#math/linear-algebra #flashcards/math A vector space, $V$, is a set over $F$, along with an addition function on $V$, $V \times V \to V$, and scalar multiplication function on $V$, $F \times V \to V$ that satisfies the following properties: commutativity, associativity, additive identity, additive inverse, multiplicative identity and distributive properties. A vector space is **not** a field since it does not have **internal** multiplication or addition properties, i.e. a vector space (times or plus) a vector space = a vector space. **A Vector Space can be a set of functions** If $S$ is a set, then $F^S$ is a vector space that is the set of functions from $S \to F$. - Let $x \in{} S; f,g \in{} F^s, \lambda \in{} F$. - Sum of $f+g \in{} F^S$ is the function $(f+g)(x) = f(x) + g(x)$ - Product of $\lambda f \in{} F^s$ is the function $(\lambda f)(x) = \lambda f(x)$ The remaining vector space axioms can be demonstrated subsequently. - The zero function, $0 : S \to F$ is $0(x) = 0$ - The inverse function, $-f : S \to F$ is $(-f)(x) = -f(x)$ **F^n and F^inf are cases of F^S** Since $F^n$ is a list of numbers in $F$ of length $n$, it can be reimagined as a discrete function ${1,\dots,n} \to F$. The same logic extends to $F^\infty$. - $(42, 3, 6) \in{} \mathbb{R}^3$, represented in $\mathbb{R}^{\{1,2,3\}}$ would be the element (a function) $f : \{1,2,3\} \to R$ where $f(1) = 42, f(2) = 3, f(3)=6$. Now extend this for all $(x_{1}, x_{2}, x_{3}) \in{} \mathbb{R}^3$. - $\mathbb{R}^{[0,1]}$ is the set of all functions $f : [0,1] \to \mathbb{R}$. **Additional Properties** From the vector space axioms, you can prove the following attributes: - Unique additive identity - Unique additive inverse - $0v = 0$ - $a0 =0$