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

To prove that \(\text{span}(v_1, \dots, v_m) = \text{span}(w_1, \dots, w_m)\), we'll need to show two things:

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

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

Let \(v \in \text{span}(v_1, \dots, v_m)\). This means that:
\[
v = c_1v_1 + c_2v_2 + \dots + c_mv_m
\]
for some scalars \(c_1, c_2, \dots, c_m \in \mathbb{F}\) (where \(\mathbb{F}\) is the field over which the vector space \(V\) is defined).

We will express each \(v_k\) in terms of \(w_k\):

\[
v_k = w_k - w_{k-1}
\]
where \(w_0 = 0\) by convention.

Substituting this into the linear combination for \(v\), we get:
\[
v = c_1(v_1) + c_2(v_2) + \dots + c_m(v_m) = c_1(w_1) + c_2(w_2 - w_1) + \dots + c_m(w_m - w_{m-1})
\]

Expanding and simplifying:
\[
v = (c_1 - c_2)w_1 + (c_2 - c_3)w_2 + \dots + (c_{m-1} - c_m)w_{m-1} + c_mw_m
\]

Thus, \(v\) is a linear combination of \(w_1, w_2, \dots, w_m\), meaning \(v \in \text{span}(w_1, \dots, w_m)\).

Since \(v\) was arbitrary, it follows that:
\[
\text{span}(v_1, \dots, v_m) \subseteq \text{span}(w_1, \dots, w_m)
\]

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

Let \(w \in \text{span}(w_1, \dots, w_m)\). This means that:
\[
w = d_1w_1 + d_2w_2 + \dots + d_mw_m
\]
for some scalars \(d_1, d_2, \dots, d_m \in \mathbb{F}\).

We know each \(w_k\) is defined as:
\[
w_k = v_1 + v_2 + \dots + v_k
\]
So, substitute this into the linear combination:
\[
w = d_1(v_1) + d_2(v_1 + v_2) + \dots + d_m(v_1 + v_2 + \dots + v_m)
\]

Expanding this expression:
\[
w = (d_1 + d_2 + \dots + d_m)v_1 + (d_2 + \dots + d_m)v_2 + \dots + d_mv_m
\]

This shows that \(w\) is a linear combination of \(v_1, v_2, \dots, v_m\), meaning \(w \in \text{span}(v_1, \dots, v_m)\).

Since \(w\) was arbitrary, it follows that:
\[
\text{span}(w_1, \dots, w_m) \subseteq \text{span}(v_1, \dots, v_m)
\]

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

Do you want details or have any questions?

Here are five related questions:
1. How would this proof change if the vector space \(V\) were infinite-dimensional?
2. What are the implications of this proof for understanding linear dependence and independence?
3. Can you construct an example of vectors \(v_1, v_2, v_3\) in \(\mathbb{R}^2\) and their corresponding \(w_1, w_2, w_3\)?
4. How does this result relate to the concept of a basis in a vector space?
5. What happens to the span equality if one of the vectors \(v_k\) is the zero vector?

**Tip:** When dealing with spans of vectors, remember that any linear combination of the vectors in a span can generate other vectors within the same span.

Understand and prove the equivalence of spans span(v1, ..., vm) and span(w1, ..., wm) in vector spaces. Explore the linear combinations of vectors and their spans through detailed proofs and examples.

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