Suppose 𝑣1
, … , 𝑣𝑚 is a list of vectors in 𝑉. For 𝑘 ∈ {1, … , 𝑚}, let
𝑤𝑘 = 𝑣1 + ⋯ + 𝑣𝑘
.
Show that span(𝑣1
, … , 𝑣𝑚) = span(𝑤1
, … , 𝑤𝑚).

We are tasked with proving that \(\text{span}(v_1, v_2, \dots, v_m) = \text{span}(w_1, w_2, \dots, w_m)\), where \(w_k = v_1 + v_2 + \cdots + v_k\) for each \(k \in \{1, 2, \dots, m\}\).

### Step 1: Recall Definitions
- \(\text{span}(v_1, v_2, \dots, v_m)\) refers to the set of all possible linear combinations of the vectors \(v_1, v_2, \dots, v_m\).
  
  In other words:
  \[
  \text{span}(v_1, v_2, \dots, v_m) = \left\{ \alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_m v_m \mid \alpha_1, \alpha_2, \dots, \alpha_m \in \mathbb{R} \right\}.
  \]
  
- Similarly, \(\text{span}(w_1, w_2, \dots, w_m)\) refers to the set of all possible linear combinations of the vectors \(w_1, w_2, \dots, w_m\), where each \(w_k = v_1 + v_2 + \cdots + v_k\).

  So, \[
  \text{span}(w_1, w_2, \dots, w_m) = \left\{ \beta_1 w_1 + \beta_2 w_2 + \cdots + \beta_m w_m \mid \beta_1, \beta_2, \dots, \beta_m \in \mathbb{R} \right\}.
  \]

### Step 2: Show \(\text{span}(v_1, \dots, v_m) \subseteq \text{span}(w_1, \dots, w_m)\)
Each vector \(v_k\) can be expressed as a linear combination of \(w_1, \dots, w_k\) because:
- \(w_1 = v_1\),
- \(w_2 = v_1 + v_2\), so \(v_2 = w_2 - w_1\),
- \(w_3 = v_1 + v_2 + v_3\), so \(v_3 = w_3 - w_2\),
- In general, for any \(k\), we can express:
  \[
  v_k = w_k - w_{k-1}.
  \]

This shows that each vector \(v_k\) can be written as a linear combination of the vectors \(w_1, w_2, \dots, w_k\), implying that any linear combination of \(v_1, v_2, \dots, v_m\) can also be written as a linear combination of \(w_1, w_2, \dots, w_m\). Thus:
\[
\text{span}(v_1, v_2, \dots, v_m) \subseteq \text{span}(w_1, w_2, \dots, w_m).
\]

### Step 3: Show \(\text{span}(w_1, \dots, w_m) \subseteq \text{span}(v_1, \dots, v_m)\)
Next, we show that each \(w_k\) is a linear combination of \(v_1, v_2, \dots, v_k\). Specifically, by definition:
\[
w_k = v_1 + v_2 + \cdots + v_k.
\]
This is already a linear combination of the vectors \(v_1, v_2, \dots, v_k\), meaning that any linear combination of \(w_1, w_2, \dots, w_m\) is also a linear combination of \(v_1, v_2, \dots, v_m\). Thus:
\[
\text{span}(w_1, w_2, \dots, w_m) \subseteq \text{span}(v_1, v_2, \dots, v_m).
\]

### Step 4: Conclude the Equality
Since we have shown both:
\[
\text{span}(v_1, v_2, \dots, v_m) \subseteq \text{span}(w_1, w_2, \dots, w_m)
\]
and
\[
\text{span}(w_1, w_2, \dots, w_m) \subseteq \text{span}(v_1, v_2, \dots, v_m),
\]
we can conclude that:
\[
\text{span}(v_1, v_2, \dots, v_m) = \text{span}(w_1, w_2, \dots, w_m).
\]

### Final Result:
Thus, we have shown that \(\text{span}(v_1, \dots, v_m) = \text{span}(w_1, \dots, w_m)\).

Would you like further clarifications, or have any questions about this proof?

Here are 5 related questions you might find interesting:
1. What is the geometric interpretation of spanning sets in \(\mathbb{R}^2\) and \(\mathbb{R}^3\)?
2. How does the dimensionality of a vector space relate to its spanning set?
3. Can two different sets of vectors span the same subspace in a vector space?
4. What role do linear independence and dependence play in spanning sets?
5. How does the concept of a basis relate to the span of a set of vectors?

**Tip:** To check if a set of vectors spans a space, determine if their linear combinations can generate any vector in the space.

Suppose v1, … , vm is a list of vectors in V. For k ∈ {1, … , m}, let wk = v1 + ⋯ + vk. Show that span(v1, … , vm) = span(w1, … , wm).

Explore how to prove that the span of vectors v1, v2, ..., vm is equal to the span of vectors w1, w2, ..., wm, where each wk is a cumulative sum of the vectors v1 through vk. This proof involves concepts from linear algebra, including vector spaces and spanning sets, suitable for undergraduate students.

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