1. Solution: Suppose $w \in \{v_1, \dots, v_m\}^\perp$. Let $v = \in \operatorname{span}(v_1, \dots, v_m)$. We have that

$$ v = a_1 v_1 + \dots a_m v_m $$ for some $a_1, \dots, a_m \in \mathbb{F}$. Moreover

$$ \langle v, w \rangle = \langle a_1 v_1 + \dots a_m v_m, w \rangle = a_1 \langle v_1, w \rangle + \dots + a_m \langle v_m, w \rangle = 0. $$ Thus $w \in (\operatorname{span}(v_1, \dots, v_m))^\perp$ and so $\{v_1, \dots, v_m\}^\perp \subset (\operatorname{span}(v_1, \dots, v_m))^\perp$.

Now suppose $w \in (\operatorname{span}(v_1, \dots, v_m))^\perp$. Since each $v_j$ is in $\operatorname{span}(v_1, \dots, v_m)$, it follows that $w$ is orthogonal to each $v_j$. Therefore $w \in \{v_1, \dots, v_m\}^\perp$ and thus $(\operatorname{span}(v_1, \dots, v_m))^\perp \subset \{v_1, \dots, v_m\}^\perp$.

2. Solution: Suppose $U^\perp = \{0\}$. Since $V = U \oplus U^\perp$ (by 6.47), it follows that $U = V$.

Conversely, suppose $U = V$. Then $U^\perp = V^\perp = \{0\}$.

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 list, it follows that \[\m{span}(f_1,\cdots,f_n)\subset U^{\perp}\]by definition. Oh the other hand, $V=U\oplus U^{\perp}$ implies \[\dim V=\dim U+\dim U^{\perp}\Longrightarrow m+n=m+\dim U^{\perp}.\]This means $\dim U^{\perp}=n$. Note that \[\dim \m{span}(f_1,\cdots,f_n)=n\]and\[\m{span}(f_1,\cdots,f_n)\subset U^{\perp},\]we conclude\[\m{span}(f_1,\cdots,f_n)= U^{\perp}.\]Thus $f_1,\cdots,f_n$ is an orthonormal basis of $U^{\perp}$.

5. Solution: By 6.47, we have $V=U\oplus U^{\perp}$. For any $v\in V$, $v$ can be written as $v=u+w$, where $u\in U$ and $w\in U^{\perp}$. By the definition of orthogonal projection, we have $P_U(v)=u$. By 6.51, we have $(U^{\perp})^{\perp}=U$, hence $P_{U^{\perp}}(v)=w$ as $w\in U^{\perp}$ and $v\in U=(U^{\perp})^{\perp}$. Hence \[ P_{U^{\perp}}(v)=w=(u+w)-u=I(v)-P_U(v)=(I-P_U)(v). \]As $v$ is chosen arbitrarily, it follows that $P_{U^{\perp}}=I-P_{U}$.

6. Solution: Suppose $P_UP_W=0$. For any $w\in W$, we have $P_UP_Ww=P_Uw=0$. It follows that $w\in U^\perp$ by Proposition 6.55 (c). Therefore $\langle w,u\rangle$ for all $u\in U$. Note that $w\in W$ is chosen arbitrarily, one has $\langle u,w\rangle =0$ for all $u\in U$ and all $w\in W$.

Conversely, if $\langle u,w\rangle =0$ for all $u\in U$ and all $w\in W$, then $W\subset U^\perp$. Now again by Proposition 6.55 (c), $P_UW=0$, therefore $P_UP_W=0$ since for all $x\in V$ we have $P_Wx\in W$.

7. Solution: Define $U = \operatorname{range} P$. Suppose $u \in U$. There exists $v \in V$ such that $Pv = u$. Then

$$ u = Pv = P^2 v = P(Pv) = Pu. $$ We’ve shown that $Pu = u$ for any $u \in U$. Let $v \in V$. By Exercise 4 in section 5B, we can write $v = u + w$ for some $u \in U$ and $w \in \operatorname{null} P$. Note that $\operatorname{null} P \subset U^\perp$. Thus

$$ Pv = P(u + w) = Pu = u = P_U(u + w) = P_Uv. $$ Therefore $P = P_U$.

8. Solution: By Exercise 5B.4, we have \[V=\m{null}P\oplus\m{range}P.\]Hence for any $u\in \m{null}P$ and $w\in \m{range}P$. By assumption, we have \[ \|w\|=\|P(\lambda u+ w)\|\le\|\lambda u+w\| \]for any $\lambda\in\mb F$. By Exercise 6A.6, it follows that $\langle u,w\rangle=0$. As $u,w$ are chosen arbitrarily, we deduce that $\m{null}P\perp\m{range}P$. Now we can choose $U=\m{null}P$ and it is obvious that $P=P_U$ now.

Solution: See Linear Algebra Done Right Solution Manual Chapter 6 Problem 18.

9. Solution: If $U$ is invariant under $T$. Then for any $v\in V$, we can express it as $v=u+w$, where $u\in U$ and $w\in U^\perp$. Since $U$ is invariant under $T$, it follows that $Tu\in U$. By definition of orthogonal projection, we have \[ P_UTP_U(v)=P_UT(u)=T(u) \]and $TP_U(v)=Tu$. Hence $P_UTP_U(v)=TP_U(v)$ for any $v\in V$, i.e. $P_UTP_U=TP_U$.

Suppose $P_UTP_U=TP_U$, then for any $u\in U$, we have $P_U(u)=u$. Hence \[P_UTP_U(u)=P_U(Tu)\]and $TP_U(u)=Tu$. That implies $P_U(Tu)=Tu$, hence $Tu\in U$ by 6.55 (d). Therefore $U$ is invariant under $T$.

Solution: See Linear Algebra Done Right Solution Manual Chapter 6 Problem 19.

10. Solution: See Linear Algebra Done Right Solution Manual Chapter 6 Problem 20.

11. Solution: See Linear Algebra Done Right Solution Manual Chapter 6 Problem 21.

12. Solution: See Linear Algebra Done Right Solution Manual Chapter 6 Problem 22.

13. Solution: See Linear Algebra Done Right Solution Manual Chapter 6 Problem 23.

14. Solution: (a) Let $\vp\in U^{\perp}$, by continuity it suffices to show that $\vp(x)=0$ for all $x\in (-1,0)\cup (0,1)$. Suppose $\vp(x_0)=\xi\ne 0$ for some $x_0\in (-1,0)\cup (0,1)$. Then there exist a $\delta>0$ such that $(x_0-\delta,x_0+\delta)\subset (-1,0)\cup (0,1)$ and $\vp(x)\ge \xi/2$ for all $x\in [x_0-\delta,x_0+\delta]$. Define $f\in C_{\R}[-1,1]$ by \[ f(x)=\left\{ \begin{array}{ll} 0, & \hbox{$x\in[-1,x_0-\delta]\cup [x_0+\delta,1]$;} \\ (x-x_0+\delta)/\delta, & \hbox{$x\in [x_0-\delta,x_0]$;} \\ (-x+x_0+\delta)/\delta, & \hbox{$x\in [x_0,x_0+\delta]$.} \end{array} \right. \]Then $f\in U$ and $f(x)\ge 1/2$ for $x\in [x_0-\delta/2,x_0+\delta/2]$. Hence \[ \langle f,\vp\rangle =\int_{-1}^1 f(x)\vp(x)dx\ge \int_{x_0-\delta/2}^{x_0+\delta/2} f(x)\vp(x)dx\ge \delta\xi/4>0. \]Hence we get a contradiction, thus completing the proof.

(b) Use $U$ as a counterexample.