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).

To show that \( \text{span}(v_1, \dots, v_m) = \text{span}(w_1, \dots, w_m) \), we need to prove that each span is contained in the other:

### Definitions:
- Given vectors \( v_1, \dots, v_m \in V \), define \( w_k = v_1 + \dots + v_k \) for each \( k \in \{1, \dots, m\} \).
- We aim to show that \( \text{span}(v_1, \dots, v_m) \subseteq \text{span}(w_1, \dots, w_m) \) and \( \text{span}(w_1, \dots, w_m) \subseteq \text{span}(v_1, \dots, v_m) \).

### 1. Show \( \text{span}(v_1, \dots, v_m) \subseteq \text{span}(w_1, \dots, w_m) \):

Take any vector in the span of \( v_1, \dots, v_m \). A generic vector \( v \) in this span can be written as:
\[
v = a_1 v_1 + a_2 v_2 + \dots + a_m v_m
\]
We want to express this vector \( v \) as a linear combination of \( w_1, \dots, w_m \).

Observe the following:
\[
v_1 = w_1
\]
\[
v_2 = w_2 - w_1
\]
\[
v_3 = w_3 - w_2
\]
\[
\vdots
\]
\[
v_m = w_m - w_{m-1}
\]
Using these, we can express \( v \) as:
\[
v = a_1 v_1 + a_2 v_2 + \dots + a_m v_m
\]
\[
v = a_1 w_1 + a_2 (w_2 - w_1) + a_3 (w_3 - w_2) + \dots + a_m (w_m - w_{m-1})
\]
\[
v = (a_1 - a_2) w_1 + (a_2 - a_3) w_2 + \dots + (a_{m-1} - a_m) w_{m-1} + a_m w_m
\]
This shows that \( v \) is a linear combination of \( w_1, \dots, w_m \), proving that \( v \in \text{span}(w_1, \dots, w_m) \).

Therefore, \( \text{span}(v_1, \dots, v_m) \subseteq \text{span}(w_1, \dots, w_m) \).

### 2. Show \( \text{span}(w_1, \dots, w_m) \subseteq \text{span}(v_1, \dots, v_m) \):

Now take any vector in the span of \( w_1, \dots, w_m \). A generic vector \( w \) in this span can be written as:
\[
w = b_1 w_1 + b_2 w_2 + \dots + b_m w_m
\]
We want to express this vector \( w \) as a linear combination of \( v_1, \dots, v_m \).

Substitute \( w_k = v_1 + v_2 + \dots + v_k \) into the expression for \( w \):
\[
w = b_1(v_1) + b_2(v_1 + v_2) + \dots + b_m(v_1 + v_2 + \dots + v_m)
\]
\[
w = (b_1 + b_2 + \dots + b_m)v_1 + (b_2 + \dots + b_m)v_2 + \dots + b_m v_m
\]
This shows that \( w \) is a linear combination of \( v_1, \dots, v_m \), proving that \( w \in \text{span}(v_1, \dots, v_m) \).

Therefore, \( \text{span}(w_1, \dots, w_m) \subseteq \text{span}(v_1, \dots, v_m) \).

### Conclusion:
Since \( \text{span}(v_1, \dots, v_m) \subseteq \text{span}(w_1, \dots, w_m) \) and \( \text{span}(w_1, \dots, w_m) \subseteq \text{span}(v_1, \dots, v_m) \), we conclude that:
\[
\text{span}(v_1, \dots, v_m) = \text{span}(w_1, \dots, w_m)
\]

Would you like more details or have any further questions?

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Here are five related questions for further exploration:
1. How can this result be used in the context of basis transformations?
2. What is the relationship between the spans if the vectors \( v_1, \dots, v_m \) are linearly dependent?
3. How does the span change if we modify the definition of \( w_k \) to \( w_k = c_1v_1 + c_2v_2 + \dots + c_kv_k \) with arbitrary coefficients?
4. What happens to the span equality if \( v_1, \dots, v_m \) are eigenvectors of a linear operator?
5. How would this result apply to vectors in an infinite-dimensional vector space?

**Tip**: When manipulating spans, carefully consider how you can express each vector in one span as a linear combination of vectors in the other span to establish equality.

Explore the equality of spans in vector spaces with a detailed proof showing that span(v1, ..., vm) equals span(w1, ..., wm). This problem covers fundamental concepts in linear algebra, including vector spaces, linear span, and the spanning set theorem. Suitable for advanced undergraduate students and beyond.

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