Category Archives: Chapter 1

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 and Closed under scalar multiplication. Additive identity: it is clear … Continue reading

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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 $a\ne 0$, then $a$ has inverse $a^{-1}$ such that $a^{-1}a=1$. … Continue reading

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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 $\dfrac{-1+\sqrt{3}i}{2}$ is a root of the quadratic equation $x^2+x+1=0$. For \[\frac{-1+\sqrt{3}i}{2}+\frac{-1-\sqrt{3}i}{2}=-1\] … Continue reading

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