--- Commentary: $\textbf{Exercise 9.}$ Prove or give a counterexample: If $v_1, v_2, \ldots, v_m$ is a linearly independent list of vectors in $V$, then $5v_1 - 4v_2, v_2, v_3, \ldots, v_m$ is linearly independent. $\textbf{Solution 9.}$ Suppose $v_1, v_2, \ldots, v_m$ is a linearly independent list of vectors in $V$. Suppose $a_1, a_2, \ldots, a_m \in \mathbb{F}$ are such that $a_1(5v_1 - 4v_2) + a_2v_2 + \cdots + a_mv_m = 0.$ Then $5a_1v_1 + (a_2 - 4a_1)v_2 + a_3v_3 + \cdots + a_mv_m = 0.$ Because $v_1, v_2, \ldots, v_m$ is linearly independent, we have $a_1 = a_2 - 4a_1 = a_3 = \cdots = a_m = 0.$ Thus $a_1 = a_2 = a_3 = \cdots = a_m = 0$. Hence $5v_1 - 4v_2, v_2, v_3, \ldots, v_m$ is linearly independent. $\textit{Commentary:}$ This exercise asks to prove that a specific linear transformation applied to the first two vectors in a linearly independent list preserves linear independence. The proof assumes that a linear combination of the transformed vectors equals zero, and then rewrites this as a linear combination of the original vectors. Since the original list is linearly independent, all coefficients in this linear combination must be zero. This includes the coefficient of $v_2$, which is $a_2 - 4a_1$. Since $a_1$ is also zero, this implies that $a_2$ is zero. Thus, all coefficients in the original linear combination are zero, proving that the transformed list is linearly independent. This exercise illustrates how the linear independence of a list of vectors is preserved under certain types of linear transformations. It also provides practice in working with the definition of linear independence and manipulating linear combinations. $\textit{Additional Examples:}$ If $v_1, v_2, \ldots, v_m$ is a linearly independent list of vectors in $V$, then $3v_1 + 2v_2, v_2, v_3, \ldots, v_m$ is linearly independent. If $v_1, v_2, \ldots, v_m$ is a linearly independent list of vectors in $V$, then $v_1, 6v_2 - 5v_1, v_3, \ldots, v_m$ is linearly independent. If $f_1, f_2, \ldots, f_m$ is a linearly independent list of functions in $C([0,1], \mathbb{R})$, then $2f_1 - 3f_2, f_2, f_3, \ldots, f_m$ is linearly independent. If $p_1, p_2, \ldots, p_m$ is a linearly independent list of polynomials in $\mathbb{Q}[x]$, then $7p_1 + 4p_2, p_2, p_3, \ldots, p_m$ is linearly independent. These examples demonstrate that the result holds for various types of linear transformations applied to the first two vectors in the list, and in different vector spaces, including function spaces and polynomial spaces.