counterexample: needs explanation of what w1 is--in the list of w_i, set each one to be equal to -1 of the v in the same position in the list of v's, then adding yields 0 for each vector in the list  --- Commentary: $\textbf{Exercise 11.}$ Prove or give a counterexample: If $v_1, \ldots, v_m$ and $w_1, \ldots, w_m$ are linearly independent lists of vectors in $V$, then the list $v_1 + w_1, \ldots, v_m + w_m$ is linearly independent. $\textbf{Solution 11.}$ The statement above is not true. For example, take $v_1, \ldots, v_m$ to be any linearly independent lists of vectors in $V$, and then let $w_1 = -v_1, \ldots, w_m = -v_m.$ The list $v_1 + w_1, \ldots, v_m + w_m$ will then consist of all 0's and thus will not be linearly independent. $\textit{Commentary:}$ This exercise presents a false statement about linear independence and asks to either prove it or provide a counterexample. The statement claims that if two lists of vectors are linearly independent, then the list obtained by adding corresponding vectors from each list is also linearly independent. The counterexample shows that this is not always the case. The key idea is to choose the second list $w_1, \ldots, w_m$ to be the negatives of the vectors in the first list $v_1, \ldots, v_m$. Then, when corresponding vectors are added, they cancel each other out, resulting in a list of all zero vectors, which is definitely not linearly independent. This counterexample illustrates a common pitfall in linear algebra: the sum of two linearly independent sets is not necessarily linearly independent. In fact, even if the two sets are disjoint and linearly independent, their union may not be linearly independent. The exercise highlights the importance of carefully examining statements about linear independence and being prepared to construct counterexamples. It also demonstrates the value of considering simple, extreme cases when trying to disprove a general statement. $\textit{Additional Examples:}$ In $\mathbb{R}^2$, let $v_1 = (1, 0)$, $v_2 = (0, 1)$, and $w_1 = (-1, 0)$, $w_2 = (0, -1)$. Both ${v_1, v_2}$ and ${w_1, w_2}$ are linearly independent, but ${v_1 + w_1, v_2 + w_2} = {(0, 0), (0, 0)}$ is not linearly independent. In $\mathbb{C}^3$, let $v_1 = (1, 0, 0)$, $v_2 = (0, 1, 0)$, $v_3 = (0, 0, 1)$, and $w_1 = (-1, 0, 0)$, $w_2 = (0, -1, 0)$, $w_3 = (0, 0, -1)$. Both ${v_1, v_2, v_3}$ and ${w_1, w_2, w_3}$ are linearly independent, but ${v_1 + w_1, v_2 + w_2, v_3 + w_3} = {(0, 0, 0), (0, 0, 0), (0, 0, 0)}$ is not linearly independent. In the space of polynomials $\mathbb{Q}[x]$, let $v_1 = 1$, $v_2 = x$, and $w_1 = -1$, $w_2 = -x$. Both ${v_1, v_2}$ and ${w_1, w_2}$ are linearly independent, but ${v_1 + w_1, v_2 + w_2} = {0, 0}$ is not linearly independent. In the space of continuous functions $C([0,1], \mathbb{R})$, let $v_1 = \sin(x)$, $v_2 = \cos(x)$, and $w_1 = -\sin(x)$, $w_2 = -\cos(x)$. Both ${v_1, v_2}$ and ${w_1, w_2}$ are linearly independent, but ${v_1 + w_1, v_2 + w_2} = {0, 0}$ (where 0 represents the zero function) is not linearly independent. These examples demonstrate the same principle in various vector spaces, including $\mathbb{R}^n$, $\mathbb{C}^n$, polynomial spaces, and function spaces. In each case, adding corresponding vectors from two linearly independent lists results in a list that is not linearly independent, because the vectors cancel each other out.