Instructor's Solutions Manual, Section 1B Exercise 7 **7** Suppose S is a nonempty set. Let $V^S$ denote the set of functions from S to V. Define a natural addition and scalar multiplication on $V^S$ , and show that $V^S$ is a vector space with these definitions. SOLUTION For f,g $\in$ $V^S$ and $\lambda$ $\in$ F, let f + g and $\lambda$ f be the functions from S to V defined by (f + g) (x) = f(x) + g(x) and ($\lambda$f)(x) = $\lambda$f(x) for x $\in$ S. It is straightforward to verify that with these definitions of addition and scalar multiplication, $V^S$ is a vector space, where the additive identity is the function from S to V that is identically 0 and the additive inverse of f $\in$ $V^S$ is the function from S to V that takes x $\in$ S to - f(x). Edward Frenkel ![[EF-1B-7 2024-03-03 11.02.32.excalidraw]]  --- Commentary: $\textbf{Exercise 7.}$ Suppose $X$ is a nonempty set. Let $V_X$ denote the set of functions from $X$ to $V$. Define a natural addition and scalar multiplication on $V_X$, and show that $V_X$ is a vector space with these definitions. $\textbf{Solution 7.}$ For $f, g\in V_X$ and $\alpha \in \mathbb{F}$, let $f + g$ and $\alpha f$ be the functions from $X$ to $V$ defined by $(f + g)(x) = f(x) + g(x) \text{ and } (\alpha f)(x) = \alpha f(x)$ for $x\in X$. It is straightforward to verify that with these definitions of addition and scalar multiplication, $V_X$ is a vector space, where the additive identity is the function from $X$ to $V$ that is identically $0$ and the additive inverse of $f\in V_X$ is the function from $X$ to $V$ that takes $x\in X$ to $-f(x)$. --- $\textbf{Solution 7.}$ For $f, g\in V_X$ and $\alpha \in \mathbb{F}$, define $(f + g)(x) = f(x) + g(x)$ and $(\alpha f)(x) = \alpha(f(x))$ for all $x\in X$. It can be verified that with these operations, $V_X$ is a vector space: 1. Addition is commutative: $(f + g)(x) = f(x) + g(x) = g(x) + f(x) = (g + f)(x)$. 2. Addition is associative: $((f + g) + h)(x) = (f + g)(x) + h(x) = (f(x) + g(x)) + h(x) = f(x) + (g(x) + h(x)) = f(x) + (g + h)(x) = (f + (g + h))(x)$. 3. The zero vector is the function $0(x) = 0_V$ for all $x\in X$, where $0_V$ is the zero vector in $V$: $(f + 0)(x) = f(x) + 0(x) = f(x) + 0_V = f(x)$. 4. The additive inverse of $f$ is $(-f)(x) = -(f(x))$: $(f + (-f))(x) = f(x) + (-f)(x) = f(x) + (-(f(x))) = 0_V$. 5. Scalar multiplication is compatible with field multiplication: $((\alpha\beta)f)(x) = (\alpha\beta)(f(x)) = \alpha(\beta(f(x))) = \alpha((\beta f)(x)) = (\alpha(\beta f))(x)$. 6. $1f = f$: $(1f)(x) = 1(f(x)) = f(x)$. 7. Scalar multiplication is distributive over vector addition: $((\alpha + \beta)f)(x) = (\alpha + \beta)(f(x)) = \alpha(f(x)) + \beta(f(x)) = (\alpha f)(x) + (\beta f)(x) = (\alpha f + \beta f)(x)$. 8. Scalar multiplication is distributive over field addition: $(\alpha(f + g))(x) = \alpha((f + g)(x)) = \alpha(f(x) + g(x)) = \alpha(f(x)) + \alpha(g(x)) = (\alpha f)(x) + (\alpha g)(x) = (\alpha f + \alpha g)(x)$. Therefore, $V_X$ is a vector space. $\textit{Commentary:}$ This exercise introduces the concept of a function space, which is a vector space whose elements are functions. The addition and scalar multiplication operations are defined pointwise, meaning they are applied to the functions' outputs for each input. Function spaces are fundamental in functional analysis and have numerous applications. They provide a way to study functions using the tools and techniques of linear algebra. The exercise also highlights how new vector spaces can be constructed from existing ones. Here, given a vector space $V$ and a set $X$, we construct a new vector space $V_X$ consisting of functions from $X$ to $V$. This construction is known as the product space or the space of vector-valued functions. $\textit{Example:}$ Let $X = {1, 2, 3}$ and $V = \mathbb{R}^2$. Then $V_X$ is the set of functions from ${1, 2, 3}$ to $\mathbb{R}^2$. An element of $V_X$ can be represented as a tuple of three 2D vectors, e.g., $((1, 2), (-1, 0), (0, 1))$. If $f = ((1, 2), (-1, 0), (0, 1))$ and $g = ((0, 1), (1, -1), (2, 1))$ are elements of $V_X$, then: - $(f + g)(1) = (1, 2) + (0, 1) = (1, 3)$ - $(f + g)(2) = (-1, 0) + (1, -1) = (0, -1)$ - $(f + g)(3) = (0, 1) + (2, 1) = (2, 2)$ So $f + g = ((1, 3), (0, -1), (2, 2))$. If $\alpha = -1$, then: - $(\alpha f)(1) = -1(1, 2) = (-1, -2)$ - $(\alpha f)(2) = -1(-1, 0) = (1, 0)$ - $(\alpha f)(3) = -1(0, 1) = (0, -1)$ So $\alpha f = ((-1, -2), (1, 0), (0, -1))$. Another example is the vector space of polynomials with real coefficients, $\mathbb{R}[x]$. This can be seen as the function space from $\mathbb{R}$ to $\mathbb{R}$, where each polynomial $p(x)$ is a function that maps a real number $x$ to another real number $p(x)$. ---