--- COMMENTARY: $\textbf{Exercise 16.}$ Suppose $V$ and $W$ are both finite-dimensional. Prove that there exists an injective linear map from $V$ to $W$ if and only if $\dim V \leq \dim W$. $\textbf{Solution 16.}$ First suppose that there exists an injective linear map $T$ from $V$ to $W$. Then by the fundamental theorem of linear maps (3.21), we have $\dim V = \dim \operatorname{null} T + \dim \operatorname{range} T = \dim{0} + \dim \operatorname{range} T = \dim \operatorname{range} T \leq \dim W.$ Conversely, suppose $\dim V \leq \dim W$. Let $v_1, \ldots, v_m$ be a basis of $V$ and let $w_1, \ldots, w_n$ be a basis of $W$. Thus $m \leq n$. Use the linear map lemma (3.4) to define a linear map $T \in \mathcal{L}(V, W)$ such that $Tv_k = w_k \text{ for } k = 1, \ldots, m.$ Suppose $v \in V$ and $Tv = 0$. There exist $c_1, \ldots, c_m \in \mathbb{F}$ such that $v = c_1v_1 + \cdots + c_mv_m.$ Thus $0 = Tv = c_1Tv_1 + \cdots + c_mTv_m = c_1w_1 + \cdots + c_mw_m.$ Because $w_1, \ldots, w_m$ is linearly independent, we have $c_1 = \cdots = c_m = 0$. Thus $v = 0$. Thus $T$ is injective, as desired. $\textit{Commentary:}$ This exercise characterizes the existence of an injective linear map between two finite-dimensional vector spaces in terms of their dimensions. The first part of the proof assumes that an injective linear map $T : V \to W$ exists and uses the Fundamental Theorem of Linear Maps to show that $\dim V \leq \dim W$. The key insight is that if $T$ is injective, then $\operatorname{null} T = {0}$, so $\dim \operatorname{null} T = 0$. Therefore, $\dim V = \dim \operatorname{range} T$, 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 $\dim V \leq \dim W$ and constructs an injective linear map $T : V \to W$. The construction chooses bases for $V$ and $W$ and defines $T$ to map the first $m$ basis vectors of $V$ to the first $m$ basis vectors of $W$, where $m = \dim V$. This ensures that $T$ is injective, because if $Tv = 0$, then the representation of $v$ in terms of the basis of $V$ must have all coefficients zero, implying $v = 0$. Geometrically, this result says that we can "embed" a vector space into a larger vector space via an injective linear map. The injectivity ensures that the map preserves the distinctness of vectors. The condition $\dim V \leq \dim W$ ensures that the larger space has enough "room" to accommodate an embedded copy of the smaller space. This result is a cornerstone of linear algebra and has many applications. For example, it's used in the study of isomorphisms between vector spaces, in the construction of dual spaces, and in the theory of tensor products. $\textit{Examples:}$ 1. There exists an injective linear map from $\mathbb{R}^3$ to $\mathbb{R}^4$, because $\dim \mathbb{R}^3 = 3 \leq 4 = \dim \mathbb{R}^4$. One such map is $T(x, y, z) = (x, y, z, 0)$. 2. There does not exist an injective linear map from $\mathbb{C}^5$ to $\mathbb{C}^4$, because $\dim \mathbb{C}^5 = 5 > 4 = \dim \mathbb{C}^4$. 3. There exists an injective linear map from $\mathbb{F}_2^2$ to $\mathbb{F}_2^3$, because $\dim \mathbb{F}_2^2 = 2 \leq 3 = \dim \mathbb{F}_2^3$. One such map is $T(x, y) = (x, y, 0)$. 4. There exists an injective linear map from $\mathcal{P}_2(\mathbb{R})$ to $\mathcal{P}_4(\mathbb{R})$, because $\dim \mathcal{P}_2(\mathbb{R}) = 3 \leq 5 = \dim \mathcal{P}_4(\mathbb{R})$. One such map is $T(a + bx + cx^2) = a + bx + cx^2 + 0x^3 + 0x^4$. These examples illustrate the principle in various vector spaces over different fields, including function spaces like polynomial spaces. In each case, the existence of an injective linear map is determined by comparing the dimensions of the domain and codomain. When an injective map exists, we can construct one by mapping basis vectors of the domain to linearly independent vectors in the codomain.