--- COMMENTARY: $\textbf{Exercise 14.}$ Prove that there does not exist a linear map from $\mathbb{F}^5$ to $\mathbb{F}^2$ whose null space equals ${(x_1, x_2, x_3, x_4, x_5) \in \mathbb{F}^5 : x_1 = 3x_2 \text{ and } x_3 = x_4 = x_5}.$ $\textbf{Solution 14.}$ Suppose $U$ is the subspace of $\mathbb{F}^5$ displayed above. Then $(3, 1, 0, 0, 0)$, $(0, 0, 1, 1, 1)$ is a basis of $U$. Hence $\dim U = 2$. If $T \in \mathcal{L}(\mathbb{F}^5, \mathbb{F}^2)$ then from the fundamental theorem of linear maps we have $\dim \operatorname{null} T = \dim \mathbb{F}^5 - \dim \operatorname{range} T = 5 - \dim \operatorname{range} T \geq 3 > \dim U,$ where the first inequality holds because $\operatorname{range} T \subseteq \mathbb{F}^2$. The inequality above shows that if $T \in \mathcal{L}(\mathbb{F}^5, \mathbb{F}^2)$, then $\operatorname{null} T \neq U$, as desired. $\textit{Commentary:}$ This exercise asks us to prove that there is no linear map from $\mathbb{F}^5$ to $\mathbb{F}^2$ whose null space is a specific 2-dimensional subspace $U$. The proof proceeds by contradiction. We suppose that such a linear map $T$ exists, and then use the Fundamental Theorem of Linear Maps to get a contradiction. First, we find a basis for $U$ and calculate its dimension, which is 2. Then, we consider a hypothetical linear map $T$ from $\mathbb{F}^5$ to $\mathbb{F}^2$ with $\operatorname{null} T = U$. Using the Fundamental Theorem, we calculate: $\dim \operatorname{null} T = \dim \mathbb{F}^5 - \dim \operatorname{range} T \geq 5 - 2 = 3,$ because $\operatorname{range} T$ is a subspace of $\mathbb{F}^2$, so its dimension is at most 2. But this contradicts the fact that $\dim U = 2$. Therefore, our assumption that such a $T$ exists must be false. Geometrically, this result says that a 2-dimensional subspace of $\mathbb{F}^5$ cannot be the null space of a linear map to $\mathbb{F}^2$. This is because the null space and the range "complement" each other in the domain, and a 2-dimensional null space would leave "too much" dimension for the range, which cannot fit in $\mathbb{F}^2$. This exercise demonstrates the use of the Fundamental Theorem of Linear Maps to prove the non-existence of certain linear maps. It also shows how the dimensions of the domain, codomain, and a hypothetical null space can constrain the possible linear maps between two spaces. $\textit{Examples:}$ 1. There does not exist a linear map from $\mathbb{R}^6$ to $\mathbb{R}^3$ whose null space is ${(x, y, z, w, t, u) \in \mathbb{R}^6 : x = y, z = w, t = u}$, because this null space is 3-dimensional, but the Fundamental Theorem would imply that it's at least 6 - 3 = 3-dimensional. 2. There does not exist a linear map from $\mathbb{C}^7$ to $\mathbb{C}^2$ whose null space is ${(z_1, z_2, z_3, z_4, z_5, z_6, z_7) \in \mathbb{C}^7 : z_1 = z_2 = z_3, z_4 = z_5, z_6 = z_7}$, because this null space is 3-dimensional, but the Fundamental Theorem would imply that it's at least 7 - 2 = 5-dimensional. 3. There does not exist a linear map from $\mathbb{F}_3^4$ to $\mathbb{F}_3$ whose null space is ${(x, y, z, w) \in \mathbb{F}_3^4 : x = 2y, z = w}$, because this null space is 2-dimensional, but the Fundamental Theorem would imply that it's at least 4 - 1 = 3-dimensional. These examples illustrate the principle in various vector spaces over different fields. In each case, we have a specific subspace that cannot be the null space of a linear map to a certain smaller space, because its dimension is too small compared to what the Fundamental Theorem requires. The calculations involve the dimensions of the domain, the codomain, and the given subspace.