Math Problem Statement

Let V = R^2 and define two binary operations on V, the vector addition and scalar multiplication, respectively, and for some scalar s, t, then determine s, t for which (V, ⊕, ⊙) is a vector space.

Solution

We are given a vector space V=R2V = \mathbb{R}^2 with two binary operations: vector addition \oplus and scalar multiplication \odot as follows:

  • Vector addition: (u1,v1)(u2,v2)=(u1+u2+3,v1+v25)(u_1, v_1) \oplus (u_2, v_2) = (u_1 + u_2 + 3, v_1 + v_2 - 5)
  • Scalar multiplication: a(u,v)=(ausa+s,avta+t)a \odot (u, v) = (au - sa + s, av - ta + t)

We are tasked with determining the scalars ss and tt for which (V,,)(V, \oplus, \odot) is a vector space. To be a vector space, the operations must satisfy several axioms such as:

  1. Associativity of addition: \oplus should be associative.
  2. Commutativity of addition: \oplus should be commutative.
  3. Identity element for addition: There should exist a zero vector 0=(0,0)\mathbf{0} = (0,0) such that (u,v)0=(u,v)(u, v) \oplus \mathbf{0} = (u, v).
  4. Additive inverse: For each (u,v)(u, v), there should be an element (u,v)-(u, v) such that (u,v)(u,v)=0(u, v) \oplus -(u, v) = \mathbf{0}.
  5. Distributivity of scalar multiplication over vector addition: a((u1,v1)(u2,v2))=(a(u1,v1))(a(u2,v2))a \odot ((u_1, v_1) \oplus (u_2, v_2)) = (a \odot (u_1, v_1)) \oplus (a \odot (u_2, v_2)).
  6. Distributivity of scalar multiplication over field addition: (a+b)(u,v)=(a(u,v))(b(u,v))(a + b) \odot (u, v) = (a \odot (u, v)) \oplus (b \odot (u, v)).
  7. Compatibility of scalar multiplication: a(b(u,v))=(ab)(u,v)a \odot (b \odot (u, v)) = (ab) \odot (u, v).
  8. Identity element of scalar multiplication: 1(u,v)=(u,v)1 \odot (u, v) = (u, v).

Step 1: Check the identity element for addition

For the identity element, we require that: (u,v)(0,0)=(u,v)(u, v) \oplus (0, 0) = (u, v) Substituting into the definition of \oplus: (u,v)(0,0)=(u+0+3,v+05)=(u+3,v5)(u, v) \oplus (0, 0) = (u + 0 + 3, v + 0 - 5) = (u + 3, v - 5) For this to equal (u,v)(u, v), we must have: u+3=uandv5=vu + 3 = u \quad \text{and} \quad v - 5 = v This leads to contradictions: 3=0and5=03 = 0 \quad \text{and} \quad -5 = 0 which are impossible. Thus, there is no identity element for addition unless the operation is modified. Therefore, for no values of ss and tt can this system form a vector space, since it fails a fundamental requirement.

Would you like to delve into any of the vector space axioms further or explore other aspects of this problem?

Here are 5 related questions you might find helpful:

  1. What happens if we change the addition operation to include different constants?
  2. How does scalar multiplication affect other vector space properties?
  3. Can this space still be useful in other mathematical contexts even if it isn’t a vector space?
  4. Could this space form a group instead of a vector space?
  5. What are the general criteria for an operation to define a vector space?

Tip: Whenever you are testing if a space is a vector space, always start by checking for the identity element and the distributive properties. If these fail, the space cannot be a vector space.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Vector Spaces
Binary Operations
Scalar Multiplication

Formulas

(u1, v1) ⊕ (u2, v2) = (u1 + u2 + 3, v1 + v2 - 5)
a ⊙ (u, v) = (au - sa + s, av - ta + t)

Theorems

Vector Space Axioms
Associativity of Addition
Commutativity of Addition
Distributivity of Scalar Multiplication

Suitable Grade Level

College Level (Linear Algebra)