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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.

5. Solution: We have
$$ \begin{aligned} \operatorname{dim} \operatorname{null} T^* &= \operatorname{dim} (\operatorname{range} T)^\perp\\ &= \operatorname{dim} W – \operatorname{dim} \operatorname{range} T\\ &= \operatorname{dim} W + \operatorname{dim} \operatorname{null} T – \operatorname{dim} V \end{aligned} $$ where the first line follows from 7.7 (a), the second from 6.50 and the third from 3.22. We alse have
$$ \begin{aligned} \operatorname{dim} \operatorname{range} T^* &= \operatorname{dim} (\operatorname{null} T)^\perp\\ &= \operatorname{dim} V – \operatorname{dim} \operatorname{null} T\\ &= \operatorname{dim} \operatorname{range} T \end{aligned} $$ where the first line follows from 7.7 (b), the second from 6.50 and the third from 3.22.

6. Solution: (a) If $T$ were self-adjoint, we would have
$$ \langle Tp, q \rangle = \langle p, T^*q \rangle = \langle p, Tq \rangle. $$ However, let $p(x) = a_0 + a_1 x + a_2 x^2$ and $q(x) = b_0 + b_1 x + b_2 x^2$. We have
$$ \begin{aligned} \langle Tp, q \rangle &= \langle a_1 x, q \rangle\\ &= a_1 \int_0^1 b_0x + b_1x^2 + b_2x^3\:dx\\ &= a_1 \left(\frac{b_0}{2}x^2 + \frac{b_1}{3}x^3 + \frac{b_2}{4}x^4\right)\biggr\rvert_0^1\\ &= a_1 \left(\frac{b_0}{2} + \frac{b_1}{3} + \frac{b_2}{4}\right). \end{aligned} $$ Similarly
$$ \langle p, Tq \rangle = \langle p, b_1x \rangle = b_1 \left(\frac{a_0}{2} + \frac{a_1}{3} + \frac{a_2}{4}\right). $$ Thus, taking $a_1 = 0$ and $b_1, a_0, a_2 > 0$ clearly shows $\langle Tp, q \rangle \neq \langle p, Tq \rangle$.
(b) 7.10 requires the chosen basis to be orthonormal.

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.

13. Solution: Define $T \in L(\mathbb{C}^4)$ by
$$ T(z_1, z_2, z_3, z_4) = (z_4, z_1, z_2, z_3). $$ We have
$$ \begin{aligned} \langle (z_1, z_2, z_3, z_4), T^*(x_1, x_2, x_3, x_4) \rangle &= \langle T(z_1, z_2, z_3, z_4), (x_1, x_2, x_3, x_4) \rangle\\ &= \langle (z_4, z_1, z_2, z_3), (x_1, x_2, x_3, x_4) \rangle\\ &= z_4\overline{x_1} + z_1\overline{x_2} + z_2\overline{x_3} + z_3\overline{x_4}\\ &= \langle (z_1, z_2, z_3, z_4), (x_2, x_3, x_4, x_1) \rangle. \end{aligned} $$ Thus $T^*(z_1, z_2, z_3, z_4) = (z_2, z_3, z_4, z_1)$. Note that $T$ is normal ($T^*T$ and $TT^*$ equal the identity), however $T \neq T^*$.

14. Solution: Since $v$ and $w$ are eigenvectors corresponding to distinct eigenvalues, by 7.22 they are orthogonal. Thus
$$ \begin{aligned} ||T(v + w)||^2 &= ||Tv + Tw||^2\\ &= ||3v + 4w||^2\\ &= ||3v||^2 + ||4w||^2\\ &= 9||v||^2 + 16||w||^2\\ &= 100, \end{aligned} $$ where the third line follows from the Pythagorean Theorem.

15. Solution: Let $w_1, w_2 \in V$. We have
$$ \begin{aligned} \langle w_1, T^*w_2 \rangle &= \langle Tw_1, w_2 \rangle\\ &= \langle \langle w_1, u \rangle x, w_2 \rangle\\ &= \langle w_1, u \rangle \langle x, w_2 \rangle\\ &= \langle w_1, \overline{\langle x, w_2 \rangle} u \rangle\\ &= \langle w_1, \langle w_2, x \rangle u \rangle\\ \end{aligned} $$ Hence $T^*v = \langle v, x \rangle u$.

(a) Suppose $T$ is selft-adjoint. Then
$$ \langle v, u \rangle x – \langle v, x \rangle u = Tv – T^*v = 0, $$ for all $v \in V$. We can assume $u$ and $x$ are non-zero (otherwise there is nothing to prove). Taking $v = u$ forces $\langle v, u \rangle \neq 0$, showing that $x$ and $u$ are linearly dependent.

Conversely, suppose $x$ and $u$ are linearly dependent. We can assume $x$ and $u$ are non-zero, otherwise $T$ would equal $0$, which already is self-adjoint. Then $u = cx$, for some non-zero $c \in \mathbb{R}$. Thus
$$ \begin{aligned} Tv &= \langle v, u \rangle x\\ &= \langle v, cx \rangle \frac{1}{c}u\\ &= \langle v, x \rangle u\\ &= T^* v. \end{aligned} $$ Therefore $T = T^*$.

