#math/linear-algebra **Problem 1** Prove that $-(-v) = v$ for every $v \in{} V$. $ \begin{align} -(-v) + (-v) & = -1(-v) + 1(-v) \\ & = (-1 + 1)(-v) \\ & = 0 (-v) \\ & = 0 \end{align} $ Thus, $-(-v)$ is the inverse of $-v$, and since we know the additive inverse is unique, $-(-v) = v$. **** **Problem 2** Suppose $a \in{} F, v \in{} V \text{ and } av=0.$ Prove that $a=0 \text{ or } v=0$. Assumes towards a contradiction that neither $a = 0 \text{ nor } v = 0; \text{ i.e. } a \neq 0 \text{ and } v \neq 0.$ $ v = 1v = (a^{-1}a)v=a^{-1}(av) = 0 $ Thus is must be that either $a=0 \text{ or } v = 0$. **** **Problem 3** Suppose $v, w \in{} V$. Explain why there exists a unique $x \in{} V$ such that $v + 3x = w$. Let $x = \frac{1}{3}(-v + w)$. $ \begin{align} v + 3x &= v + 3\frac{1}{3}(-v + w) \\ & = v + -v + w \\ & = 0 + w \\ & = w \end{align} $ To show uniqueness, let $x,y \in{} V$ such that $v + 3x = w \text{ and } v + 3y = w$. $ \begin{align} v + 3x & = w \\ & = v + 3y \\ -v + v + 3x & = v + 3y + -v \\ 3x & = 3y \\ x &= y \end{align} $ Thus there exists a unique $x \in{} V$ such that $v + 3x = w$. **** **Problem 4** The empty set is not a vector space. The empty set fails to satisfy only one of the requirements listed in 1.19. Which one? The 1.19 requirements are: Commutativity, Associativity, Additive Identity, Additive Inverse, Multiplicative identity, distributive properties. It fails the additive identity, since it requirements that there exist an element $0 \in{} V$. However, the empty set, $\{\}$ has no elements. **** **Problem 5** Show that in the definition of a vector space (1.19), the additive inverse condition can be replaced with the condition that $ 0v = 0 \text{ for all }v \in{} V $ Here the 0 on the left side is the number 0, and the 0 on the right side is the additive identity of $V$. The 1.19 Additive Inverse condition is: "For every $v \in{} V$, there exists $w \in{} V$ such that $v + w = 0$." Assume that $0v = 0 \text{ for all } v \in{} V$. $ \begin{align} 0 & = 0v & \text{Assumption} \\ & = (1 + -1)v \\ & =1v + -1v \\ & = v + -v \end{align} $ Thus if $0v = 0 \text{ for all } v \in{} V$ $\implies$ for every $v \in{} V$, there exists $w \in{} V$ such that $v + w = 0$. Assume for every $v \in{} V$, there exists $w \in{} V$ such that $v + w = 0$. $ \begin{align} 0v & = 0v \\ & =(0 + 0)v \\ & = 0v + 0v \\ -(0v) + 0v & =0v + 0v + -(0v) & \text{Assumption}\\ 0 & = 0v \end{align} $ Thus if for every $v \in{} V$, there exists $w \in{} V$ such that $v + w = 0$ $\implies$ $0v = 0 \text{ for all }v \in{} V$. **Problem 6** Let $\infty$ and $-\infty$ denote two distinct objects, neither of which is in $\mathbb{R}$. Define an addition and scalar multiplication on $\mathbb{R} \cup \{\infty\} \cup \{-\infty\}$ as you could guess from the notation. Specifically, the sum and product of two real numbers is as usual and for $t \in{} \mathbb{R}$ define $ t\infty = \begin{cases} -\infty & \text{ if } t < 0.\\ 0 & \text{ if } t = 0. \\ \infty & \text{ if } t > 0. \end{cases} \quad \quad t(-\infty) = \begin{cases} \infty & \text{ if } t < 0.\\ 0 & \text{ if } t = 0. \\ -\infty & \text{ if } t > 0. \end{cases} $ $ \begin{align} t + \infty &= \infty + t = \infty & \text{ (1)}\\ t + (-\infty) &= (-\infty) + t = -\infty & \text{ (2)}\\ \infty + \infty &= \infty & \text{ (3)}\\ (-\infty) + (-\infty) &= -\infty & \text{ (4)}\\ \infty + (-\infty) &= 0& \text{ (5)} \end{align} $ Is $\mathbb{R} \cup \{\infty\} \cup \{-\infty\}$ a vector space over $\mathbb{R}$? $ \begin{align} (\infty + \infty) + -(\infty) & = \infty + -(\infty) & \text{Rule 3}\\ & = \infty + -(\infty) \\ & = 0 \end{align} $ $ \begin{align} \infty + (\infty + -(\infty)) & = \infty + (0) & \text{Rule 5}\\ & = \infty \end{align} $ Thus $(\infty + \infty) + -(\infty) \neq \infty + (\infty + -(\infty))$. So, this cannot be a vector space.