Prove that a non-empty subset W is a subspace of a vector space V over
F, if and only if 𝑐1 𝛼 + 𝑐2 𝛽 ∈ 𝑊, ∀ 𝛼, 𝛽 ∈ 𝑊, 𝑐1, 𝑐2 ∈ 𝐹. Show that
the subset 𝑊 = {(𝑥, 𝑦, 𝑧) | 𝑥 + 𝑦 + 𝑧 = 0} of the vector space 𝑉3
(𝑅) is
a subspace of 𝑉3
(𝑅).

### Part 1: Proving the Subspace Condition

Let \( V \) be a vector space over a field \( F \), and let \( W \) be a non-empty subset of \( V \).

**Theorem:** \( W \) is a subspace of \( V \) if and only if for any \( \alpha, \beta \in W \) and any scalars \( c_1, c_2 \in F \), the linear combination \( c_1 \alpha + c_2 \beta \) is also in \( W \).

#### **Proof:**

**(\(\Rightarrow\) Direction):** 

Assume that \( W \) is a subspace of \( V \). To prove that \( W \) satisfies the given condition:

1. **Closure under addition:** Since \( W \) is a subspace, for any \( \alpha, \beta \in W \), \( \alpha + \beta \in W \).
2. **Closure under scalar multiplication:** Since \( W \) is a subspace, for any \( \alpha \in W \) and any scalar \( c_1 \in F \), \( c_1 \alpha \in W \).
3. **Linear combination:** Let \( c_1, c_2 \in F \) and \( \alpha, \beta \in W \). Then:
   \[
   c_1 \alpha + c_2 \beta = c_1 \alpha + c_2 \beta \in W
   \]
   because \( W \) is closed under addition and scalar multiplication. Hence, \( c_1 \alpha + c_2 \beta \in W \).

**(\(\Leftarrow\) Direction):**

Assume that \( W \) is a non-empty subset of \( V \) and for any \( \alpha, \beta \in W \) and any scalars \( c_1, c_2 \in F \), \( c_1 \alpha + c_2 \beta \in W \).

We need to show that \( W \) is a subspace, i.e., it is closed under addition, scalar multiplication, and contains the zero vector.

1. **Non-emptiness:** By assumption, \( W \) is non-empty, so there exists at least one element \( \alpha \in W \).
   
2. **Closure under addition:** Take \( c_1 = c_2 = 1 \). Then for any \( \alpha, \beta \in W \):
   \[
   \alpha + \beta = 1\alpha + 1\beta \in W
   \]
   which proves that \( W \) is closed under addition.

3. **Closure under scalar multiplication:** Take \( \beta = 0 \) (the zero vector in \( V \)). Then for any \( c_1 \in F \) and \( \alpha \in W \):
   \[
   c_1 \alpha = c_1 \alpha + 0\beta \in W
   \]
   which proves that \( W \) is closed under scalar multiplication.

4. **Zero vector:** Since \( W \) is closed under scalar multiplication, for \( c_1 = 0 \), we have:
   \[
   0 \cdot \alpha = 0 \in W
   \]
   Thus, \( W \) contains the zero vector.

Since \( W \) is non-empty, closed under addition, and closed under scalar multiplication, \( W \) is a subspace of \( V \).

### Part 2: Showing \( W = \{(x, y, z) \mid x + y + z = 0\} \) is a Subspace of \( \mathbb{R}^3 \)

Let \( V = \mathbb{R}^3 \), and consider

Learn how to prove that a non-empty subset W is a subspace of a vector space V over field F if and only if c1 alpha + c2 beta ∈ W for all alpha, beta ∈ W and scalars c1, c2 ∈ F. Explore the subset W = {(x, y, z) | x + y + z = 0} of the vector space V3(R) and its properties.

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