Prove that ℒ(𝑉, 𝑊) is a vector space, as was asserted in 3.6. solution Suppose 𝑆, 𝑇 ∈ ℒ(𝑉, 𝑊). Then (𝑆 + 𝑇)(𝑣) = 𝑆𝑣 + 𝑇𝑣 = 𝑇𝑣 + 𝑆𝑣 = (𝑇 + 𝑆)𝑣 for every 𝑣 ∈ 𝑉. Thus 𝑆 + 𝑇 = 𝑇 + 𝑆 . In other words, addition is commutative on ℒ(𝑉, 𝑊) . Suppose 𝑅, 𝑆, 𝑇 ∈ ℒ(𝑉, 𝑊). Then ((𝑅 + 𝑆) + 𝑇)(𝑣) = (𝑅 + 𝑆)𝑣 + 𝑇𝑣 = (𝑅𝑣 + 𝑆𝑣) + 𝑇𝑣 = 𝑅𝑣 + (𝑆𝑣 + 𝑇𝑣) = 𝑅𝑣 + (𝑆 + 𝑇)𝑣 = (𝑅 + (𝑆 + 𝑇))(𝑣) for every 𝑣 ∈ 𝑉. Thus (𝑅 + 𝑆) + 𝑇 = 𝑅 + (𝑆 + 𝑇) . In other words, addition is associative on ℒ(𝑉, 𝑊) . Suppose 𝑎, 𝑏 ∈ 𝐅 and 𝑇 ∈ ℒ (𝑉, 𝑊) . Then ((𝑎𝑏)𝑇)(𝑣) = (𝑎𝑏)(𝑇𝑣) = 𝑎(𝑏𝑇𝑣) = (𝑎(𝑏𝑇))(𝑣) for every 𝑣 ∈ 𝑉. Thus (𝑎𝑏)𝑇 = 𝑎(𝑏𝑇) . The linear map 0 ∈ ℒ(𝑉, 𝑊) defined by 0𝑣 = 0 (where the 0 on the right is the additive identity for 𝑊 ) is clearly an additive identity for ℒ(𝑉, 𝑊) . For 𝑇 ∈ ℒ(𝑉, 𝑊) , define −𝑇 ∈ ℒ(𝑉, 𝑊) by (−𝑇)(𝑣) = −(𝑇𝑣) for all 𝑣 ∈ 𝑉. It is easy to verify that −𝑇 is an additive inverse of 𝑇. Suppose 𝑇 ∈ ℒ (𝑉, 𝑊). Then (1𝑇)(𝑣) = 1(𝑇𝑣) = 𝑇𝑣 for every 𝑣 ∈ 𝑉. Thus 1𝑇 = 𝑇. Suppose 𝑎 ∈ 𝐅 and 𝑆, 𝑇 ∈ ℒ(𝑉, 𝑊). Then (𝑎(𝑆 + 𝑇))(𝑣) = 𝑎((𝑆 + 𝑇)(𝑣)) >[!Q?] What does this mean? >how have the parentheses changed anything? = 𝑎(𝑆𝑣 + 𝑇𝑣) = 𝑎(𝑆𝑣) + 𝑎(𝑇𝑣) = (𝑎𝑆)(𝑣) + (𝑎𝑇)(𝑣) = (𝑎𝑆 + 𝑎𝑇)(𝑣) for every 𝑣 ∈ 𝑉. Thus 𝑎(𝑆 + 𝑇) = 𝑎𝑆 + 𝑎𝑇. Suppose 𝑎, 𝑏 ∈ 𝐅 and 𝑇 ∈ ℒ (𝑉, 𝑊) . Then ((𝑎 + 𝑏)𝑇)(𝑣) = (𝑎 + 𝑏)(𝑇𝑣) = 𝑎(𝑇𝑣) + 𝑏(𝑇𝑣) = (𝑎𝑇)(𝑣) + (𝑏𝑇)(𝑣) = (𝑎𝑇 + 𝑏𝑇)(𝑣) for every 𝑣 ∈ 𝑉. Thus (𝑎 + 𝑏)𝑇 = 𝑎𝑇 + 𝑏𝑇. We have now verified that ℒ(𝑉, 𝑊) is a vector space with the operations of addition and scalar multiplication that were defined on ℒ(𝑉, 𝑊) . Edward Frenkel --- Commentary: $\textbf{Exercise 5.}$ Prove that $\mathcal{L}(V, W)$ is a vector space, as was asserted in 3.6. $\textbf{Solution 5.}$ Suppose $S, T \in \mathcal{L}(V, W)$. Then $(S + T)(v) = Sv + Tv = Tv + Sv = (T + S)v$ for every $v \in V$. Thus $S + T = T + S$. In other words, addition is commutative on $\mathcal{L}(V, W)$. Suppose $R, S, T \in \mathcal{L}(V, W)$. Then $\begin{align*} ((R + S) + T)(v) &= (R + S)v + Tv \\ &= (Rv + Sv) + Tv \\ &= Rv + (Sv + Tv) \\ &= Rv + (S + T)v \\ &= (R + (S + T))(v) \end{align*}$ for every $v \in V$. Thus $(R + S) + T = R + (S + T)$. In other words, addition is associative on $\mathcal{L}(V, W)$. Suppose $a, b \in \mathbb{F}$ and $T \in \mathcal{L}(V, W)$. Then $((ab)T)(v) = (ab)(Tv) = a(bTv) = (a(bT))(v)$ for every $v \in V$. Thus $(ab)T = a(bT)$. The linear map $0 \in \mathcal{L}(V, W)$ defined by $0v = 0$ (where the 0 on the right is the additive identity for $W$) is clearly an additive identity for $\mathcal{L}(V, W)$. For $T \in \mathcal{L}(V, W)$, define $-T \in \mathcal{L}(V, W)$ by $(-T)(v) = -(Tv)$ for all $v \in V$. It is easy to verify that $-T$ is an additive inverse of $T$. Suppose $T \in \mathcal{L}(V, W)$. Then $(1T)(v) = 1(Tv) = Tv$ for every $v \in V$. Thus $1T = T$. Suppose $a \in \mathbb{F}$ and $S, T \in \mathcal{L}(V, W)$. Then $\begin{align*} (a(S + T))(v) &= a((S + T)(v)) \\ &= a(Sv + Tv) \\ &= a(Sv) + a(Tv) \\ &= (aS)(v) + (aT)(v) \\ &= (aS + aT)(v) \end{align*}$ for every $v \in V$. Thus $a(S + T) = aS + aT$. Suppose $a, b \in \mathbb{F}$ and $T \in \mathcal{L}(V, W)$. Then $\begin{align*} ((a + b)T)(v) &= (a + b)(Tv) \\&= a(Tv) + b(Tv) \\ &= (aT)(v) + (bT)(v) \\ &= (aT + bT)(v) \end{align*}$ for every $v \in V$. Thus $(a + b)T = aT + bT$. We have now verified that $\mathcal{L}(V, W)$ is a vector space with the operations of addition and scalar multiplication that were defined on $\mathcal{L}(V, W)$. $\textit{Commentary:}$ This exercise proves that the set of linear maps from a vector space $V$ to a vector space $W$, denoted $\mathcal{L}(V, W)$, forms a vector space under the operations of pointwise addition and scalar multiplication. The proof verifies each of the vector space axioms for $\mathcal{L}(V, W)$: 1. Commutativity of addition: $(S + T)(v) = (T + S)(v)$ for all $v \in V$. 2. Associativity of addition: $((R + S) + T)(v) = (R + (S + T))(v)$ for all $v \in V$. 3. Compatibility of scalar multiplication with field multiplication: $((ab)T)(v) = (a(bT))(v)$ for all $v \in V$. 4. Existence of additive identity: The zero map $0$ defined by $0v = 0$ for all $v \in V$ is the additive identity in $\mathcal{L}(V, W)$. 5. Existence of additive inverses: For each $T \in \mathcal{L}(V, W)$, the map $-T$ defined by $(-T)(v) = -(Tv)$ for all $v \in V$ is the additive inverse of $T$. 6. Scalar multiplication identity: $(1T)(v) = Tv$ for all $v \in V$. 7. Distributivity of scalar multiplication over vector addition: $(a(S + T))(v) = (aS + aT)(v)$ for all $v \in V$. 8. Distributivity of scalar multiplication over field addition: $((a + b)T)(v) = (aT + bT)(v)$ for all $v \in V$. Each of these properties is verified by showing that the corresponding equality holds when the maps are applied to an arbitrary vector $v \in V$. This is sufficient because linear maps are completely determined by their action on vectors. This result is fundamental in linear algebra, as it allows us to treat collections of linear maps as vector spaces in their own right. This opens up the possibility of applying linear algebraic techniques to study linear maps, which is a key idea in many areas of mathematics, including functional analysis and representation theory. $\textit{Examples:}$ 1. $\mathcal{L}(\mathbb{R}^n, \mathbb{R}^m)$ is a vector space over $\mathbb{R}$. Its elements are the $m \times n$ real matrices, and its dimension is $mn$. 2. $\mathcal{L}(\mathcal{P}_n(\mathbb{F}), \mathbb{F})$ is a vector space over $\mathbb{F}$. Its elements are the linear functionals on $\mathcal{P}_n(\mathbb{F})$, and its dimension is $n+1$. 3. $\mathcal{L}(V, V)$, often denoted $\operatorname{End}(V)$, is a vector space over $\mathbb{F}$ for any vector space $V$. Its elements are the linear operators on $V$, and its dimension is $(\dim V)^2$. 4. $\mathcal{L}(V, \mathbb{F})$, often denoted $V^*$ and called the dual space of $V$, is a vector space over $\mathbb{F}$ for any vector space $V$. Its elements are the linear functionals on $V$, and its dimension is equal to $\dim V$. These examples illustrate some of the most common and important spaces of linear maps that arise in linear algebra. They also demonstrate how the dimension of $\mathcal{L}(V, W)$ is related to the dimensions of $V$ and $W$. The proofs that these are indeed vector spaces follow the same pattern as the general proof in the solution, but with the specific definitions of addition and scalar multiplication for matrices, linear functionals, and linear operators.