To determine whether a set is a vector space, it must satisfy the following axioms:

1. Closure under addition: If u and v are vectors in V, then u + v is also in V.
2. Closure under scalar multiplication: If u is a vector in V and c is a scalar, then cu is also in V.
3. Commutativity of addition: u + v = v + u.
4. Associativity of addition: (u + v) + w = u + (v + w).
5. Additive identity: There exists a vector 0 in V such that u + 0 = u for all u in V.
6. Additive inverse: For every vector u in V, there exists a vector -u in V such that u + (-u) = 0.
7. Distributivity of scalar multiplication with respect to vector addition: c(u + v) = cu + cv.
8. Distributivity of scalar multiplication with respect to scalar addition: (c + d)u = cu + du.
9. Scalar multiplication identity: 1u = u for all u in V.

Now, let's check these axioms for the given set of vectors in ℝ2 with xy ≥ 0.

1. Closure under addition: If we take two vectors u = (x1, y1) and v = (x2, y2) both in the first quadrant (x1y1 ≥ 0, x2y2 ≥ 0), their sum u + v = (x1 + x2, y1 + y2) will also be in the first quadrant, so the set is closed under addition. However, if we take one vector in the first quadrant and one in the third quadrant, their sum might not satisfy xy ≥ 0, so the set is not closed under addition.

2. Closure under scalar multiplication: If we take a vector u = (x, y) in the first quadrant and a positive scalar c, then cu = (cx, cy) will also be in the first quadrant. However, if we take a negative scalar, cu might not satisfy xy ≥ 0, so the set is not closed under scalar multiplication.

Therefore, the given set is not a vector space because it does not satisfy the closure under addition and scalar multiplication axioms.

Question

To determine whether a set is a vector space, it must satisfy the following axioms:

1. Closure under addition: If u and v are vectors in V, then u + v is also in V.
2. Closure under scalar multiplication: If u is a vector in V and c is a scalar, then cu is also in V.
3. Commutativity of addition: u + v = v + u.
4. Associativity of addition: (u + v) + w = u + (v + w).
5. Additive identity: There exists a vector 0 in V such that u + 0 = u for all u in V.
6. Additive inverse: For every vector u in V, there exists a vector -u in V such that u + (-u) = 0.
7. Distributivity of scalar multiplication with respect to vector addition: c(u + v) = cu + cv.
8. Distributivity of scalar multiplication with respect to scalar addition: (c + d)u = cu + du.
9. Scalar multiplication identity: 1u = u for all u in V.

Now, let's check these axioms for the given set of vectors in ℝ2 with xy ≥ 0.

1. Closure under addition: If we take two vectors u = (x1, y1) and v = (x2, y2) both in the first quadrant (x1y1 ≥ 0, x2y2 ≥ 0), their sum u + v = (x1 + x2, y1 + y2) will also be in the first quadrant, so the set is closed under addition. However, if we take one vector in the first quadrant and one in the third quadrant, their sum might not satisfy xy ≥ 0, so the set is not closed under addition.

2. Closure under scalar multiplication: If we take a vector u = (x, y) in the first quadrant and a positive scalar c, then cu = (cx, cy) will also be in the first quadrant. However, if we take a negative scalar, cu might not satisfy xy ≥ 0, so the set is not closed under scalar multiplication.

Therefore, the given set is not a vector space because it does not satisfy the closure under addition and scalar multiplication axioms.

Knowee AI · Accepted Answer

To determine whether a set is a vector space, it must satisfy the following axioms:

1. Closure under addition: If u and v are vectors in V, then u + v is also in V.
2. Closure under scalar multiplication: If u is a vector in V and c is a scalar, then cu is also in V.
3. Commutativity of addition: u + v = v + u.
4. Associativity of addition: (u + v) + w = u + (v + w).
5. Additive identity: There exists a vector 0 in V such that u + 0 = u for all u in V.
6. Additive inverse: For every vector u in V, there exists a vector -u in V such that u + (-u) = 0.
7. Distributivity of scalar multiplication with respect to vector addition: c(u + v) = cu + cv.
8. Distributivity of scalar multiplication with respect to scalar addition: (c + d)u = cu + du.
9. Scalar multiplication identity: 1u = u for all u in V.

Now, let's check these axioms for the given set of vectors in ℝ2 with xy ≥ 0.

1. Closure under addition: If we take two vectors u = (x1, y1) and v = (x2, y2) both in the first quadrant (x1y1 ≥ 0, x2y2 ≥ 0), their sum u + v = (x1 + x2, y1 + y2) will also be in the first quadrant, so the set is closed under addition. However, if we take one vector in the first quadrant and one in the third quadrant, their sum might not satisfy xy ≥ 0, so the set is not closed under addition.

2. Closure under scalar multiplication: If we take a vector u = (x, y) in the first quadrant and a positive scalar c, then cu = (cx, cy) will also be in the first quadrant. However, if we take a negative scalar, cu might not satisfy xy ≥ 0, so the set is not closed under scalar multiplication.

Therefore, the given set is not a vector space because it does not satisfy the closure under addition and scalar multiplication axioms.

Question

Solution

Similar Questions

Upgrade your grade with Knowee