To prove that $\text{span}(v_1, \dots, v_m) = \text{span}(w_1, \dots, w_m)$, we'll show the following two inclusions:

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

Take any vector $v \in \text{span}(v_1, \dots, v_m)$. By definition, this means that: $v = c_1v_1 + c_2v_2 + \dots + c_mv_m$ for some scalars $c_1, c_2, \dots, c_m$.

Now, we know that each vector $w_k$ is defined as: $w_k = v_1 + v_2 + \dots + v_k$

We can express $v_k$ in terms of the $w_k$ vectors as: $v_k = w_k - w_{k-1}$ where $w_0 = 0$ by convention.

Substituting this into the expression for $v$, we get: $v = c_1v_1 + c_2v_2 + \dots + c_mv_m = c_1w_1 + c_2(w_2 - w_1) + \dots + c_m(w_m - w_{m-1})$

Expanding this: $v = c_1w_1 + c_2w_2 - c_2w_1 + \dots + c_mw_m - c_mw_{m-1}$

Grouping like terms, we get: $v = (c_1 - c_2)w_1 + (c_2 - c_3)w_2 + \dots + (c_{m-1} - c_m)w_{m-1} + c_mw_m$

Thus, $v$ is expressed as a linear combination of the $w_k$ vectors. Hence, $v \in \text{span}(w_1, \dots, w_m)$, which proves: $\text{span}(v_1, \dots, v_m) \subseteq \text{span}(w_1, \dots, w_m)$

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

Take any vector $w \in \text{span}(w_1, \dots, w_m)$. By definition, this means that: $w = d_1w_1 + d_2w_2 + \dots + d_mw_m$ for some scalars $d_1, d_2, \dots, d_m$.

Recall that each $w_k$ is defined as: $w_k = v_1 + v_2 + \dots + v_k$

Substituting this back into the expression for $w$: $w = d_1(v_1) + d_2(v_1 + v_2) + \dots + d_m(v_1 + v_2 + \dots + v_m)$

Expanding this expression: $w = (d_1 + d_2 + \dots + d_m)v_1 + (d_2 + d_3 + \dots + d_m)v_2 + \dots + d_mv_m$

This shows that $w$ is a linear combination of the $v_1, v_2, \dots, v_m$ vectors. Therefore, $w \in \text{span}(v_1, \dots, v_m)$, which proves: $\text{span}(w_1, \dots, w_m) \subseteq \text{span}(v_1, \dots, v_m)$

Conclusion

Since we have established both inclusions: $\text{span}(v_1, \dots, v_m) \subseteq \text{span}(w_1, \dots, w_m)$ 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. What happens if one or more of the vectors $v_1, \dots, v_m$ are linearly dependent?
2. How does this proof extend if the vectors $v_1, \dots, v_m$ belong to a subspace of $V$?
3. Can you find an example where $\text{span}(v_1, \dots, v_m)$ is not equal to $\text{span}(w_1, \dots, w_m)$ in a different context (e.g., using different definitions)?
4. How would the proof change if $w_k$ were defined as $w_k = v_k + v_{k+1} + \dots + v_m$?
5. How does the concept of span relate to the idea of basis vectors in a vector space?

Tip: When proving span equalities, try expressing the vectors of one span as linear combinations of the vectors in the other span. This method is often key to establishing the necessary inclusions.