Tag Archives: Exercise A

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. If $\ca M(T,(v_1,\cdots,v_n))$ is invertible, then there exists $B\in \mb … Continue reading

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Chapter 9 Exercise A

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Chapter 8 Exercise A

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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: (This solution works for $\dim V<\infty$, I am not sure … Continue reading

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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, we have $\langle u,v\rangle=\langle v,u\rangle$. Hence \begin{align*} \langle u+v,u-v\rangle&=\langle u,u\rangle-\langle … Continue reading

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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 $Tu\in\m{range} T \subset U$, hence $U$ is invariant under $T$. 2. Solution: See Linear … Continue reading

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Chapter 3 Exercise A

1. Suppose $b,c\in \R$. Define $T: \R^3 \to \R^2$ by \[T(x, y, z)= (2x-4y +3z + b,6x +cxyz).\] Show that $T$ is linear if and only if $b=c=0$. Solution: If $T$ is linear, then \[(0,0)=T(0,0,0)=(b,0)\]by 3.11, hence $b=0$. We also … Continue reading

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Chapter 2 Exercise A

1. Suppose $v_1$, $v_2$, $v_3$, $v_4$ spans $V$. Prove that the list\[v_1-v_2,v_2-v_3,v_3-v_4,v_4\] also spans $V$. 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 … Continue reading

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Chapter 1 Exercise A

1. Suppose $a$ and $b$ are real numbers, not both 0. Find real numbers $c$ and $d$ such that\[\frac{1}{a+bi}=c+di.\]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. Show that \[\frac{-1+\sqrt{3}i}{2}\] is a cube root of 1 (meaning that its cube equals 1). Soltion1:From … Continue reading

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