# Chapter 7 Exercise C

1. Solution: We give a counterexample. Define $T \in \mathcal{L}(\mathcal{R}^2)$ by

\begin{aligned} Te_1 = e_1\\ Te_2 = -e_2 \end{aligned}

where $e_1, e_2$ is the standard basis of $\mathbb{R}^2$. The matrix of $T$ with respect to this same basis is

$$\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix},$$

which equals its transpose, therefore $T$ is self-adjoint. Moreover, the basis $\frac{1}{\sqrt{2}}(e_1 + e_2), \frac{1}{\sqrt{2}}(e_1 – e_2)$ is orthonormal and

\begin{aligned} \left\langle T\left(\frac{1}{\sqrt{2}}(e_1 + e_2)\right), \frac{1}{\sqrt{2}}\left(e_1 + e_2\right) \right\rangle &= 0\\ \left\langle T\left(\frac{1}{\sqrt{2}}(e_1 – e_2)\right), \frac{1}{\sqrt{2}}\left(e_1 – e_2\right) \right\rangle &= 0, \end{aligned}

but $T$ is not positive because

$$\langle Te_2, e_2 \rangle = \langle -e_2, e_2 \rangle = -1.$$

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

3. Solution: For all $u \in U$, we have

$$\langle T|_U u, u \rangle = \langle Tu, u \rangle = \langle u, Tu \rangle = \langle u, T|_U u \rangle.$$

Thus, $T|_U$ is self-adjoint. Furthermore,

$$\langle T|_U u, u \rangle = \langle Tu, u \rangle \ge 0,$$

which shows that $T|_U$ is positive.

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).

11. Solution: Let $e_1, e_2, e_3$ and $f_1, f_2, f_3$ be orthonormal bases of $\mathbb{F}^3$ consisting of eigenvectors of $T_1$ and $T_2$, respectively, corresponding to the eigenvalues $2, 5, 7$. Define $S$ by

$$Se_j = f_j$$

for $j = 1, 2, 3$. One easily checks that $S$ is an isometry (using the Pythagorean Theorem). Then, because $S^{-1} = S^*$ (by 7.42), we have $S^*f_j = e_j$. Thus

$$T_1e_1 = 2e_1 = S^*(2f_1) = S^*(T_2f_1) = S^*T_2Se_1.$$

Similarly $T_1e_2 = S^*T_2Se_2$ and $T_1e_3 = S^*T_2Se_3$. Therefore $T_1 = S^*T_2S$.

12. Solution: Let $e_1, e_2, e_3, e_4$ denote an orthonormal basis of $\mathbb{F}^4$. Define $T_1, T_2 \in \mathcal{L}(\mathbb{F}^4)$ by

\begin{aligned} T_1e_1 &= 2e_1\\ T_1e_2 &= 2e_2\\ T_1e_3 &= 5e_3\\ T_1e_4 &= 7e_4\\ \\ T_2e_1 &= 2e_1\\ T_2e_2 &= 5e_2\\ T_2e_3 &= 5e_3\\ T_2e_4 &= 7e_4. \end{aligned}

Then both $T_1$ and $T_2$ are self-adjoint (the matrices equal their transposes) and $2, 5, 7$ are their eigenvalues. Suppose by contradiction that $S$ is an isometry on $V$ such that $T_1 = S^*T_2S$. Let $v \in V$ be the vector that $S$ maps to $e_2$. Then

$$T_1v = S^*T_2Sv = S^*T_2e_2 = 5S^*e_2 = 5v$$

Therefore $v \in E(T_1, 5) = \operatorname{span}(e_3)$. Let also $w \in V$ be the vector that $S$ maps to $e_3$. Note that $v, w$ is linearly independent, because $e_2, e_3$ is linearly independent. Then

$$T_1w = S^*T_2Sw = S^*T_2e_3 = 5S^*e_3 = 5w.$$

Therefore $w \in E(T_1, 5) = \operatorname{span}(e_3)$. But this is a contradiction, because we can’t have a linearly independent list of length $2$, $v, w$, in a $1$-dimensional vector space, $\operatorname{span}(e_3)$. Hence, there does not exist such $S$.

Notice that it wasn’t necessary to require $S$ to be an isometry, we just needed to suppose, by contradiction, the existence of an invertible $S$ such that $T_1 = S^{-1}T_2S$. This $S$ does not exist. Since the desired isometry must satisfy the same property (because the adjoint of an isometry equals its inverse), it follows that there cannot exist such isometry. The key idea here is that the eigenspaces of $T_1$ and $T_2$ don’t fit.

13. Solution: It is false. Let $e_1, \dots, e_n$ be an orthonormal basis of $V$ and define $Se_j = e_1$ for $j = 1, \dots, n$. Then $||Se_j|| = 1$ for each $e_j$, but obviously $S$ is not invertible, therefore $S$ is not an isometry (7.42 requires isometries to be invertible).

14. Solution: In the exercise, $T$ was already shown to be self-adjoint. So $-T$ is also self-adjoint. Note that $1, \cos x, \cos 2x, \dots, \cos nx, \sin x, \sin 2x, \dots, \sin nx$ is a basis of $V$ consisting of eigenvectors of $-T$ whose corresponding eigenvalues are all nonnegative. Thus by 7.35 $-T$ is positive.