--- Commentary: $\textbf{Exercise 3.}$ Suppose $v_1, \ldots, v_m$ is a list of vectors in $V$. For $k \in {1, \ldots, m}$, let $w_k = v_1 + \cdots + v_k.$ Show that $\operatorname{span}(v_1, \ldots, v_m) = \operatorname{span}(w_1, \ldots, w_m)$. $\textbf{Solution 3.}$ Suppose $k \in {1, \ldots, m}$. If $k = 1$, then $v_k = w_k$. If $k > 1$, then $v_k = w_k - w_{k-1}.$ Thus we see that $v_k \in \operatorname{span}(w_1, \ldots, w_m)$. The paragraph above implies that $\operatorname{span}(v_1, \ldots, v_m) \subseteq \operatorname{span}(w_1, \ldots, w_m)$. To prove the inclusion in the other direction, note that for each $k \in {1, \ldots, m}$ we have $w_k \in \operatorname{span}(v_1, \ldots, v_k) \subseteq \operatorname{span}(v_1, \ldots, v_m).$ Thus $\operatorname{span}(w_1, \ldots, w_m) \subseteq \operatorname{span}(v_1, \ldots, v_m)$. Hence $\operatorname{span}(v_1, \ldots, v_m) = \operatorname{span}(w_1, \ldots, w_m),$ as desired. $\textit{Commentary:}$ This exercise demonstrates that the span of a list of vectors is unchanged if each vector is replaced by the sum of all preceding vectors in the list. The forward inclusion is shown by expressing each $v_k$ as a difference of $w_k$ and $w_{k-1}$ (or just $w_k$ if $k=1$). The reverse inclusion follows from the fact that each $w_k$ is a sum of $v_1, \ldots, v_k$. This property is useful in simplifying or transforming spanning sets. It shows that there's often flexibility in choosing a spanning set for a given subspace. The cumulative sums $w_k$ can be easier to work with in some situations than the original vectors $v_k$. $\textit{Example:}$ In $\mathbb{R}^3$, let $v_1 = (1, 0, 0)$, $v_2 = (0, 1, 0)$, $v_3 = (0, 0, 1)$. Then $w_1 = (1, 0, 0)$, $w_2 = (1, 1, 0)$, $w_3 = (1, 1, 1)$. The span of ${v_1, v_2, v_3}$ and ${w_1, w_2, w_3}$ is the same, namely $\mathbb{R}^3$. In the space of polynomials $\mathbb{P}(\mathbb{R})$, let $v_1 = 1$, $v_2 = x$, $v_3 = x^2$. Then $w_1 = 1$, $w_2 = 1 + x$, $w_3 = 1 + x + x^2$. The span of ${v_1, v_2, v_3}$ and ${w_1, w_2, w_3}$ is the same, namely the subspace of polynomials of degree at most 2. In the space of continuous functions $C([0,1], \mathbb{R})$, let $v_1 = \sin(x)$, $v_2 = \cos(x)$, $v_3 = \sin(2x)$. Then $w_1 = \sin(x)$, $w_2 = \sin(x) + \cos(x)$w_3 = \sin(x) + \cos(x) + \sin(2x)$. The span of ${v_1, v_2, v_3}$ and ${w_1, w_2, w_3}$ is the same, namely the subspace of linear combinations of these trigonometric functions. In the space of $p$-adic numbers $\mathbb{Q}_p^n$, let $v_1 = (1, 0, \ldots, 0)$, $v_2 = (0, p, 0, \ldots, 0)$, $v_3 = (0, 0, p^2, 0, \ldots, 0)$, and so on. Then $w_1 = (1, 0, \ldots, 0)$, $w_2 = (1, p, 0, \ldots, 0)$, $w_3 = (1, p, p^2, 0, \ldots, 0)$, and so on. The span of ${v_1, v_2, \ldots, v_n}$ and ${w_1, w_2, \ldots, w_n}$ is the same, namely the space of $p$-adic vectors where each coordinate has a different power of $p$. In the space of modular polynomials $(\mathbb{Z}/m\mathbb{Z})[x]$, let $v_1 = 1$, $v_2 = x$, $v_3 = x^2$. Then $w_1 = 1$, $w_2 = 1 + x$, $w_3 = 1 + x + x^2$. The span of ${v_1, v_2, v_3}$ and ${w_1, w_2, w_3}$ is the same, namely the subspace of polynomials of degree at most 2, with coefficients in $\mathbb{Z}/m\mathbb{Z}$. These examples show that the property holds across a wide variety of vector spaces, including spaces over fields like $\mathbb{Q}_p$ and $\mathbb{Z}/m\mathbb{Z}$ that have different algebraic structures than the real or complex numbers. The key ingredient is the linearity of the vector space operations, which allows the span to be preserved under these transformations of the spanning vectors. --- $\textbf{Exercise 3.