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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.…

## 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:…

## Chapter 6 Exercise A

2. Solution: It does not satisfy definiteness. For the function takes $(0,1,0)$, $(0,1,0)$ to $0$, but $(0,1,0)\ne 0$. 4. Solution: (a) Note that $V$ is a real inner product space,…

## Chapter 5 Exercise A

1. Solution: (a) For any $u\in U$, then $Tu=0\in U$ since $U\subset \m{null} T$, hence $U$ is invariant under $T$. (b) For any $u\in U$, then $Tu\in\m{range} T \subset U$,…

## 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…

## Chapter 3 Exercise A

1. Solution: If $T$ is linear, then $(0,0)=T(0,0,0)=(b,0)$by 3.11, hence $b=0$. We also have $T(1,1,1)=T(1,1,0)+T(0,0,1),$it is equivalent to $(1+b,6+c)=(b-2,6)+(3+b,0)=(1+2b,6).$Thus $6+c=6$ implies $c=0$. Conversely, if $b=c=0$, $T$ is obviously linear. See…

## Chapter 2 Exercise A

1. Solution: We just need to show that $v_1$, $v_2$, $v_3$, $v_4$ can be expressed as linear combination of $v_1-v_2$, $v_2-v_3$, $v_3-v_4$, $v_4$. Note that $v_1=(v_1-v_2)+(v_2-v_3)+(v_3-v_4)+v_4,$$v_2=(v_2-v_3)+(v_3-v_4)+v_4,$$v_3=(v_3-v_4)+v_4,\quad v_4=v_4.$ 2. Solution: (a)…

## 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 $\dfrac{-1+\sqrt{3}i}{2}$…