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Chapter 10 Exercise A

1. Solution: If $T$ is invertible, then there exists $S\in\ca L(V)$ such that $TS=ST=I$. Then it follows from 10.4 that\[\ca M(S,(v_1,\cdots,v_n))\ca M(T,(v_1,\cdots,v_n))=\ca M(ST,(v_1,\cdots,v_n))=I\]\[\ca M(T,(v_1,\cdots,v_n))\ca M(S,(v_1,\cdots,v_n))=\ca M(TS,(v_1,\cdots,v_n))=I.\]Hence $\ca M(T,(v_1,\cdots,v_n))$ is invertible.…

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Chapter 7 Exercise A

1. Solution: By definition, we have \[\begin{align*}\langle (z_1,\cdots,z_n),T^*(w_1,\cdots,w_n)\rangle=&\langle T(z_1,\cdots,z_n),(w_1,\cdots,w_n) \rangle \\=& z_1w_2+\cdots+z_{n-1}w_{n}=\langle (z_1,\cdots,z_n),(w_2,\cdots,w_n,0)\rangle.\end{align*}\]Therefore $T^*(w_1,\cdots,w_n)=(w_2,\cdots,w_n,0)$ or $T^*(z_1,\cdots,z_n)=(z_2,\cdots,z_n,0)$. See also Linear Algebra Done Right Solution Manual Chapter 6 Problem 27. 2. Solution:…

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Chapter 4 Exercise

1. Empty 2. Solution: False. Consider $1=(z^m+1)+(-z^m)\notin \{0\}\cup\{p\in\ca P(\mb F):\deg p=m\}$. Note that \[(z^m+1)\in \{0\}\cup\{p\in\ca P(\mb F):\deg p=m\}\]and\[-z^m\in \{0\}\cup\{p\in\ca P(\mb F):\deg p=m\},\]it follows that $\{0\}\cup\{p\in\ca P(\mb F):\deg p=m\}$ is not…

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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 $\dfrac{-1+\sqrt{3}i}{2}$…

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