Chapter 7 Exercise C

2. Solution: Note that $T$ is a positive operator on $V$, we have \begin{equation}\label{7CP2.1} \langle T(v-w),v-w\rangle\ge 0. \end{equation}On the other hand, \[Tv=w\quad\text{ and } \quad Tw=v\]imply that $T(v-w)=w-v$, hence \begin{equation}\label{7CP2.2}\langle T(v-w),v-w\rangle=-\langle v-w,v-w\rangle\le 0.\end{equation}Therefore $\langle v-w,v-w\rangle=0$ by (\ref{7CP2.1}) and (\ref{7CP2.2}), i.e. $v=w$.

4. Solution: By 7.6 (c) and (e), we have $$(TT^*)*=(T^*)*T^*=TT^*,\quad (T^*T)^*=T^*(T^*)^*=T^*T.$$Hence both $TT^*$ and $T^*T$ are self-adjoint.

On the other hand, for any $v\in V$, we have\[\langle T^*Tv,v\rangle =\langle Tv,(T^*)^*v\rangle=\langle Tv,Tv\rangle\geqslant 0.\]Hence $T^*T$ is a positive operator. Similarly, for any $w\in W$, we have\[\langle TT^*v,v\rangle =\langle T^*v,T^*v\rangle\geqslant 0.\]Hence $TT^*$ is a positive operator.

5. Solution: Suppose $T$ and $S$ are positive operators on $V$, then $T^*=T$ and $S^*=S$. Therefore, we have $$(T+S)^*=T^*+S^*=T+S.$$Hence $T+S$ is self-adjoint.

Again since $T$ and $S$ are positive operators on $V$,, for any $v\in V$, $\langle Tv,v\rangle \geqslant 0$ and $\langle Sv,v\rangle \geqslant 0$. Thus we have\[\langle(T+S)v,v \rangle=\langle Tv,v\rangle+\langle Sv,v\rangle\geqslant 0.\]Therefore $T+S$ is a positive operator.

6. Solution: Since $T$ is positive, it follows from 7.35 (a) $\iff$ (d) that there exists a self-adjoint operator $S$ such that $S^2=T$. For any positive integer $k$, we have $(S^k)^*=(S^*)^k=S^k$ by 7.6 (e) and the equality $S=S^*$. Hence $S^k$ is self-adjoint.

Note that we also have $(S^k)^2=(S^2)^k=T^k$, hence $T^k$ has a self-adjoint square root. It follows from 7.35 (a) $\iff$ (d) again that $T^k$ is positive.

7. Solution: Suppose $\langle Tv,v\rangle >0$ for every $v\in V$ with $v\ne 0$. If $T$ is not invertible, there must exist a nonzero $u\in V$ such that $Tu=0$, hence $\langle Tu,u\rangle =0$ for $u\ne 0$. Therefore we get a contradiction, which in turn implies that $T$ is invertible.

Conversely, suppose $T$ is invertible. Since $T$ is positive, it follows from 7.35 (a) $\iff$ (d) that there exists a self-adjoint operator $S$ such that $S^2=T$. Because $T$ is injective, so is $S$. Hence for every $v\in V$ with $v\ne 0$, we have $Sv\ne 0$. Moreover, since $S$ is self-adjoint (and $Sv\ne 0$), we have$$\langle Tv,v\rangle =\langle S^2v,v\rangle =\langle Sv,Sv\rangle >0.$$

8. Solution: If $\langle \cdot,\cdot\rangle_T$ is an inner product on $V$, then for any $v\in V$ we have $$\langle Tv,v\rangle =\langle v,v\rangle_T\geqslant 0.$$Hence $T$ is positive. Moreover, for any nonzero $v\in V$ we have $$\langle Tv,v\rangle =\langle v,v\rangle_T>0.$$It follows from Problem 7 that $T$ is invertible.

Conversely, suppose that $T$ is an invertible positive operator. We show that $\langle \cdot,\cdot\rangle_T$ is an inner product on $V$ by checking definition 6.3.

Positivity: Note that $T$ is positive, we have $$\langle v,v\rangle_T=\langle Tv,v\rangle \geqslant 0.$$ Definiteness: If $v=0$, then $$\langle v,v\rangle_T=\langle Tv,v\rangle \geqslant 0.$$If $\langle v,v\rangle_T=0$, since $T$ is invertible positive operator on $V$, it follows from Problem 7 that $v=0$.

