--- Commentary: $\textbf{Exercise 13.}$ Suppose $v_1, \ldots, v_m$ is linearly independent in $V$ and $w \in V$. Show that $v_1, \ldots, v_m, w \text{ is linearly independent} \iff w \notin \operatorname{span}(v_1, \ldots, v_m).$ $\textbf{Solution 13.}$ First suppose $v_1, \ldots, v_m, w$ is linearly independent. Then no vector in that list is a linear combination of the other vectors in the list. In particular, $w \notin \operatorname{span}(v_1, \ldots, v_m)$. Conversely, suppose $w \notin \operatorname{span}(v_1, \ldots, v_m)$. Because $v_1, \ldots, v_m$ is linearly independent, no $v_k$ is in the span of $v_1, \ldots, v_{k-1}$. Thus the linear dependence lemma (2.19) implies that $v_1, \ldots, v_m, w$ is linearly independent. $\textit{Commentary:}$ This exercise characterizes when a vector $w$ can be added to a linearly independent set ${v_1, \ldots, v_m}$ to obtain a larger linearly independent set. The condition is that $w$ is not in the span of ${v_1, \ldots, v_m}$. The proof of the forward direction is straightforward: if ${v_1, \ldots, v_m, w}$ is linearly independent, then by definition, no vector in this set is a linear combination of the others. In particular, $w$ is not a linear combination of ${v_1, \ldots, v_m}$, so $w$ is not in their span. The proof of the reverse direction uses the linear dependence lemma. Because ${v_1, \ldots, v_m}$ is linearly independent, no $v_k$ is in the span of the preceding vectors. The linear dependence lemma then implies that if $w$ is also not in the span of ${v_1, \ldots, v_m}$, then the entire set ${v_1, \ldots, v_m, w}$ must be linearly independent. This result is useful when constructing linearly independent sets incrementally. Given a linearly independent set, we can add any vector that is not in the span of the set to obtain a larger linearly independent set. The exercise also reinforces the connection between linear independence and the span of a set of vectors. It provides practice in using the linear dependence lemma and in reasoning about linear combinations and spans. $\textit{Additional Examples:}$ In $\mathbb{R}^3$, let $v_1 = (1, 0, 0)$, $v_2 = (0, 1, 0)$. The set ${v_1, v_2}$ is linearly independent. If $w = (0, 0, 1)$, then $w \notin \operatorname{span}(v_1, v_2)$, and indeed, ${v_1, v_2, w}$ is linearly independent. However, if $w = (1, 1, 0)$, then $w \in \operatorname{span}(v_1, v_2)$, and indeed, ${v_1, v_2, w}$ is linearly dependent. In $\mathbb{C}^4$, let $v_1 = (1, i, 0, 0)$, $v_2 = (0, 1, i, 0)$, $v_3 = (0, 0, 1, i)$. The set ${v_1, v_2, v_3}$ is linearly independent. If $w = (1, 1, 1, 1)$, then $w \notin \operatorname{span}(v_1, v_2, v_3)$, and indeed, ${v_1, v_2, v_3, w}$ is linearly independent. However, if $w = (1, 2i, 1, i)$, then $w \in \operatorname{span}(v_1, v_2, v_3)$ (specifically, $w = v_1 + v_2 + v_3$), and indeed, ${v_1, v_2, v_3, w}$ is linearly dependent. In the space of polynomials $\mathbb{Q}[x]$, let $v_1 = 1$, $v_2 = x$. The set ${v_1, v_2}$ is linearly independent. If $w = x^2$, then $w \notin \operatorname{span}(v_1, v_2)$, and indeed, ${v_1, v_2, w}$ is linearly independent. However, if $w = 2 - 3x$, then $w \in \operatorname{span}(v_1, v_2)$ (specifically, $w = 2v_1 - 3v_2$), and indeed, ${v_1, v_2, w}$ is linearly dependent. In the space of continuous functions $C([0,1], \mathbb{R})$, let $v_1 = \sin(x)$, $v_2 = \cos(x)$. The set ${v_1, v_2}$ is linearly independent. If $w = x$, then $w \notin \operatorname{span}(v_1, v_2)$, and indeed, ${v_1, v_2, w}$ is linearly independent. However, if $w = \sin(x) + \cos(x)$, then $w \in \operatorname{span}(v_1, v_2)$ (specifically, $w = v_1 + v_2$), and indeed, ${v_1, v_2, w}$ is linearly dependent. These examples demonstrate the result in various vector spaces, including $\mathbb{R}^n$, $\mathbb{C}^n$, polynomial spaces, and function spaces. In each case, we start with a linearly independent set and consider adding a new vector $w$. If $w$ is not in the span of the original set, then the augmented set is linearly independent. If $w$ is in the span of the original set, then the augmented set is linearly dependent. This illustrates the tight connection between linear independence and the span of a set of vectors. ---