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?

---

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.