--- Commentary: $\textbf{Exercise 12.}$ Suppose $v_1, \ldots, v_m$ is linearly independent in $V$ and $w \in V$. Prove that if $v_1 + w, \ldots, v_m + w$ is linearly dependent, then $w \in \operatorname{span}(v_1, \ldots, v_m)$. $\textbf{Solution 12.}$ Suppose $v_1 + w, \ldots, v_m + w$ is linearly dependent. Then there exist scalars $a_1, \ldots, a_m$, not all 0, such that $a_1(v_1 + w) + \cdots + a_m(v_m + w) = 0.$ Rearranging this equation, we have $a_1v_1 + \cdots + a_mv_m = -(a_1 + \cdots + a_m)w.$ If $a_1 + \cdots + a_m$ were 0, then the equation above would contradict the linear independence of $v_1, \ldots, v_m$. Thus $a_1 + \cdots + a_m \neq 0$. Hence we can divide both sides of the equation above by $-(a_1 + \cdots + a_m)$, showing that $w \in \operatorname{span}(v_1, \ldots, v_m)$. $\textit{Commentary:}$ This exercise explores a condition under which a vector $w$ must be in the span of a linearly independent set ${v_1, \ldots, v_m}$. The condition is that the set obtained by adding $w$ to each vector in the linearly independent set is linearly dependent. The proof proceeds by assuming that ${v_1 + w, \ldots, v_m + w}$ is linearly dependent, so there is a non-trivial linear combination of these vectors that equals zero. Rearranging the terms in this linear combination, we obtain a linear combination of ${v_1, \ldots, v_m}$ that equals a multiple of $w$. If this multiple were zero, it would contradict the linear independence of ${v_1, \ldots, v_m}$. Therefore, we can divide by this multiple to express $w$ as a linear combination of ${v_1, \ldots, v_m}$, showing that $w$ is in their span. This result is useful in situations where we want to determine whether a given vector is in the span of a linearly independent set. Instead of trying to express the vector directly as a linear combination of the set, we can check whether adding the vector to each element of the set results in a linearly dependent set. The exercise also reinforces the concept of linear dependence and its relationship to the span of a set of vectors. It provides practice in manipulating linear combinations and using the properties of linearly independent sets. $\textit{Additional Examples:}$ In $\mathbb{R}^3$, let $v_1 = (1, 0, 0)$, $v_2 = (0, 1, 0)$, and $w = (1, 1, 1)$. The set ${v_1, v_2}$ is linearly independent, and ${v_1 + w, v_2 + w} = {(2, 1, 1), (1, 2, 1)}$ is linearly dependent. Indeed, $w = 1 \cdot v_1 + 1 \cdot v_2$, so $w \in \operatorname{span}(v_1, v_2)$. In $\mathbb{C}^4$, let $v_1 = (1, i, 0, 0)$, $v_2 = (0, 1, i, 0)$, $v_3 = (0, 0, 1, i)$, and $w = (1, 1, 1, 1)$. The set ${v_1, v_2, v_3}$ is linearly independent, and ${v_1 + w, v_2 + w, v_3 + w} = {(2, 1+i, 1, 1), (1, 2, 1+i, 1), (1, 1, 2, 1+i)}$ is linearly dependent. Indeed, $w = 1 \cdot v_1 + 1 \cdot v_2 + 1 \cdot v_3$, so $w \in \operatorname{span}(v_1, v_2, v_3)$. In the space of polynomials $\mathbb{Q}[x]$, let $v_1 = 1$, $v_2 = x$, and $w = x^2$. The set ${v_1, v_2}$ is linearly independent, and ${v_1 + w, v_2 + w} = {1 + x^2, x + x^2}$ is linearly dependent. However, $w \notin \operatorname{span}(v_1, v_2)$, because $x^2$ cannot be expressed as a linear combination of 1 and $x$. In the space of continuous functions $C([0,1], \mathbb{R})$, let $v_1 = \sin(x)$, $v_2 = \cos(x)$, and $w = x$. The set ${v_1, v_2}$ is linearly independent, and ${v_1 + w, v_2 + w} = {\sin(x) + x, \cos(x) + x}$ is linearly dependent. However, $w \notin \operatorname{span}(v_1, v_2)$, because $x$ cannot be expressed as a linear combination of $\sin(x)$ and $\cos(x)$. These examples illustrate the result in various vector spaces, including $\mathbb{R}^n$, $\mathbb{C}^n$, polynomial spaces, and function spaces. The first two examples confirm the result: $w$ is in the span of the linearly independent set, and adding $w$ to each vector in the set results in a linearly dependent set. The last two examples show that the converse is not true: adding $w$ to each vector in a linearly independent set can result in a linearly dependent set even if $w$ is not in the span of the original set.