--- Commentary: $\textbf{Exercise 10.}$ Prove or give a counterexample: If $v_1, v_2, \ldots, v_m$ is a linearly independent list of vectors in $V$ and $\lambda \in \mathbb{F}$ with $\lambda \neq 0$, then $\lambda v_1, \lambda v_2, \ldots, \lambda v_m$ is linearly independent. $\textbf{Solution 10.}$ Suppose $v_1, v_2, \ldots, v_m$ is a linearly independent list of vectors in $V$ and $\lambda \in \mathbb{F}$ with $\lambda \neq 0$. Suppose $a_1, a_2, \ldots, a_m \in \mathbb{F}$ are such that $a_1\lambda v_1 + \cdots + a_m\lambda v_m = 0.$ Because $v_1, v_2, \ldots, v_m$ is linearly independent, we have $a_1\lambda = \cdots = a_m\lambda = 0.$ Because $\lambda \neq 0$, we have $a_1 = \cdots = a_m = 0$. Hence $\lambda v_1, \lambda v_2, \ldots, \lambda v_m$ is linearly independent. $\textit{Commentary:}$ This exercise proves that multiplying each vector in a linearly independent list by the same non-zero scalar preserves linear independence. The proof assumes that a linear combination of the scaled vectors equals zero. Since $\lambda$ is non-zero, this implies that the same linear combination of the original vectors equals zero. But since the original list is linearly independent, all coefficients in this linear combination must be zero. Thus, all coefficients in the linear combination of the scaled vectors are zero, proving that the scaled list is linearly independent. This result is a consequence of the fact that scaling vectors doesn't change their "direction", only their "magnitude". So if the original vectors are "pointing in different directions" (which is what linear independence means geometrically), then the scaled vectors are also "pointing in different directions". This exercise provides practice in working with the definition of linear independence and illustrates a fundamental property of linearly independent lists. $\textit{Additional Examples:}$ If $v_1, v_2, \ldots, v_m$ is a linearly independent list of vectors in a real vector space $V$ and $\lambda$ is a non-zero real number, then $\lambda v_1, \lambda v_2, \ldots, \lambda v_m$ is linearly independent. If $v_1, v_2, \ldots, v_m$ is a linearly independent list of vectors in a complex vector space $V$ and $\lambda$ is a non-zero complex number, then $\lambda v_1, \lambda v_2, \ldots, \lambda v_m$ is linearly independent. If $f_1, f_2, \ldots, f_m$ is a linearly independent list of functions in $C([0,1], \mathbb{R})$ and $\lambda$ is a non-zero real number, then $\lambda f_1, \lambda f_2, \ldots, \lambda f_m$ is linearly independent. If $p_1, p_2, \ldots, p_m$ is a linearly independent list of polynomials in $\mathbb{F}_q[x]$ and $\lambda$ is a non-zero element of $\mathbb{F}_q$, then $\lambda p_1, \lambda p_2, \ldots, \lambda p_m$ is linearly independent. These examples demonstrate that the result holds over any field, including the real numbers, complex numbers, and finite fields, and in various types of vector spaces, including function spaces and polynomial spaces.