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

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 \( V \), and we need to prove two properties:

1. For every \( k \in \mathbb{R} \) and every \( u, v \in V \), we have:
   \[
   k(u - v) = ku - kv
   \]
2. For every vector \( v \in V \), we have:
   \[
   v + v = 2v
   \]

---

### Part 1: \( k(u - v) = ku - kv \)

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:
   \[
   u - v = u + (-v)
   \]
   where \( -v \) is the additive inverse of \( v \) (i.e., \( v + (-v) = 0 \)).

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

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

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

---

### Part 2: \( v + v = 2v \)

This property is based on the definition of scalar multiplication.

#### Step-by-Step Justification:
1. **Express \( 2v \) as scalar multiplication**:  
   By definition, the scalar \( 2 \) applied to a vector \( v \) is:
   \[
   2v = v + v
   \]

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

Hence, we have shown that \( v + v = 2v \), as required.

---

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

1. \( k(u - v) = ku - kv \) by using the distributive property of scalar multiplication over addition and the property of scalar multiplication with inverses.
2. \( v + v = 2v \) 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 \( \mathbb{R} \), such as \( \mathbb{C} \)?
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.

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.

Explore the step-by-step solution to proving vector space properties including the distributive property of scalar multiplication and vector addition using axioms. This is suitable for college-level linear algebra students.

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