--- COMMENTARY: $\textbf{Exercise 9.}$ Suppose $T \in \mathcal{L}(V, W)$ is injective and $v_1, \ldots, v_n$ is linearly independent in $V$. Prove that $Tv_1, \ldots, Tv_n$ is linearly independent in $W$. $\textbf{Solution 9.}$ To show that $Tv_1, \ldots, Tv_n$ is linearly independent, suppose that $a_1, \ldots, a_n \in \mathbb{F}$ are such that $a_1Tv_1 + \cdots + a_nTv_n = 0.$ Because $T$ is a linear map, this equation can be rewritten as $T(a_1v_1 + \cdots + a_nv_n) = 0.$ Because $T$ is injective, this implies that $a_1v_1 + \cdots + a_nv_n = 0.$ Because $v_1, \ldots, v_n$ is linearly independent, the equation above implies that $a_1 = \cdots = a_n = 0$. Thus $Tv_1, \ldots, Tv_n$ is linearly independent. $\textit{Commentary:}$ This exercise shows that injective linear maps preserve linear independence. If $T$ is injective and $v_1, \ldots, v_n$ is linearly independent, then $Tv_1, \ldots, Tv_n$ is also linearly independent. The proof follows directly from the definitions of linear independence and injectivity. We assume that $Tv_1, \ldots, Tv_n$ is linearly dependent, so there are scalars $a_1, \ldots, a_n$, not all zero, such that $a_1Tv_1 + \cdots + a_nTv_n = 0$. Using the linearity of $T$, we can rewrite this as $T(a_1v_1 + \cdots + a_nv_n) = 0$. But since $T$ is injective, this implies that $a_1v_1 + \cdots + a_nv_n = 0$. This contradicts the linear independence of $v_1, \ldots, v_n$, so our assumption must be false, and $Tv_1, \ldots, Tv_n$ must be linearly independent. Geometrically, this result says that injective linear maps preserve the "spread-outness" of vectors. If a set of vectors is "spread-out" (linearly independent), then their images under an injective map are also "spread-out". This is because injective maps don't "collapse" any vectors together. This property of injective maps is fundamental in linear algebra. It's used, for example, in the proof that a linear map is injective if and only if its null space is ${0}$. It's also a key step in many proofs about bases and dimension. $\textit{Examples:}$ 1. Let $T : \mathbb{R}^3 \to \mathbb{R}^4$ be defined by $T(x, y, z) = (x, y, z, 0)$. $T$ is injective. If $v_1 = (1, 0, 0)$, $v_2 = (0, 1, 0)$, $v_3 = (0, 0, 1)$, then $v_1, v_2, v_3$ is linearly independent in $\mathbb{R}^3$, and $Tv_1 = (1, 0, 0, 0)$, $Tv_2 = (0, 1, 0, 0)$, $Tv_3 = (0, 0, 1, 0)$ is linearly independent in $\mathbb{R}^4$. 2. Let $T : \mathbb{C}^2 \to \mathbb{C}^3$ be defined by $T(z_1, z_2) = (z_1, z_2, z_1 + z_2)$. $T$ is injective. If $v_1 = (1, i)$, $v_2 = (i, -1)$, then $v_1, v_2$ is linearly independent in $\mathbb{C}^2$, and $Tv_1 = (1, i, 1+i)$, $Tv_2 = (i, -1, i-1)$ is linearly independent in $\mathbb{C}^3$. 3. Let $T : \mathbb{F}_2^3 \to \mathbb{F}_2^4$ be defined by $T(x, y, z) = (x, y, z, x+y+z)$. $T$ is injective. If $v_1 = (1, 0, 1)$, $v_2 = (1, 1, 0)$, $v_3 = (0, 1, 1)$, then $v_1, v_2, v_3$ is linearly independent in $\mathbb{F}_2^3$, and $Tv_1 = (1, 0, 1, 0)$, $Tv_2 = (1, 1, 0, 0)$, $Tv_3 = (0, 1, 1, 0)$ is linearly independent in $\mathbb{F}_2^4$. These examples demonstrate the principle in various vector spaces over different fields. In each case, we have an injective linear map and a linearly independent set of vectors, and the images of these vectors under the map are also linearly independent. This illustrates the general principle that injective maps preserve linear independence. ---