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 seen that \[ \langle \xi,e_i\rangle=\langle v,e_i\rangle-\langle v,e_i\rangle=0 \]for $i=1,\cdots,m$. This implies\[\langle \xi,e_1\rangle e_1 +\cdots+\langle v,e_m\rangle e_m\rangle=0.\]By 6.13, we have \begin{align*} \|v\|^2=&\|\xi\|^2+\|\langle v,e_1\rangle e_1 +\cdots+\langle v,e_m\rangle e_m\|^2\\ =&\|\xi\|^2+|\langle v,e_1\rangle|^2+\cdots+|\langle v,e_m\rangle|^2.\end{align*} It follows that $\|\xi\|^2=0$, hence $\xi=0$. Thus $v=\langle v,e_1\rangle e_1 +\cdots+\langle v,e_m\rangle e_m$, namely $v\in \m{span}(e_1,\cdots,e_m)$.

4. Solution: See Linear Algebra Done Right Solution Manual Chapter 6 Problem 9.

14. Solution: Since $e_1,\cdots,e_n$ is an orthonormal basis of $V$, we have $\dim\,V=n$. To show that $v_1,\cdots,v_n$ is a basis of $V$, it suffices to show that $v_1,\cdots,v_n$ is linearly independent. We prove it by contradiction.

Suppose $v_1,\cdots,v_n$ is linearly dependent, then there exist $a_1,\cdots,a_n\in\mathbb F$ such that $a_k\ne 0$ for some $k\in\{1,\cdots,n\}$ and \[\sum_{i=1}^na_iv_i=0.\]On one hand, by 6.25, we have $$\Big\|\sum_{i=1}^na_i(e_i-v_i)\Big\|^2=\Big\|\sum_{i=1}^na_ie_i\Big\|^2=\sum_{i=1}^n|a_i|^2.$$On the other hand, we also have\begin{align*}\Big\|\sum_{i=1}^na_i(e_i-v_i)\Big\|^2=&\,\Big\langle \sum_{i=1}^n a_i(e_i-v_i),\sum_{j=1}^n a_j(e_j-v_j)\Big\rangle\\ = &\,\sum_{i=1}^n\sum_{j=1}^n\Big\langle a_i(e_i-v_i),a_j(e_j-v_j)\Big\rangle\\ \text{by 6.15}\quad \leqslant &\,\sum_{i=1}^n\sum_{j=1}^n\|a_i(e_i-v_i)\|\|a_j(e_j-v_j)\|\\ = &\,\sum_{i=1}^n\sum_{j=1}^n|a_i||a_j|\|e_i-v_i\|\|e_j-v_j\|\\ \text{by assumption and }a_k\ne 0 \quad <&\,\sum_{i=1}^n\sum_{j=1}^n\frac{1}{n}|a_i||a_j|=\frac{1}{n}\Big(\sum_{i=1}^n|a_i|\Big)^2\\ \text{by Problem 6.A.12}\quad\leqslant &\sum_{i=1}^n|a_i|^2.\end{align*}Hence we get $$\sum_{i=1}^n|a_i|^2<\,\sum_{i=1}^n|a_i|^2,$$which is impossible, hence completing the proof.

15. Solution: Suppose there exists $g$ such that $\vp(f)=\langle f,g\rangle$ for all $f\in C_{\R}[-1,1]$. We would like to show a contradiction.

For any positive integer $n$ and integer $-n\leqslant i\leqslant n-1$, define\[f_{n,i}(x)=\begin{cases}4n^2(x-i/n),\quad &\text{if }x\in [i/n,i/n+1/(2n)]\\ 4n^2((i+1)/n-x),\quad &\text{if }x\in [i/n+1/(2n),(i+1)/n]\\ 0,\quad &\text{otherwise },\end{cases}\]then $f_{n,i}(x)\in C_{\R}[-1,1]$ and $f_{n,i}(0)=0$.

Given any $\epsilon>0$, since $g\in C_{\R}[-1,1]$, by the fact that a continuous function on a closed interval is uniformally continuous, there exists $N$ such that for any $n\geqslant N$, we have \begin{equation}\label{6B151}|g(x)-g(y)|\leqslant \epsilon\end{equation} if $|x-y|\leqslant 1/n$.

Note that \begin{equation}\label{6B152}\int_{-1}^{1}f_{n,i}(x)dx=\int_{i/n}^{(i+1)/n}f_{n,i}(x)dx=1,\end{equation} for any $y\in [i/n,(i+1)/n]$ we have \begin{align*}&\left|g\left(y\right)-\int_{-1}^1 f_{n,i}(x)g(x)dx\right|\\=& \left|\int_{i/n}^{(i+1)/n}f_{n,i}(x)\left(g\left(y\right)-g(x)\right)dx\right|\\ \leqslant& \int_{i/n}^{(i+1)/n}f_{n,i}(x)\left|g\left(y\right)-g(x)\right|dx \\ \text{by \eqref{6B151} and \eqref{6B152}}\quad \leqslant& \int_{i/n}^{(i+1)/n}f_{n,i}(x)\epsilon dx=\epsilon.\end{align*}

On the other hand, we also have$$0=f_{n,i}(0)=\vp(f_{n,i})=\langle f_{n,i},g\rangle=\int_{-1}^1 f_{n,i}(x)g(x)dx.$$ Hence we have $$|g(y)|=|g(y)-f_{n,i}(0)|\leqslant \epsilon$$ for any $y\in [i/n,(i+1)/n]$. Thus $|g(x)|\leqslant \epsilon $ by taking all $-n\leqslant i\leqslant n-1$ with $n\geqslant N$.

Since $\epsilon$ is chosen arbitrarily, we have $g(x)\equiv 0$. Hence $\vp f\equiv 0$ for all $f\in C_{\R}[-1,1]$, which is impossible. Therefore the proof is complete.

About Linearity

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