Question

Let V=R. For u,v∈V and α∈R define vector addition by u⊕v=u+v−1 and scalar multiplication by α⊙u=αu−α+1. Verify that V with the operations ⊕ and ⊙ is a vector space. Hint: you need to identify the zero vector

0
​
and the additive inverse ⊖u of u and verify that the eight conditions hold. For example associativity of scalar multiplication holds since for α,β∈R and v∈V we have
α⊙(β⊙u)
​

=α⊙(βu−β+1)
=α(βu−β+1)−α+1
=αβu−αβ+1
=(αβ)⊙u
​

Answer 1

To verify that V with the operations ⊕ and ⊙ is a vector space, we need to show that it satisfies the eight conditions of a vector space. The conditions include existence of a zero vector, existence of an additive inverse, commutativity and associativity of vector addition, distributivity of scalar multiplication, compatibility of scalar multiplication with field multiplication, and identity element of scalar multiplication. By showing that all eight conditions are satisfied, we can conclude that V with the operations ⊕ and ⊙ is a vector space.

Step-by-step explanation:

To verify that V with the operations ⊕ and ⊙ is a vector space, we need to show that it satisfies the eight conditions of a vector space.

1. Existence of a zero vector: For any v∈V, we have v ⊕ 0 = v + 0 - 1 = v - 1 + 1 - 1 = v. So, 0 is the zero vector.
2. Existence of an additive inverse: For any v∈V, we have v ⊕ ⊖v = v + ⊖v - 1 = v - v - 1 = -1, which is a scalar multiple of the zero vector.
3. Commutativity of vector addition: For any u,v∈V, we have u ⊕ v = u + v - 1 and v ⊕ u = v + u - 1. Since addition of real numbers is commutative, u + v = v + u, and therefore u ⊕ v = v ⊕ u.
4. Associativity of vector addition: For any u,v,w∈V, we have (u ⊕ v) ⊕ w = (u + v - 1) ⊕ w = (u + v - 1) + w - 1 = u + (v + w) - 2 = u ⊕ (v ⊕ w).
5. Distributivity of scalar multiplication with respect to vector addition: For any α∈R and u,v∈V, we have α ⊙ (u ⊕ v) = α ⊙ (u + v - 1) = α(u + v - 1) - α + 1 = αu + αv - α - α + 1 = (α ⊙ u) ⊕ (α ⊙ v).
6. Distributivity of scalar multiplication with respect to scalar addition: For any α,β∈R and u∈V, we have (α + β) ⊙ u = (α + β)u - (α + β) + 1 = αu - α + 1 + βu - β + 1 = (α ⊙ u) ⊕ (β ⊙ u).
7. Compatibility of scalar multiplication with field multiplication: For any α,β∈R and u∈V, we have (αβ) ⊙ u = (αβ)u - αβ + 1 = α(βu - β + 1) - α + 1 = α ⊙ (β ⊙ u).
8. Identity element of scalar multiplication: For any u∈V, we have 1 ⊙ u = 1u - 1 + 1 = u - 1 + 1 = u.

Since all eight conditions hold, we can conclude that V with the operations ⊕ and ⊙ is a vector space.