x, y, z) such that x = y and z = 2y forms a subspace.
Show that the following sets form subspaces
The set of all (x, y) in such that x = y
The set of all (x, y) in such that x – y = 0
The set of all (x, y, z) in such that x + y = 3z
The set of all (x, y, z) in such that x = y and z = 2y
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