# Category Archives: Chapter 7

## Chapter 7 Exercise D

11. Solution: It follows from 7.45 that $T=S\sqrt{T^*T}$ for an isometry $S\in\ca L(V)$. Note that $\sqrt{T^*T}$ is self-adjoint and $S^{-1}=S^*$ (7.42 (e), (f)), we have $TT^*=S\sqrt{T^*T}(S\sqrt{T^*T})^*=S\sqrt{T^*T}\sqrt{T^*T}S^*=S(T^*T)S^{-1}.$It follows from Problem 15 of Exercise 5A that $TT^*$ and $T^*T$ have … Continue reading

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## Chapter 7 Exercise C

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. … Continue reading

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## Chapter 7 Exercise B

1. Solution: It is true. Consider the  standard orthonormal basis $e_1,e_2,e_3$ of $\mb R^3$. Define $T\in \ca L(\mb R^3)$ by the rule:$Te_1=e_1,\quad Te_2=2e_2+e_1,\quad Te_3=3e_3.$Since we have$\langle Te_1,e_2\rangle =\langle e_1,e_2\rangle =0,$$\langle e_1,Te_2\rangle =\langle e_1,e_1+2e_2\rangle =1,$it follows that $T$ is not self-adjoint. … Continue reading

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## Chapter 7 Exercise A

1. Solution: By definition, we have \begin{align*}\langle (z_1,\cdots,z_n),T^*(w_1,\cdots,w_n)\rangle=&\langle T(z_1,\cdots,z_n),(w_1,\cdots,w_n) \rangle \\=& z_1w_2+\cdots+z_{n-1}w_{n}=\langle (z_1,\cdots,z_n),(w_2,\cdots,w_n,0)\rangle.\end{align*}Therefore $T^*(w_1,\cdots,w_n)=(w_2,\cdots,w_n,0)$ or $T^*(z_1,\cdots,z_n)=(z_2,\cdots,z_n,0)$. See also Linear Algebra Done Right Solution Manual Chapter 6 Problem 27. 2. Solution: (This solution works for $\dim V<\infty$, I am not sure … Continue reading

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