--- COMMENTARY: $\textbf{Exercise 18.}$ Suppose $U$ and $V$ are finite-dimensional and that $U$ is a subspace of $V$. Prove that there exists $T \in \mathcal{L}(V, W)$ such that $\operatorname{null} T = U$ if and only if $\dim U \geq \dim V - \dim W$. $\textbf{Solution 18.}$ First suppose there exists $T \in \mathcal{L}(V, W)$ such that $\operatorname{null} T = U$. Then $\dim U = \dim \operatorname{null} T = \dim V - \dim \operatorname{range} T \geq \dim V - \dim W,$ where the second equality comes from the fundamental theorem of linear maps. To prove the other direction, now suppose $\dim U \geq \dim V - \dim W$. Let $u_1, \ldots, u_m$ be a basis of $U$. Extend $u_1, \ldots, u_m$ to a basis $u_1, \ldots, u_m, v_1, \ldots, v_n$ of $V$. Let $w_1, \ldots, w_p$ be a basis of $W$. For $a_1, \ldots, a_m, b_1, \ldots, b_n \in \mathbb{F}$ define $T(a_1u_1 + \cdots + a_mu_m + b_1v_1 + \cdots + b_nv_n) = b_1w_1 + \cdots + b_nw_n.$ Because $\dim W \geq \dim V - \dim U$, we have $p \geq n$ and so $w_n$ on the right side of the equation above makes sense. Clearly $T \in \mathcal{L}(V, W)$ and $\operatorname{null} T = U$. $\textit{Commentary:}$ This exercise characterizes when a given subspace $U$ of a vector space $V$ can be the null space of a linear map from $V$ to another vector space $W$, in terms of the dimensions of $U$, $V$, and $W$. The first part of the proof assumes that a linear map $T : V \to W$ exists with $\operatorname{null} T = U$ and uses the Fundamental Theorem of Linear Maps to show that $\dim U \geq \dim V - \dim W$. The key insight is that $\dim \operatorname{null} T = \dim V - \dim \operatorname{range} T$, and since $\operatorname{range} T$ is a subspace of $W$, we have $\dim \operatorname{range} T \leq \dim W$. The second part of the proof assumes that $\dim U \geq \dim V - \dim W$ and constructs a linear map $T : V \to W$ with $\operatorname{null} T = U$. The construction chooses a basis for $U$, extends it to a basis for $V$, and also chooses a basis for $W$. It then defines $T$ to map the vectors in the basis of $V$ that are not in the basis of $U$ to the first $n$ vectors in the basis of $W$, where $n = \dim V - \dim U$. This ensures that $\operatorname{null} T = U$. Geometrically, this result says that for a subspace $U$ of $V$ to be the null space of a map from $V$ to $W$, it must be "large enough" relative to the dimensions of $V$ and $W$. Specifically, the codimension of $U$ in $V$ (which is $\dim V - \dim U$) must be at most $\dim W$. This is because the null space and the range of a linear map "complement" each other in the domain, and the range is a subspace of the codomain. This result is useful in various contexts where we want to construct linear maps with prescribed null spaces, or where we want to understand the constraints that the existence of such maps places on the dimensions of the spaces involved. $\textit{Examples:}$ 1. Let $V = \mathbb{R}^4$, $W = \mathbb{R}^3$, and $U = {(x, y, z, w) \in \mathbb{R}^4 : x = y}$. Then $\dim U = 3$, and $\dim U = 3 \geq 4 - 3 = \dim V - \dim W$. Indeed, the map $T(x, y, --- Okay, let's continue with the requested problems from Section 3B. $\textbf{Exercise 18.}$ Suppose $V$ and $W$ are finite-dimensional and that $U$ is a subspace of $V$. Prove that there exists $T \in \mathcal{L}(V, W)$ such that null $T = U$ if and only if $\text{dim } U \geq \text{dim } V - \text{dim } W$. $\textbf{Solution 18.