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Chapter 6 Exercise B

1. Solution:

(a) One can easily check that each of the four vectors has norm $\sin^2 \theta + \cos^2 \theta$, which equals $1$. Moreover, we have
$$ \begin{aligned} \langle (\cos\theta, \sin\theta), (-\sin\theta, \cos\theta) \rangle &= -\cos\theta \sin\theta + \sin\theta \cos\theta = 0\\ \langle (\cos\theta, \sin\theta), (\sin\theta, -\cos\theta) \rangle &= \cos\theta \sin\theta – \sin\theta \cos\theta = 0, \end{aligned} $$ which shows that they are orthogonal.

(b) Clearly, for any $v$ and $u$ in $\mathbb{R}^2$ with $||v|| = ||u|| = 1$, we can write $v = (\cos \theta, \sin \theta)$ and $u = (\cos \alpha, \sin \alpha)$ for some angles $\theta$ and $\alpha$. If $v, u$ is an orthonormal basis, then we must have
$$ 0 = \langle v, u \rangle = \langle (\cos \theta, \sin \theta), (\cos \alpha, \sin \alpha) \rangle = \cos\theta \cos\alpha + \sin\theta \sin\alpha = \cos(\theta – \alpha). $$ One solution is to take choose $\theta$ and $\alpha$ such that $\alpha = \theta + \frac{\pi}{2}$. Then
$$ \begin{aligned} (\cos \alpha, \sin \alpha) &= (\cos(\theta + \frac{\pi}{2}), \sin(\theta + \frac{\pi}{2}))\\ &= (\cos\theta \cos\frac{\pi}{2} – \sin\theta\sin\frac{\pi}{2}, \sin\theta \cos\frac{\pi}{2} + \sin\frac{\pi}{2} \cos\theta)\\ &= (-\sin\theta, \cos\theta).\\ \end{aligned} $$ Which shows that $v, u$ is of the first form given in part (a).

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 $v\in \m{span}(e_1,\cdots,e_m)$.

3. Solution: Applying the Gram-Schmidt Procedure to the given basis, we get the following basis
$$ (1, 0, 0),\:\frac{1}{\sqrt{2}}(0, 1, 1),\:\frac{1}{\sqrt{2}}(0, -1, 1). $$ As in the proof of 6.37, we see that the matrix of $T$ with respect to this basis is upper triangular.

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

5. Solution: Applying the Gram-Schmidt Procedure, we get the following basis
$$ 1, \: 2\sqrt{3}(x – \frac{1}{2}), \: 6\sqrt{5}(x^2 – x + \frac{1}{6}). $$

6. Solution: Let $D$ denote the differential operator. Note that $D$ is already upper-triangular with respect to the standard basis of $\mathcal{P}_2{\mathbb{R}}$. Therefore, by the same reasoning used in the proof of 6.37, $\mathcal{M}(D)$ is upper-triangular with respect to the basis found in Exercise 5.

7. Solution:Defining $\varphi(p) = p(\frac{1}{2})$ and $\langle p, q \rangle = \int_{0}^{1} p(x)q(x)\ dx$ and using the formula from 6.43 together with the basis found in Exercise 5, we find that
$$ q(x) = -15x^2 + 15x – \frac{3}{2}. $$

8. Solution: Using the orthonormal basis found in Exercise 5 and the formula in 6.43, we get $q(x) = \dfrac{-24}{\pi^2}\left(x – \dfrac{1}{2}\right)$.

9. Solution: Suppose $v_1, \dots, v_m$ is a lienarly dependent list in $V$. Let $k$ be the smallest integer such that $v_k \in \operatorname{span}(v_1, \dots, v_{k-1})$. Then $v_1, \dots, v_{k-1}$ is linearly independent and we can apply the Gram-Schmidt Procedure to produce an orthonormal list $e_1, \dots, e_{k-1}$ whose span is the same. Therefore $v_k \in \operatorname{span}(e_1, \dots, e_{k-1})$ and, by 6.30,
$$ v_k = \langle v_k, e_1 \rangle e_1 + \dots + \langle v_k, e_{k-1} \rangle e_{k-1}. $$ But the right hand side is exactly what we subtract from $v_k$ when calculating $e_k$, hence the Gram-Schmidt Procedure cannot continue because we can’t divide by $0$. If, however, you discard $v_k$ (and every other vector to which happens the same thing), you end up producing an orthonormal basis whose span equals $\operatorname{span}(v_1, \dots, v_m)$.

10. Solution: Just apply the Gram-Schmidt Procedure once on $v_1, \dots, v_m$ to get the orthonormal an $e_1, \dots, e_m$. Note that, if we change $e_i$ to $-e_i$ without relabeling the vectors, $e_1, \dots, e_m$ is still orthonormal and we still have $\operatorname{span}(v_1, \dots, v_j) = \operatorname{span}(e_1, \dots, e_j)$ for all $j \in \{1, \dots, m\}$. Because we have $m$ vectors and for each vector we are free to choose a $\pm$ sign. There are $2^m$ such lists.

