Math Problem Statement

answer this problem according to the rules, theorems and procedures of the textbook Schaum’s outlines: Linear Algebra, by Lipschutz and Lipson

Let V be a vector space. Carefully show the following, justifying every step of your solution: 1. For every k ∈ R and every u, v ∈ V, we have k(u - v) = ku - kv. 2. For every vector v ∈ V, we have v + v = 2v.

Solution

To solve the problem using the rules and theorems from Schaum’s Outline of Linear Algebra by Lipschutz and Lipson, I'll go through each part carefully, justifying the solution step by step.

Problem Statement:

We are given a vector space VV, and we need to prove two properties:

  1. For every kRk \in \mathbb{R} and every u,vVu, v \in V, we have: k(uv)=kukvk(u - v) = ku - kv
  2. For every vector vVv \in V, we have: v+v=2vv + v = 2v

Part 1: k(uv)=kukvk(u - v) = ku - kv

This is a distributive property of scalar multiplication over vector subtraction, and we can justify each step using the axioms of vector spaces as outlined in Schaum’s Outline:

Step-by-Step Justification:

  1. Vector subtraction in terms of addition:
    Recall that in a vector space, subtraction is defined as adding the additive inverse: uv=u+(v)u - v = u + (-v) where v-v is the additive inverse of vv (i.e., v+(v)=0v + (-v) = 0).

  2. Distribute scalar kk over the sum:
    By the distributive property of scalar multiplication over vector addition (vector space axiom): k(u+(v))=ku+k(v)k(u + (-v)) = ku + k(-v)

  3. Simplify using scalar multiplication and inverses:
    By the property of scalar multiplication, we know that k(v)=(kv)k(-v) = -(kv) (the scalar multiplied by the additive inverse is the additive inverse of the scalar product): ku+k(v)=kukvku + k(-v) = ku - kv

Thus, we have shown that k(uv)=kukvk(u - v) = ku - kv, as required.


Part 2: v+v=2vv + v = 2v

This property is based on the definition of scalar multiplication.

Step-by-Step Justification:

  1. Express 2v2v as scalar multiplication:
    By definition, the scalar 22 applied to a vector vv is: 2v=v+v2v = v + v

  2. Interpretation in terms of addition:
    In a vector space, multiplying a vector vv by a scalar 22 means adding vv to itself: v+v=2vv + v = 2v This is derived directly from the definition of scalar multiplication in the vector space axioms.

Hence, we have shown that v+v=2vv + v = 2v, as required.


Conclusion:

Using vector space properties from Schaum's Outline, we've carefully proven both parts:

  1. k(uv)=kukvk(u - v) = ku - kv by using the distributive property of scalar multiplication over addition and the property of scalar multiplication with inverses.
  2. v+v=2vv + v = 2v by the definition of scalar multiplication as repeated addition.

Would you like to dive deeper into any of these concepts or have more detailed explanations?

Here are 5 related questions you might find useful to expand on these ideas:

  1. How can we use the distributive property for more complex scalar and vector expressions?
  2. What are the axioms of a vector space, and how do they ensure properties like these?
  3. How would these proofs change if we were working in a field other than R\mathbb{R}, such as C\mathbb{C}?
  4. How does the concept of linearity relate to the distributive properties of scalar multiplication?
  5. Can you generalize these proofs for matrices and matrix multiplication?

Tip: Understanding the axioms of a vector space is crucial as they form the foundation for all properties and operations within linear algebra.

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Math Problem Analysis

Mathematical Concepts

Vector Spaces
Scalar Multiplication
Vector Addition

Formulas

k(u - v) = ku - kv
v + v = 2v

Theorems

Distributive property of scalar multiplication over vector addition
Definition of scalar multiplication

Suitable Grade Level

College-level Linear Algebra