# Category Archives: Chapter 6

## Chapter 6 Exercise C

3. Solution: By 6.31, we have $\m{span}(e_1,\cdots e_m)=\m{span}(u_1,\cdots,u_m)=U.$Note that $e_1,\cdots e_m$ is an orthonormal list, it follows that $e_1,\cdots e_m$, is an orthonormal basis of $U$. By 6.47, we have $V=U\oplus U^{\perp}$. As $e_1,\cdots e_m,f_1,\cdots,f_n$ is an orthonormal … Continue reading

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## Chapter 6 Exercise B

2. Solution: If $v\in \m{span}(e_1,\cdots,e_m)$, then $e_1$, $\cdots$, $e_m$ is an orthonormal basis of $\m{span}(e_1,\cdots,e_m)$ by 6.26. By 6.30, it follows that$\|v\|^2=|\langle v,e_1\rangle|^2+\cdots+|\langle v,e_m\rangle|^2.$ If $\|v\|^2=|\langle v,e_1\rangle|^2+\cdots+|\langle v,e_m\rangle|^2$, we denote $\xi=v-(\langle v,e_1\rangle e_1 +\cdots+\langle v,e_m\rangle e_m).$It is easily … 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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