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.

About Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

This entry was posted in Chapter 1 and tagged .