--- Commentary: $\textbf{Exercise 12.}$ Suppose $T$ is a linear map from $\mathbb{F}^4$ to $\mathbb{F}^2$ such that $\operatorname{null} T = {(x_1, x_2, x_3, x_4) \in \mathbb{F}^4 : x_1 = 5x_2 \text{ and } x_3 = 7x_4}.$ Prove that $T$ is surjective. $\textbf{Solution 12.}$ The hypothesis implies that $(5, 1, 0, 0)$, $(0, 0, 7, 1)$ is a basis of $\operatorname{null} T$. Hence $\dim \operatorname{null} T = 2$. From the fundamental theorem of linear maps we have $\dim \operatorname{range} T = \dim \mathbb{F}^4 - \dim \operatorname{null} T = 4 - 2 = 2.$ Because $\operatorname{range} T$ is a two-dimensional subspace of $\mathbb{F}^2$, we have $\operatorname{range} T = \mathbb{F}^2$. In other words, $T$ is surjective. $\textit{Commentary:}$ This exercise gives a specific description of the null space of a linear map $T$ from $\mathbb{F}^4$ to $\mathbb{F}^2$, and asks us to prove that $T$ is surjective. The key observation is that the given description of $\operatorname{null} T$ allows us to find a basis for it, and hence to calculate its dimension. Specifically, the vectors $(5, 1, 0, 0)$ and $(0, 0, 7, 1)$ form a basis for $\operatorname{null} T$, so $\dim \operatorname{null} T = 2$. We then use the Fundamental Theorem of Linear Maps, which states that for a linear map $T : V \to W$ between finite-dimensional vector spaces, we have $\dim V = \dim \operatorname{null} T + \dim \operatorname{range} T.$ In this case, $\dim \mathbb{F}^4 = 4$ and $\dim \operatorname{null} T = 2$, so we can solve for $\dim \operatorname{range} T$ and get $\dim \operatorname{range} T = 2$. But $\operatorname{range} T$ is a subspace of $\mathbb{F}^2$, and the only 2-dimensional subspace of $\mathbb{F}^2$ is $\mathbb{F}^2$ itself. So $\operatorname{range} T = \mathbb{F}^2$, which means that $T$ is surjective. Geometrically, this result says that if the null space of a linear map from $\mathbb{F}^4$ to $\mathbb{F}^2$ is a 2-dimensional subspace, then the map must be surjective. This is because the null space and the range "complement" each other in the domain, and if the null space is "half" of the domain, then the range must be "all" of the codomain. This exercise demonstrates the power of the Fundamental Theorem of Linear Maps in deducing properties of linear maps from information about their null spaces or ranges. It also shows how the dimension of a subspace can determine its geometric properties. $\textit{Examples:}$ 1. Let $T : \mathbb{R}^5 \to \mathbb{R}^2$ be a linear map such that $\operatorname{null} T = {(x, y, z, w, t) \in \mathbb{R}^5 : x = 2y = 3z = 4w = 5t}$. Then $\dim \operatorname{null} T = 1$, so $\dim \operatorname{range} T = 5 - 1 = 4 > 2$. Thus, $T$ is not surjective. 2. Let $T : \mathbb{C}^6 \to \mathbb{C}^3$ be a linear map such that $\operatorname{null} T = {(z_1, z_2, z_3, z_4, z_5, z_6) \in \mathbb{C}^6 : z_1 = iz_2, z_3 = -z_4, z_5 = 2iz_6}$. Then $\dim \operatorname{null} T = 3$, so $\dim \operatorname{range} T = 6 - 3 = 3$. Thus, $T$ is surjective. 3. Let $T : \mathbb{F}_2^5 \to \mathbb{F}_2^3$ be a linear map such that $\operatorname{null} T = {(x, y, z, w, t) \in \mathbb{F}_2^5 : x = y, z = w + t}$. Then $\dim \operatorname{null} T = 3$, so $\dim \operatorname{range} T = 5 - 3 = 2 < 3$. Thus, $T$ is not surjective. These examples illustrate the principle in various vector spaces over different fields. In each case, we use the given description of the null space to find its dimension, and then use the Fundamental Theorem of Linear Maps to deduce the dimension of the range. If the dimension of the range equals the dimension of the codomain, the map is surjective; otherwise, it's not.