Chapter 1 Exercise B


1. Prove that $-(-v)=v$ for every $v\in V$.
Solution: By definition, we have\[(-v)+(-(-v))=0\quad\text{and}\quad v+(-v)=0.\]This implies both $v$ and $-(-v)$ are additive inverses of $-v$, by the uniqueness of additive inverse, it follows that $-(-v)=v$.

2. Suppose $a\in\mathbb F$, $v\in V$, and $av=0$. Prove that $a=0$ or $v=0$.

Solution: If $a\ne 0$, then $a$ has inverse $a^{-1}$ such that $a^{-1}a=1$. Hence \[v=1\cdot v=(a^{-1}a)v=a^{-1}(av)=a^{-1}\cdot 0=0.\]Here we use associativity in 1.19 and and 1.30.


3. Suppose $v,w\in V$. Explain why there exists a unique $x\in V$ such that $v+3x=w$.

Solution: Let $x=\dfrac{1}{3}(w-v)$, then \[v+3x=v+3\cdot \dfrac{1}{3}(w-v)=v+(w-v)=w.\]This shows existence. Now we show uniqueness. Suppose, we have another vector $x’$ such that $v+3x’=w$. Then $v+3x’=w$ implies $3x’=w-v$. Similarly, $3x=w-v$. Hence \[3(x-x’)=3x-3x’=(w-v)-(w-v)=0.\]By Problem 2, it follows that $x-x’=0$. This shows uniqueness.

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?

Solution: Additive identity: there exists an element $0\in V$ such that $v+0=v$ for all $v\in V$ ; This means $V$ cannot be empty.

5. Show that in the definition of a vector space (1.19), the additive inverse condition can be replaced with the condition that\begin{equation}\label{1B51}0v=0\quad \text{for all}~v\in V.\end{equation} 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 phrase “a condition can be replaced” in a definition means that the collection of objects satisfying the definition is unchanged if the original condition is replaced with the new condition.)

Solution: If we assume the additive inverse condition, we already show $(\ref{1B51})$ in 1.29. Now we assume $(\ref{1B51})$ and then show additive inverse condition. Since we have $(\ref{1B51})$, we have \[ v+((-1)v)=1v+((-1)v)=(1+(-1))v=0v=0, \]this means the existence of additive inverse, i.e. the additive inverse condition.

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 \[ t\infty=\left\{ \begin{array}{ll} -\infty, & \hbox{if $t<0$,} \\ 0, & \hbox{if $t=0$,} \\ \infty, & \hbox{if $t>0$,} \end{array} \right. \quad t(-\infty)=\left\{ \begin{array}{ll} \infty, & \hbox{if $t<0$,} \\ 0, & \hbox{if $t=0$,} \\ -\infty, & \hbox{if $t>0$,} \end{array} \right. \] \[t+\infty=\infty+t=\infty,\quad t+(-\infty)=(-\infty)+t=-\infty,\] \[\infty+\infty=\infty,\quad (-\infty)+(-\infty)=-\infty,\quad \infty+(-\infty)=0.\]Is $\mathbb R\cup \{\infty\}\cup\{\infty\}$ a vector space over $\mathbb R$? Explain.

Solution: This is not a vector space over $\mathbb R$. Consider the distributive properties in 1.19. If this is a vector space over $\mathbb R$, we will have \[\infty=(2+(-1))\infty=2\infty+(-1)\infty=\infty+(-\infty)=0.\]Hence for any $t\in\mathbb R$, one has\[t=0+t=\infty+t=\infty=0.\]We get a contradiction since zero vector is unique.


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