Additivity, homogeneity, and conjugate symmetry can be checked directly with out any difficulties.

9. Solution: Let $e_1,e_2$ be an orthonormal basis of $\mb F^2$ and $\theta\in[0,2\pi)$, define $T_\theta\in\ca L(\mb F^2)$ by\[T_\theta e_1=\cos\theta e_1+\sin\theta e_2,\quad T_\theta e_2=\sin \theta e_1-\cos\theta e_2.\]Note that $$\cos\theta=\langle T_\theta e_1,e_1\rangle=\langle e_1,(T_\theta)*e_1\rangle,$$ $$\sin\theta=\langle T_\theta e_2,e_1\rangle=\langle e_2,(T_\theta)^*e_1\rangle,$$ we have $(T_\theta)^*e_1=\cos\theta e_1+\sin\theta e_2$. Similarly, note that $$\sin\theta=\langle T_\theta e_1,e_2\rangle=\langle e_1,(T_\theta)^*e_2\rangle,$$ $$-\cos\theta=\langle T_\theta e_2,e_2\rangle=\langle e_2,(T_\theta)^*e_2\rangle,$$ we have $(T_\theta)^*e_2=\sin\theta e_1-\cos\theta e_2$. Hence $T_\theta=(T_\theta)^*$, which implies that $T_\theta$ is self-adjoint. Also note that \begin{align*}& (T_\theta)^2e_1=T_\theta(\cos\theta e_1+\sin\theta e_2)\\ =&\cos\theta(\cos\theta e_1+\sin\theta e_2)+\sin\theta(\sin \theta e_1-\cos\theta e_2)\\=&(\cos^2\theta+\sin^2\theta)e_1=e_1,\end{align*} \begin{align*}& (T_\theta)^2e_2=T_\theta(\sin\theta e_1-\cos\theta e_2)\\ =&\sin\theta(\cos\theta e_1+\sin\theta e_2)-\cos\theta(\sin \theta e_1-\cos\theta e_2)\\=&(\cos^2\theta+\sin^2\theta)e_2=e_2,\end{align*}we have $(T_\theta)^2=\mathrm{id}$. Therefore we have infinitely many self-adjoint operators as the square root of $\mathrm{id}$.

The construction comes from the following idea. Let $T$ be self-adjoint such that $T^2=\mathrm{id}$. Let $\begin{pmatrix}a & b\\ c & d\end{pmatrix}$ be the matrix with respect to an orthonormal basis of $\mb F^2$. Since $T$ is self-adjoint, we have $$\begin{pmatrix}a & b\\ c & d\end{pmatrix}=\begin{pmatrix}\bar a & \bar c\\ \bar b & \bar d\end{pmatrix}.$$ Hence $a,d\in\mb R$ and $b=\bar c$. If $T^2=\mathrm{id}$, then we have $a^2+bc=1$, $ab+bd=0$, and $d^2+bc=1$. Hence we can take $a=-d$ and $b=c$. Then $a^2+b^2=1$. Take $a=\cos\theta$ and $b=\sin\theta$, we get the construction.

10. Solution: “(a)$\Longrightarrow$(b)” If $S$ is an isometry, so is $S^*$ by 7.42 (g)$\iff$(a). It follows from 7.42 (a)$\iff$(b) that $$\langle S^*u,S^*v\rangle =\langle u,v\rangle$$ for all $u,v\in V$.

“(b)$\Longrightarrow$(c)” If $e_1,\cdots,e_m$ is an orthonormal basis of $V$, we have $\langle e_i,e_j\rangle =\delta_{ij}$. Since $$\langle S^*u,S^*v\rangle =\langle u,v\rangle$$ for all $u,v\in V$, we have\[\langle S^*e_i,S^*e_j\rangle=\langle e_i,e_j\rangle =\delta_{ij}.\]Hence $S^*e_1,\cdots,S^*e_m$ is an orthornormal basis of $V$.

“(c)$\Longrightarrow$(d)” is trivial.

“(d)$\Longrightarrow$(a)” It follows from 7.42 (a)$\iff$(d) that $S^*$ is an isometry. So is $S$ by 7.42 (g)$\iff$(a) since $(S^*)^*=S$ from 7.6 (c).

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