--- Commentary: $\textbf{Exercise 8.}$ Suppose $v_1, v_2, v_3, v_4$ is linearly independent in $V$. Prove that the list $v_1 - v_2$, $v_2 - v_3$, $v_3 - v_4$, $v_4$ is also linearly independent. $\textbf{Solution 8.}$ To prove that the list displayed above is linearly independent, suppose $a_1, a_2, a_3, a_4 \in \mathbb{F}$ are such that $a_1(v_1 - v_2) + a_2(v_2 - v_3) + a_3(v_3 - v_4) + a_4v_4 = 0.$ Rearranging terms, the equation above can be rewritten as $a_1v_1 + (a_2 - a_1)v_2 + (a_3 - a_2)v_3 + (a_4 - a_3)v_4 = 0.$ Because $v_1, v_2, v_3, v_4$ is linearly independent, the equation above implies that $\begin{align*} a_1 &= 0 \\ a_2 - a_1 &= 0 \\ a_3 - a_2 &= 0 \\ a_4 - a_3 &= 0. \end{align*}$ The first equation above tells us that $a_1 = 0$. That information, combined with the second equation, tells us that $a_2 = 0$. That information, combined with the third equation, tells us that $a_3 = 0$. That information, combined with the fourth equation, tells us that $a_4 = 0$. Thus $v_1 - v_2$, $v_2 - v_3$, $v_3 - v_4$, $v_4$ is linearly independent. $\textit{Commentary:}$ This exercise demonstrates how to prove the linear independence of a transformed list of vectors, given the linear independence of the original list. The key steps are setting up a linear combination equal to zero, rearranging the terms to match the original list of vectors, and then using the linear independence of the original list to conclude that all coefficients must be zero. The specific transformation used here, replacing each vector (except the last) by the difference between it and the next vector, can be thought of as a discrete derivative operation. The exercise shows that this operation preserves linear independence. This type of argument is often used in the study of differential equations, where one wants to prove properties of a set of functions and their derivatives based on properties of the original functions. $\textit{Example:}$ In $\mathbb{Q}_p^4$, if the list $(1, p, p^2, p^3)$, $(p, p^2, p^3, 1)$, $(p^2, p^3, 1, p)$, $(p^3, 1, p, p^2)$ is linearly independent, then so is the list $(1-p, p-p^2, p^2-p^3, p^3)$, $(p-p^2, p^2-p^3, p^3-1, 1)$, $(p^2-p^3, p^3-1, 1-p, p)$, $(p^3, 1, p, p^2)$. In the space of polynomials over $\mathbb{Z}/3\mathbb{Z}$, if the list $1$, $x$, $x^2$, $x^3$ is linearly independent, then so is the list $1 - x$, $x - x^2$, $x^2 - x^3$, $x^3$. In the space of matrices over $\mathbb{F}_2$, if the list $\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}$, $\begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}$, $\begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}$, $\begin{pmatrix} 1 & 0 \ 1 & 1 \end{pmatrix}$ is linearly independent, then so is the list $\begin{pmatrix} 1 & 1 \ 1 & 0 \end{pmatrix}$, $\begin{pmatrix} 1 & 0 \ 1 & 1 \end{pmatrix}$, $\begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}$, $\begin{pmatrix} 1 & 0 \ 1 & 1 \end{pmatrix}$. These examples illustrate the same principle in various vector spaces. The specific arithmetic of the field may change the details of the calculations, but the overall argument remains the same. This demonstrates the power and generality of linear algebraic techniques. --- $\textbf{Exercise 8.}$ Suppose $v_1, v_2, v_3, v_4$ is linearly independent in $V$. Prove that the list $v_1 - v_2$, $v_2 - v_3$, $v_3 - v_4$, $v_4$ is also linearly independent. $\textbf{Solution 8.}$ To prove that the list displayed above is linearly independent, suppose $a_1, a_2, a_3, a_4 \in \mathbb{F}$ are such that $a_1(v_1 - v_2) + a_2(v_2 - v_3) + a_3(v_3 - v_4) + a_4v_4 = 0.$ Rearranging terms, the equation above can be rewritten as $a_1v_1 + (a_2 - a_1)v_2 + (a_3 - a_2)v_3 + (a_4 - a_3)v_4 = 0.$ Because $v_1, v_2, v_3, v_4$ is linearly independent, the equation above implies that $\begin{align*} a_1 &= 0 \\ a_2 - a_1 &= 0 \\ a_3 - a_2 &= 0 \\ a_4 - a_3 &= 0. \end{align*}$ The first equation above tells us that $a_1 = 0$. That information, combined with the second equation, tells us that $a_2 = 0$. That information, combined with the third equation, tells us that $a_3 = 0$. That information, combined with the fourth equation, tells us that $a_4 = 0$. Thus $v_1 - v_2$, $v_2 - v_3$, $v_3 - v_4$, $v_4$ is linearly independent. $\textit{Commentary:}$ This exercise demonstrates how to prove the linear independence of a transformed list of vectors, given the linear independence of the original list. The key steps are setting up a linear combination equal to zero, rearranging the terms to match the original list of vectors, and then using the linear independence of the original list to conclude that all coefficients must be zero. The specific transformation used here, replacing each vector (except the last) by the difference between it and the next vector, can be thought of as a discrete derivative operation. The exercise shows that this operation preserves linear independence. This type of argument is often used in the study of differential equations, where one wants to prove properties of a set of functions and their derivatives based on properties of the original functions. $\textit{Additional Examples:}$ In $\mathbb{Q}^4$, if the list $(1, 2, 3, 4)$, $(2, 4, 6, 8)$, $(3, 6, 9, 12)$, $(4, 8, 12, 16)$ is linearly independent, then so is the list $(1, 0, -1, -2)$, $(1, 2, 3, 4)$, $(1, 2, 3, 4)$, $(4, 8, 12, 16)$. In $\mathbb{F}_7^4$, if the list $(1, 2, 3, 4)$, $(2, 4, 6, 1)$, $(3, 6, 2, 5)$, $(4, 1, 5, 2)$ is linearly independent, then so is the list $(6, 5, 4, 3)$, $(5, 3, 1, 6)$, $(4, 2, 6, 5)$, $(4, 1, 5, 2)$. Note that in $\mathbb{F}_7$, $-1 = 6$, $-2 = 5$, etc. In the space of polynomials $\mathbb{R}[x]$, if the list $1$, $x$, $x^2$, $x^3$ is linearly independent, then so is the list $1$, $x-1$, $x^2-x$, $x^3$. In the space of continuous functions $C([0,1], \mathbb{R})$, if the list $1$, $x$, $x^2$, $e^x$ is linearly independent, then so is the list $1$, $x-1$, $x^2-x$, $e^x$. These examples illustrate the same principle in various vector spaces. The specific arithmetic of the field may change the details of the calculations, but the overall argument remains the same. This demonstrates the power and generality of linear algebraic techniques. ---