}$ Suppose $v_1, \ldots, v_m$ is a list of vectors in $V$. For $k \in {1, \ldots, m}$, let $w_k = v_1 + \cdots + v_k.$ Show that $\operatorname{span}(v_1, \ldots, v_m) = \operatorname{span}(w_1, \ldots, w_m)$. $\textbf{Solution 3.}$ Suppose $k \in {1, \ldots, m}$. If $k = 1$, then $v_k = w_k$. If $k > 1$, then $v_k = w_k - w_{k-1}.$ Thus we see that $v_k \in \operatorname{span}(w_1, \ldots, w_m)$. The paragraph above implies that $\operatorname{span}(v_1, \ldots, v_m) \subseteq \operatorname{span}(w_1, \ldots, w_m)$. To prove the inclusion in the other direction, note that for each $k \in {1, \ldots, m}$ we have $w_k \in \operatorname{span}(v_1, \ldots, v_k) \subseteq \operatorname{span}(v_1, \ldots, v_m).$ Thus $\operatorname{span}(w_1, \ldots, w_m) \subseteq \operatorname{span}(v_1, \ldots, v_m)$. Hence $\operatorname{span}(v_1, \ldots, v_m) = \operatorname{span}(w_1, \ldots, w_m),$ as desired. $\textit{Commentary:}$ This exercise demonstrates that the span of a list of vectors is unchanged if each vector is replaced by the sum of all preceding vectors in the list. The forward inclusion is shown by expressing each $v_k$ as a difference of $w_k$ and $w_{k-1}$ (or just $w_k$ if $k=1$). The reverse inclusion follows from the fact that each $w_k$ is a sum of $v_1, \ldots, v_k$. This property is useful in simplifying or transforming spanning sets. It shows that there's often flexibility in choosing a spanning set for a given subspace. The cumulative sums $w_k$ can be easier to work with in some situations than the original vectors $v_k$. $\textit{Additional Examples:}$ In $\mathbb{Q}_p^3$, if $v_1 = (1, p, p^2)$, $v_2 = (p, p^2, 1)$, $v_3 = (p^2, 1, p)$, then $w_1 = (1, p, p^2)$, $w_2 = (1+p, p+p^2, 1+p^2)$, $w_3 = (1+p+p^2, 1+p+p^2, 1+p+p^2)$. The span of ${v_1, v_2, v_3}$ equals the span of ${w_1, w_2, w_3}$. In the space of polynomials over $\mathbb{F}_3[x]$, if $v_1 =1$, $v_2 = x$, $v_3 = x^2$, then $w_1 = 1$, $w_2 = 1 + x$, $w_3 = 1 + x + x^2$. The span of ${v_1, v_2, v_3}$ equals the span of ${w_1, w_2, w_3}$. In the space of continuous functions $C([0,1], \mathbb{C})$, if $v_1 = e^x$, $v_2 = \sin(x)$, $v_3 = \cos(x)$, then $w_1 = e^x$, $w_2 = e^x + \sin(x)$, $w_3 = e^x + \sin(x) + \cos(x)$. The span of ${v_1, v_2, v_3}$ equals the span of ${w_1, w_2, w_3}$. In the space of sequences $\ell^2(\mathbb{N})$, if $v_1 = (1, 0, 0, \ldots)$, $v_2 = (0, \frac{1}{2}, 0, 0, \ldots)$, $v_3 = (0, 0, \frac{1}{3}, 0, 0, \ldots)$, then $w_1 = (1, 0, 0, \ldots)$, $w_2 = (1, \frac{1}{2}, 0, 0, \ldots)$, $w_3 = (1, \frac{1}{2}, \frac{1}{3}, 0, 0, \ldots)$. The span of ${v_1, v_2, v_3}$ equals the span of ${w_1, w_2, w_3}$. These examples demonstrate the same principle in different vector spaces, including over the $p$-adic numbers, over a finite field, in a function space, and in a sequence space. The original vectors $v_k$ can be quite different from the cumulative sums $w_k$, but their spans are always equal. This illustrates the robustness and generality of this property of spans. ---