11. Solution: Let $w \in V$. Define $\varphi(v) = \langle v, w \rangle_1$ and $\psi(v) = \langle v, w \rangle_2$. Since $\varphi(v) = 0$ if and only if $\psi(v) = 0$, it follows that $\operatorname{null} \varphi =\operatorname{null} \psi$. By Theorem 1 in Chapter 3 notes we have
$$ \operatorname{span}(\varphi) = (\operatorname{null} \varphi)^0 = (\operatorname{null} \psi)^0 = \operatorname{span}(\psi). $$ Thus $\varphi = c\psi$ for some $c \in \mathbb{F}$. Hence, for each fixed $w$ we have $\langle v, w \rangle_1 = c\langle v, w \rangle_2$ for every $v \in V$. Chosing $v = w$ now implies that $c$ is real and positive. Fix $w_1, w_2 \in V$ and let $c_1, c_2 \in \mathbb{F}$ such that
$$ \begin{aligned} \langle v, w_1 \rangle_1 &= c_1 \langle v, w_1 \rangle_2\\ \langle v, w_2 \rangle_1 &= c_2 \langle v, w_2 \rangle_2.\\ \end{aligned} $$ Pluging $v = w_2$ in the first equation and $v = w_1$ in the second yields
$$ \begin{aligned} \langle w_2, w_1 \rangle_1 &= c_1 \langle w_2, w_1 \rangle_2\\ \langle w_1, w_2 \rangle_1 &= c_2 \langle w_1, w_2 \rangle_2.\\ \end{aligned} $$ Then
$$ c_1 \langle w_2, w_1 \rangle_2 = \langle w_2, w_1 \rangle_1 = \overline{\langle w_1, w_2 \rangle_1} = \overline{c_2 \langle w_1, w_2 \rangle_2} = \bar{c_2} \langle w_2, w_1 \rangle_2. $$ Hence $c_1 = \bar{c_2}$. Because both are real, it follows that $c_1 = c_2$. Therefore, the constant is the same for all $v, w \in V$.

13. Solution: We show it by induction on $n$. It is clearly true for $n=1$ where we can choose $w=v_1$. Since $v_1,\dots,v_m$ is a linearly independent list of vectors in $V$, by 6.31 Gram-Schmidt procedure, we get an orthonormal list $e_1,\dots,e_n$. By induction hypothesis, there exists $w'$ such that $\langle w, v_i\rangle >0$ for all $i=1,\dots,n-1$. Then we let $w=w'+k e_n$, where $k\in\mathbb F$. Clearly, we have $\langle e_n,v_i\rangle =0$ for all $i=1,\dots,n-1$. Hence $$\langle w,v_i\rangle =\langle w',v_i\rangle >0$$for $i=1,\dots,n-1$. We show that there exists $k\in \mathbb F$ such that $\langle w,v_n\rangle >0$. It suffices to show that $\langle e_n,v_n\rangle \ne 0$ (since then $\langle k e_n,v_n\rangle$ can run over any value as $k$ changes). We argue it by contradiction. Suppose $\langle e_n,v_n\rangle = 0$. Note that $\langle e_n,v_i\rangle =0$ for all $i=1,\dots,n-1$, then $\langle e_n,v\rangle=0$ for all $v\in \mathrm{span}\{v_1,\dots,v_n\}$. In particular, we have $\langle e_n,e_n\rangle =0$, which is a contradiction. Hence we are done.

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\\ \leqslant &\, \left|\sum_{i=1}^n\sum_{j=1}^n\Big\langle a_i(e_i-v_i),a_j(e_j-v_j)\Big\rangle\right|\\ \leqslant &\, \sum_{i=1}^n\sum_{j=1}^n\left|\Big\langle a_i(e_i-v_i),a_j(e_j-v_j)\Big\rangle\right| \\ \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.

17. Solution:

(a) For additivity, suppose $u_1, u_2 \in V$. Then, for $v \in V$, we have
$$ (\Phi(u_1 + u_2))(v) = \langle v, u_1 + u_2 \rangle = \langle v, u_1 \rangle + \langle v, u_2 \rangle = (\Phi u_1)(v) + (\Phi u_2)(v). $$ For homogeneity, suppose $u \in V$ and $c \in \mathbb{R}$. Then, for $v \in V$, we have
$$ (\Phi(cu))(v) = \langle v, cu \rangle = c\langle v, u \rangle = c(\Phi u)(v). $$ (b) If $\mathbb{F} = \mathbb{C}$, then the homogeneity property of linear maps is not satisfied, because we would have $(\Phi(cu))(v) = \bar{c}(\Phi u)(v)$, but $c = \bar{c}$ if and only if $c$ is a real number.

(c) This is the same as the second part in the proof of 6.42. Suppose there are $u_1$ and $u_2$ in $V$ such that $\Psi u_1 = \Psi u_2$. Then
$$ 0 = (\Psi u_1 – \Psi u_2)(v) = (\Psi(u_1 – u_2))(v) = \langle v, u_1 – u_2 \rangle $$ for all $v \in V$. Choosing $v = u_1 – u_2$ shows that $u_1 – u_2 = 0$ and thus $u_1 = u_2$.

(d) From (c), we get that $\dim\mathrm{null}~ \Phi = 0$. Thus, from 3.22, we have $$ \operatorname{dim} V = \operatorname{dim} \operatorname{null} \Phi + \operatorname{dim} \operatorname{range} \Phi = \operatorname{dim} \operatorname{range} \Phi. $$ However, $\operatorname{dim} V = \operatorname{dim} V’$. This shows that $\Phi$ also surjective. Hence $\Phi$ is invertible, that is, an isomorphism from $V$ to $V’$.


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