--- Commentary: $\textbf{Exercise 3.}$ Suppose $v_1, \ldots, v_m$ is a list of vectors in $V$. Define $T \in \mathcal{L}(\mathbb{F}^m, V)$ by $T(z_1, \ldots, z_m) = z_1v_1 + \cdots + z_mv_m.$ (a) What property of $T$ corresponds to $v_1, \ldots, v_m$ spanning $V$? (b) What property of $T$ corresponds to the list $v_1, \ldots, v_m$ being linearly independent? $\textbf{Solution 3.}$ (a) The list $v_1, \ldots, v_m$ spans $V$ if and only if $T$ is surjective. (b) The list $v_1, \ldots, v_m$ is linearly independent if and only if $T$ is injective. $\textit{Commentary:}$ This exercise connects properties of a list of vectors to properties of a specific linear map defined using these vectors. Part (a) shows that the spanning property of the list corresponds to the surjectivity of $T$. This makes sense geometrically: if $v_1, \ldots, v_m$ span $V$, then every vector in $V$ can be written as a linear combination of these vectors, which is exactly what it means for $T$ to be surjective. Part (b) shows that the linear independence of the list corresponds to the injectivity of $T$. Again, this makes geometric sense: if $v_1, \ldots, v_m$ are linearly independent, then no non-trivial linear combination of these vectors is zero, which is exactly what it means for $T$ to be injective. This exercise provides a way to translate between properties of lists of vectors and properties of linear maps. It's a key idea in the study of bases and dimension. $\textit{Examples:}$ 1. Let $V = \mathbb{R}^3$, and let $v_1 = (1, 0, 0)$, $v_2 = (0, 1, 0)$, $v_3 = (0, 0, 1)$. Define $T \in \mathcal{L}(\mathbb{R}^3, \mathbb{R}^3)$ by $T(z_1, z_2, z_3) = z_1v_1 + z_2v_2 + z_3v_3$. Then $v_1, v_2, v_3$ is a basis of $\mathbb{R}^3$, so it spans $\mathbb{R}^3$ and is linearly independent. Indeed, $T$ is bijective (both surjective and injective). 2. Let $V = \mathcal{P}_2(\mathbb{C})$, and let $v_1 = 1$, $v_2 = z$, $v_3 = z^2$. Define $T \in \mathcal{L}(\mathbb{C}^3, \mathcal{P}_2(\mathbb{C}))$ by $T(z_1, z_2, z_3) = z_1v_1 + z_2v_2 + z_3v_3$. Then $v_1, v_2, v_3$ is a basis of $\mathcal{P}_2(\mathbb{C})$, so it spans $\mathcal{P}_2(\mathbb{C})$ and is linearly independent. Indeed, $T$ is bijective. 3. Let $V = \mathbb{F}_2^3$, and let $v_1 = (1, 0, 0)$, $v_2 = (0, 1, 0)$. Define $T \in \mathcal{L}(\mathbb{F}_2^2, \mathbb{F}_2^3)$ by $T(z_1, z_2) = z_1v_1 + z_2v_2$. Then $v_1, v_2$ spans a proper subspace of $\mathbb{F}_2^3$ and is linearly independent. Indeed, $T$ is injective but not surjective. These examples demonstrate the principle in various vector spaces. In the first two examples, the given list of vectors is a basis, so $T$ is bijective. In the third example, the list spans a proper subspace and is linearly independent, so $T$ is injective but not surjective. These correspond exactly to the stated properties of spanning and linear independence.