If you find any mistakes, please make a comment! Thank you.

## Chapter 1 Exercise C

1. Solution: (a) $\{(x_1,x_2,x_3)\in\mathbb F^3:x_1+2x_2+3x_3=0\}$ is a subspace of $\mathbb F^3$. By 1.34, to show a subset is a subspace, we just need to check Additive identity, Closed under addition…

## Chapter 1 Exercise B

1. 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. Solution: If…

## Chapter 1 Exercise A

1.Solution: Because $(a+bi)(a-bi)=a^2+b^2$, one has$\frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}.$Hence$c=\frac{a}{a^2+b^2},d=-\frac{b}{a^2+b^2}.$ 2. Solution1:From direct computation, we have$\left(\frac{-1+\sqrt{3}i}{2}\right)^2=\frac{-1-\sqrt{3}i}{2},$hence $\left(\frac{-1+\sqrt{3}i}{2}\right)^3=\frac{-1-\sqrt{3}i}{2}\cdot\frac{-1+\sqrt{3}i}{2}=1.$This means $\dfrac{-1+\sqrt{3}i}{2}$ is a cube root of 1. Solution2: Note that $(a+bi)+(a-bi)=2a$ and $(a+bi)(a-bi)=a^2+b^2,$ it follows that $\dfrac{-1+\sqrt{3}i}{2}$…