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.