Math Problem Statement

the set R^2 with addition defined by [a;b] +[c;d] =[a-c;b-d] and with scalar multiplication defined as usual. is this a vector space

Solution

To determine if R2\mathbb{R}^2 with the given operations is a vector space, we must check if it satisfies the axioms of a vector space under the defined operations.

  1. Addition: Defined as [a;b]+[c;d]=[ac;bd][a; b] + [c; d] = [a - c; b - d].
  2. Scalar Multiplication: Defined as usual, so for a scalar kk and a vector [a;b][a; b], we have k[a;b]=[ka;kb]k \cdot [a; b] = [k \cdot a; k \cdot b].

Let's go through the key properties a vector space must satisfy, focusing on addition and scalar multiplication:

  1. Closure under addition: For any two vectors [a;b][a; b] and [c;d][c; d], their sum is [ac;bd][a - c; b - d]. Since subtraction is closed in R2\mathbb{R}^2, this is satisfied.

  2. Associativity of addition: We need ([a;b]+[c;d])+[e;f]=[a;b]+([c;d]+[e;f])( [a; b] + [c; d] ) + [e; f] = [a; b] + ( [c; d] + [e; f] ). However, given that addition is defined as [a;b]+[c;d]=[ac;bd][a; b] + [c; d] = [a - c; b - d], associativity may not hold as expected in usual vector spaces due to the unusual subtraction operation.

  3. Existence of an additive identity: In standard vector spaces, the zero vector [0;0][0; 0] serves as the additive identity. But with this unusual addition, there’s no [0;0][0; 0] that works as an identity in the traditional sense since [a;b]+[0;0]=[a;b][0;0]=[a;b][a; b] + [0; 0] = [a; b] - [0; 0] = [a; b] would not yield an identity under subtraction.

  4. Existence of an additive inverse: We require each vector [a;b][a; b] to have an inverse [c;d][c; d] such that [a;b]+[c;d]=[0;0][a; b] + [c; d] = [0; 0]. This is complicated by the fact that addition is defined as subtraction.

Given these issues, particularly with associativity and identity, R2\mathbb{R}^2 under these operations does not satisfy the axioms of a vector space.

Would you like a deeper dive into any specific vector space axiom or clarification on why this structure fails?

Here are some related questions for further exploration:

  1. How does the requirement for an additive identity impact vector space structure?
  2. Can non-standard addition operations create valid vector spaces?
  3. What properties must scalar multiplication satisfy in vector spaces?
  4. Are there examples of non-standard vector spaces in other mathematical fields?
  5. What happens if we change the addition operation to be traditional addition?

Tip: Always verify the existence of an additive identity and inverses when assessing vector space validity.

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Math Problem Analysis

Mathematical Concepts

Vector Spaces
Linear Algebra
Addition Operations
Scalar Multiplication

Formulas

[a; b] + [c; d] = [a - c; b - d]
k * [a; b] = [k * a; k * b]

Theorems

Axioms of Vector Spaces
Closure under addition
Associativity of addition
Existence of additive identity
Additive inverse

Suitable Grade Level

Grades 11-12