}$ First suppose there exists $T \in \mathcal{L}(V, W)$ such that null $T = U$. Then $\text{dim } U = \text{dim null } T = \text{dim } V - \text{dim range } T \geq \text{dim } V - \text{dim } W,$ where the second equality comes from the fundamental theorem of linear maps. To prove the other direction, now suppose $\text{dim } U \geq \text{dim } V - \text{dim } W$. Let $u_1, \ldots, u_m$ be a basis of $U$. Extend to a basis $u_1, \ldots, u_m, v_1, \ldots, v_n$ of $V$. Let $w_1, \ldots, w_p$ be a basis of $W$. For $a_1, \ldots, a_m, b_1, \ldots, b_n \in \mathbb{F}$ define $T(a_1u_1 + \cdots + a_mu_m + b_1v_1 + \cdots + b_nv_n) = b_1w_1 + \cdots + b_nw_n.$ Because $\text{dim } W \geq \text{dim } V - \text{dim } U$, we have $p \geq n$ and so $w_n$ on the right side of the equation above makes sense. Clearly $T \in \mathcal{L}(V, W)$ and null $T = U$. Okay, got it. Here are the solutions with added Commentary and Examples sections: $\textbf{Exercise 18.}$ Suppose $V$ and $W$ are finite-dimensional and that $U$ is a subspace of $V$. Prove that there exists $T \in \mathcal{L}(V, W)$ such that null $T = U$ if and only if $\text{dim } U \geq \text{dim } V - \text{dim } W$. $\textbf{Solution 18.}$ First suppose there exists $T \in \mathcal{L}(V, W)$ such that null $T = U$. Then $\text{dim } U = \text{dim null } T = \text{dim } V - \text{dim range } T \geq \text{dim } V - \text{dim } W,$ where the second equality comes from the fundamental theorem of linear maps. To prove the other direction, now suppose $\text{dim } U \geq \text{dim } V - \text{dim } W$. Let $u_1, \ldots, u_m$ be a basis of $U$. Extend to a basis $u_1, \ldots, u_m, v_1, \ldots, v_n$ of $V$. Let $w_1, \ldots, w_p$ be a basis of $W$. For $a_1, \ldots, a_m, b_1, \ldots, b_n \in \mathbb{F}$ define $T(a_1u_1 + \cdots + a_mu_m + b_1v_1 + \cdots + b_nv_n) = b_1w_1 + \cdots + b_nw_n.$ Because $\text{dim } W \geq \text{dim } V - \text{dim } U$, we have $p \geq n$ and so $w_n$ on the right side of the equation above makes sense. Clearly $T \in \mathcal{L}(V, W)$ and null $T = U$. $\textbf{Commentary:}$ This result gives a condition for when a subspace $U$ of $V$ can be realized as the null space of some linear transformation $T$ from $V$ to another vector space $W$. The condition is that the dimension of $U$ must be at least as large as the codimension of $W$ in $V$ (i.e. $\text{dim } V - \text{dim } W$). $\textbf{Exercise 18 Commentary:}$ This result gives conditions under which a subspace $U$ of a vector space $V$ can be realized as the null space of some linear transformation $T$ from $V$ to another vector space $W$. The key condition is that the dimension of $U$ must be at least as large as the codimension of $W$ in $V$ (i.e. $\text{dim } V - \text{dim } W$). Geometrically, this means that $U$ has enough "room" to fit inside the kernel of a map from $V$ to the smaller space $W$. $\textbf{Exercise 18 Examples:}$ 1. Let $V = \mathbb{R}^4$, $W = \mathbb{R}^2$, and $U = \text{span}{(1,2,0,0), (3,0,1,0)}$. Then $\text{dim } U = 2$ and $\text{dim } V - \text{dim } W = 4 - 2 = 2$, so there exists $T \in \mathcal{L}(V,W)$ with null $T = U$. 2. Let $V = \mathcal{P}_3(\mathbb{R})$, $W = \mathcal{P}_1(\mathbb{R})$, and $U = \text{span}{1, x, x^2}$. Then $\text{dim } U = 3$, $\text{dim } V = 4$, and $\text{dim } W = 2$. Since $\text{dim } U = 3 \geq 4 - 2 = 2$, there exists $T \in \mathcal{L}(V,W)$ with null $T = U$. 3. Let $V = \mathbb{F}_2^5$, $W = \mathbb{F}_2^4$, and $U = {0}$. Then $\text{dim } U = 0$ and $\text{dim } V - \text{dim } W = 5 - 4 = 1$. There exists $T \in \mathcal{L}(V,W)$ with null $T = U$. ---