(b) Again, we can assume $u$ and $x$ are both non-zero in both directions of the proof.
We have
$$ \begin{aligned} \langle \langle v, u \rangle x, x \rangle u &= T^*(\langle v, u \rangle x)\\ &= T^*Tv\\ &= TT^*v\\ &= T(\langle v, x \rangle u)\\ &= \langle \langle v, x \rangle u, u \rangle x. \end{aligned} $$ Taking $v = u$ ensures $\langle \langle v, u \rangle x, x \rangle \neq 0$, showing that $u$ and $x$ are linearly dependent.

Conversely, suppose $x$ and $u$ are linearly dependent. Then $u = cx$ for some non-zero $c \in \mathbb{F}$. Then
$$ \begin{aligned} &= TT^*v\\ &= T(\langle v, x \rangle u)\\ &= \langle \langle v, x \rangle u, u \rangle x\\ &= \langle \langle v, x \rangle x, cx \rangle cx\\ &= \langle \langle v, cx \rangle x, x \rangle cx\\ &= \langle \langle v, u \rangle x, x \rangle u\\ &= T^*(\langle v, u \rangle x)\\ &= T^*Tv. \end{aligned} $$ Hence $TT^* = T^*T$.

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.

18. Solution: We give a counterexample. Let $V = \mathbb{R}^2$ and $T$ defined by
$$ Te_1 = e_1 + e_2, Te_2 = – e_1 – e_2 $$ where $e_1, e_2$ is the standard basis of $\mathbb{R}^2$. Its matrix with respect to the same basis is
$$ \begin{pmatrix} 1 & -1\\ 1 & -1 \end{pmatrix}. $$ Taking the transpose, we see that $T^*$ is defined by
$$ T^*e_1 = e_1 – e_2, T^*e_2 = e_1 – e_2. $$ Note that $||Te_1|| = ||T^*e_1||$ and $||Te_2|| = ||T^*e_2||$. However $\mathcal{M}(T^*)\mathcal{M}(T) \neq \mathcal{M}(T)\mathcal{M}(T^*)$, thus $T$ is not normal.

19. Solution: We saw in Exercise 16 that $\operatorname{null} T = \operatorname{null} T^*$ (for $T$ normal). Thus $(z_1, z_2, z_3) \in \operatorname{null} T$ and we have $$ \begin{aligned} 0 &= \langle T^*(z_1, z_2, z_3), (1, 1, 1) \rangle\\ &= \langle (z_1, z_2, z_3), T(1, 1, 1) \rangle\\ &= \langle (z_1, z_2, z_3), (2, 2, 2) \rangle\\ &= 2z_1 + 2z_2 + 3z_3. \end{aligned} $$ Dividing by $2$ yields the desired result.

20. Solution: Let $v \in V$ and $w \in W$. Then
$$ \begin{aligned} ((\Phi_V \circ T^*)(w))(v) &= (\Phi_V(T^*w))(v)\\ &= \langle v, T^*v \rangle\\ &= \langle Tv, w \rangle. \end{aligned} $$ On the other hand, we have $$ \begin{aligned} ((T’ \circ \Phi_W)(w))(v) &= (T’ \circ \Phi_W(w))(v)\\ &= (\Phi_W(w) \circ T)(v)\\ &= (\Phi_W(w))(Tv)\\ &= \langle Tv, w \rangle. \end{aligned} $$ Therefore $\Phi_V \circ T^* = T’ \circ \Phi_W$.

21. Solution:

(a) Let $e_j = \dfrac{\cos jx}{\sqrt{\pi}}$ and $f_j = \dfrac{\sin jx}{\sqrt{\pi}}$. By Exercise 4 in section 6B, $\frac{1}{\sqrt{2\pi}}, e_1, \dots, e_n, f_1, \dots, f_n$ is an orthonormal basis of $V$. Note that $De_j = -jf_j$ and $Df_j = je_j$. Then, for any $v, w \in V$, we have
$$ \begin{aligned} \langle v, D^*w \rangle &= \langle Dv, w \rangle\\ &= \left\langle D\left(\left\langle v, \frac{1}{\sqrt{2\pi}} \right\rangle \frac{1}{\sqrt{2\pi}} + \sum_{j=1}^n (\langle v, e_j \rangle e_j + \langle v, f_j \rangle f_j)\right), w \right\rangle\\ &= \left\langle \sum_{j=1}^n (-j\langle v, e_j \rangle f_j + j\langle v, f_j \rangle e_j), \left\langle w, \frac{1}{\sqrt{2\pi}} \right\rangle \frac{1}{\sqrt{2\pi}} + \sum_{j=1}^n (\langle w, e_j \rangle e_j + \langle w, f_j \rangle f_j) \right\rangle\\ &= \sum_{j=1} (-j\langle v, e_j \rangle \langle w, f_j \rangle + j\langle v, f_j \rangle \langle w, e_j \rangle)\\ &= \left\langle \left\langle v, \frac{1}{\sqrt{2\pi}} \right\rangle \frac{1}{\sqrt{2\pi}} + \sum_{j=1}^n (\langle v, e_j \rangle e_j + \langle v, f_j \rangle f_j), \sum_{j=1}^n (-j\langle w, f_j \rangle e_j + j\langle w, e_j \rangle f_j)\right\rangle\\ &= \langle v, -Dw \rangle. \end{aligned} $$ Hence $D^* = -D$. Obviously $D$ is normal but not self-adjoint.

(b) Note that $T = D^2$. Thus $T^* = (DD)^* = D^*D^* = (-D)(-D) = D^2 = T$.