--- COMMENTARY: $\textbf{Exercise 17.}$ Suppose $V$ and $W$ are both finite-dimensional. Prove that there exists a surjective linear map from $V$ onto $W$ if and only if $\dim V \geq \dim W$. $\textbf{Solution 17.}$ First suppose there exists a surjective linear map $T$ from $V$ onto $W$. Then $\dim W = \dim \operatorname{range} T = \dim V - \dim \operatorname{null} T \leq \dim V,$ where the second equality comes from the fundamental theorem of linear maps. To prove the other direction, now suppose $\dim W \leq \dim V$. Let $w_1, \ldots, w_m$ be a basis of $W$ and let $v_1, \ldots, v_n$ be a basis of $V$. For scalars $a_1, \ldots, a_n \in \mathbb{F}$ define $T(a_1v_1 + \cdots + a_nv_n)$ by $T(a_1v_1 + \cdots + a_nv_n) = a_1w_1 + \cdots + a_mw_m.$ Because $\dim W \leq \dim V$, we have $m \leq n$ and so $a_m$ on the right side of the equation above makes sense. Clearly $T$ is a surjective linear map from $V$ onto $W$. $\textit{Commentary:}$ This exercise characterizes the existence of a surjective linear map between two finite-dimensional vector spaces in terms of their dimensions. The first part of the proof assumes that a surjective linear map $T : V \to W$ exists and uses the Fundamental Theorem of Linear Maps to show that $\dim V \geq \dim W$. The key insight is that if $T$ is surjective, then $\operatorname{range} T = W$, so $\dim \operatorname{range} T = \dim W$. The Fundamental Theorem then gives $\dim W = \dim V - \dim \operatorname{null} T$, implying $\dim W \leq \dim V$. The second part of the proof assumes that $\dim V \geq \dim W$ and constructs a surjective 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 basis vectors of $W$, where $m = \dim W$. This ensures that $T$ is surjective, because every vector in $W$ is a linear combination of the basis vectors of $W$, which are all in the range of $T$. Geometrically, this result says that we can "collapse" a vector space onto a smaller vector space via a surjective linear map. The surjectivity ensures that the map hits every vector in the codomain. The condition $\dim V \geq \dim W$ ensures that the domain has enough "material" to cover the entire codomain. This result is dual to the previous one about injective maps, and together they form a fundamental pair of theorems in linear algebra. They are used, for example, in the study of quotient spaces and in the theory of duality. $\textit{Examples:}$ 1. There exists a surjective linear map from $\mathbb{R}^4$ to $\mathbb{R}^3$, because $\dim \mathbb{R}^4 = 4 \geq 3 = \dim \mathbb{R}^3$. One such map is $T(x, y, z, w) = (x, y, z)$. 2. There does not exist a surjective linear map from $\mathbb{C}^4$ to $\mathbb{C}^5$, because $\dim \mathbb{C}^4 = 4 < 5 = \dim \mathbb{C}^5$. 3. There exists a surjective linear map from $\mathbb{F}_3^3$ to $\mathbb{F}_3^2$, because $\dim \mathbb{F}_3^3 = 3 \geq 2 = \dim \mathbb{F}_3^2$. One such map is $T(x, y, z) = (x, y)$. 4. There exists a surjective linear map from $\mathcal{P}_4(\mathbb{R})$ to $\mathcal{P}_2(\mathbb{R})$, because $\dim \mathcal{P}_4(\mathbb{R}) = 5 \geq 3 = \dim \mathcal{P}_2(\mathbb{R})$. One such map is $T(a + bx + cx^2 + dx^3 + ex^4) = a + bx + cx^2$. These examples illustrate the principle in various vector spaces over different fields, including function spaces like polynomial spaces. In each case, the existence of a surjective linear map is determined by comparing the dimensions of the domain and codomain. When a surjective map exists, we can construct one by mapping basis vectors of the domain onto the basis vectors of the codomain.