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 whether $V$ is finite dimensional or not)Note that $(T^*)^*=T$, it suffices to show $\lambda$ is an eigenvalue of $T$ then $\bar\lambda$ is an eigenvalue of $T^*$. Let $v$ be a eigenvectors of $T$ corresponding to $\lambda$, then $Tv=\lambda v$. We have \begin{equation}\label{7AP1}0=((T-\lambda I)v,w)=(v,(T^*-\bar\lambda I)w)\end{equation}for all $w\in V$. If $\bar\lambda$ is not an eigenvalue of $T^*$, then $T^*-\bar\lambda I$ is surjective by 5.6. It follows that there exists some $\xi\in V$ such that $(T^*-\bar\lambda I)\xi=v$. By $(\ref{7AP1})$, we have \[0=(v,(T^*-\bar\lambda I)\xi)=(v,v).\]But $v\ne 0$ since $v$ is a eigenvectors, we get a contradiction. Hence we get the conclusion.

Solution: See Linear Algebra Done Right Solution Manual Chapter 6 Problem 28.


3. Solution: Let $u\in U$ and $w\in U^\perp$, then we have \begin{equation}\label{7AP1.1} \langle Tu,w\rangle=\langle u,T^*w\rangle. \end{equation} If $U$ is invariant under $T$, then $Tu\in U$ for all $u\in U$. Hence for a fixed $w\in U^\perp$, we have \[ 0=\langle Tu,w\rangle=\langle u,T^*w\rangle \]for all $u\in U$. This implies $T^*w\in U^{\perp}$. As $w$ is chosen arbitrarily, we conclude $U^{\perp}$ is invariant under $T^*$. The other direction is similar.

Solution: See Linear Algebra Done Right Solution Manual Chapter 6 Problem 29.


4. Solution: See Linear Algebra Done Right Solution Manual Chapter 6 Problem 30.


7. Solution: We have $(ST)^*=T^*S^*$ by 7.6 (e). Hence $ST$ is self-adjoint if and only if \[T^*S^*=ST.\]Note that $S,T\in\ca L(V)$ are self-adjoint, we have $S^*=S$ and $T^*=T$. Therefore $ST$ is self-adjoint if and only if $T^*S^*=ST$, which is equivalent to $ST=TS$.


8. Solution: See Linear Algebra Done Right Solution Manual Chapter 7 Problem 3 (a).


9. Solution: See Linear Algebra Done Right Solution Manual Chapter 7 Problem 3 (b).


10. Solution: See Linear Algebra Done Right Solution Manual Chapter 7 Problem 5.


11. Solution: See Linear Algebra Done Right Solution Manual Chapter 7 Problem 4.


12. Solution: Let $u$ be a unit eigenvector (i.e. $\|u\|=1$) of $T$ corresponding to eigenvalue 3, then $Tu=3u$. Let $w$ be a unit eigenvector (i.e. $\|w\|=1$) of $T$ corresponding to eigenvalue 4, then $Tv=4w$.

By 7.22, we have $\langle u,w\rangle =0$. Let $v=au+bw$, then it follows from 6.25 that $$\|v\|^2=\|au+bw\|^2=|a|^2+|b|^2$$ and $$\|Tv\|^2=\|Tau+Tbw\|^2=\|3au+4bw\|^2=9|a|^2+16|b|^2.$$Hence we need to choose  $a$ and $b$ such that\[|a|^2+|b|^2=2,\quad 9|a|^2+16|b|^2=25.\]A simple solution is $a=1$ and $b=1$. Hence $v=u+w$ satisfies the requirement.


16. Solution: See Linear Algebra Done Right Solution Manual Chapter 7 Problem 6.


17. Solution: See Linear Algebra Done Right Solution Manual Chapter 7 Problem 7.


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