# Tag Archives: 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 Posted in Chapter 10 | Tagged ## Chapter 9 Exercise A Posted in Chapter 9 | Tagged ## Chapter 8 Exercise A Posted in Chapter 8 | Tagged ## 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*}ThereforeT^*(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 Posted in Chapter 7 | Tagged ## 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 Posted in Chapter 6 | Tagged ## Chapter 5 Exercise A 1. Solution: (a) For anyu\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 Posted in Chapter 5 | Tagged ## 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 Posted in Chapter 3 | Tagged ## 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 Posted in Chapter 2 | Tagged ## 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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