--- Commentary: $\textbf{Exercise 7.}$ Suppose $v_1, v_2, v_3, v_4$ is a basis of $V$. Prove that $v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4$ is also a basis of $V$. $\textbf{Solution 7.}$ To prove that $v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4$ is linearly independent, suppose $a, b, c, d \in \mathbb{F}$ and $a(v_1 + v_2) + b(v_2 + v_3) + c(v_3 + v_4) + dv_4 = 0.$ Then $av_1 + (a + b)v_2 + (b + c)v_3 + (c + d)v_4 = 0.$ Because $v_1, v_2, v_3, v_4$ is linearly independent, this implies that $a = a + b = b + c = c + d = 0,$ which implies that $a = b = c = d = 0$. Thus $v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4$ is linearly independent. To prove that $v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4$ spans $V$, suppose $u \in V$. Because $v_1, v_2, v_3, v_4$ is a basis of $V$, there exist $a, b, c, d \in \mathbb{F}$ such that $u = av_1 + bv_2 + cv_3 + dv_4.$ Thus $u = a(v_1 + v_2) + (b - a)(v_2 + v_3) + (c - b + a)(v_3 + v_4) + (d - c + b - a)v_4.$ Hence $v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4$ spans $V$. Because $v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4$ is linearly independent and spans $V$, it is a basis of $V$. $\textit{Commentary:}$ This exercise asks to prove that if $v_1, v_2, v_3, v_4$ is a basis of $V$, then a new list formed by adding adjacent vectors in this basis is also a basis of $V$. The proof has two parts. First, we prove that the new list is linearly independent. We assume a linear combination of the new vectors equals zero and show that this implies a linear combination of the original vectors equals zero. Since the original vectors are linearly independent, this implies that all coefficients must be zero, proving the linear independence of the new list. Second, we prove that the new list spans $V$. We take an arbitrary vector $u$ in $V$ and express it as a linear combination of the original basis vectors. Then, we rearrange this linear combination to express $u$ as a linear combination of the new vectors, proving that the new list spans $V$. Together, these two parts show that the new list is a basis of $V$. This exercise illustrates a way to transform one basis into another. The transformation here is simple: we're adding adjacent vectors. But the same principles can be used to prove that more complex transformations of bases also yield bases. $\textit{Examples:}$ 1. In $\mathbb{R}^4$, if $v_1 = (1, 0, 0, 0)$, $v_2 = (0, 1, 0, 0)$, $v_3 = (0, 0, 1, 0)$, $v_4 = (0, 0, 0, 1)$ is the standard basis, then $v_1 + v_2 = (1, 1, 0, 0)$, $v_2 + v_3 = (0, 1, 1, 0)$, $v_3 + v_4 = (0, 0, 1, 1)$, $v_4 = (0, 0, 0, 1)$ is also a basis. 2. In $\mathcal{P}_3(\mathbb{R})$, if $1, x, x^2, x^3$ is the standard basis, then $1 + x, x + x^2, x^2 + x^3, x^3$ is also a basis. 3. In the space of $2 \times 2$ matrices $M_{2 \times 2}(\mathbb{C})$, if $\begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}$ is a basis, then $\begin{pmatrix} 1 & 1 \ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \ 1 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}$ is also a basis. These examples demonstrate the principle in various vector spaces, including $\mathbb{R}^n$, polynomial spaces, and matrix spaces. In each case, adding adjacent basis vectors yields a new basis. This illustrates the flexibility in choosing bases for a vector space. ---