Exercises 7.A
(1) Suppose $n$ is a positive integer. Define $T\in\oper{\wF^n}$ by
$$ T(z_1,\dots,z_n)=(0,z_1,\dots,z_{n-1}) $$
Find a formula for $\adjt{T}(z_1,\dots,z_n)$.
Solution Suppose $n=2$. Then
$$ T(z_1,z_2)=(0,z_1) $$
and
$$ \innprd{Tz}{w}=\innprdbg{(0,z_1)}{(w_1,w_2)}=z_1\overline{w_2} $$
and
$$ z_1\overline{w_2}=\innprd{Tz}{w}=\innprd{z}{\adjt{T}w}=z_1\overline{(\adjt{T}w)_1}+z_2\overline{(\adjt{T}w)_2} $$
We know $\adjt{T}w$ is unique. So if we find a solution, then that is the only solution. Just looking at the equation, we can see that
$$\align{ &(\adjt{T}w)_1=w_2 \\ &(\adjt{T}w)_2=0 }$$
is a solution. That is, $\adjt{T}(w_1,w_2)=(w_2,0)$.
Suppose $n=3$. Then
$$ T(z_1,z_2,z_3)=(0,z_1,z_2) $$
and
$$ \innprd{Tz}{w}=\innprdbg{(0,z_1,z_2)}{(w_1,w_2,w_3)}=z_1\overline{w_2}+z_2\overline{w_3} $$
and
$$ z_1\overline{w_2}+z_2\overline{w_3}=\innprd{Tz}{w}=\innprd{z}{\adjt{T}w}=z_1\overline{(\adjt{T}w)_1}+z_2\overline{(\adjt{T}w)_2}+z_3\overline{(\adjt{T}w)_3} $$
Again looking at the equation, we can see that
$$\align{ &(\adjt{T}w)_1=w_2 \\ &(\adjt{T}w)_2=w_3 \\ &(\adjt{T}w)_3=0 }$$
is a solution and hence is the unique solution. That is, $\adjt{T}(w_1,w_2,w_3)=(w_2,w_3,0)$. More generally, we get
$$\align{ \innprdbg{(z_1,\dots,z_n)}{\adjt{T}(w_1,\dots,w_n)} &= \innprdbg{T(z_1,\dots,z_n)}{(w_1,\dots,w_n)} \\ &= \innprdbg{(0,z_1,\dots,z_{n-1})}{(w_1,\dots,w_n)} \\ &= z_1\overline{w_2}+\dots+z_{n-1}\overline{w_n} \\ &= \innprdbg{(z_1,\dots,z_{n-1})}{(w_2,\dots,w_n)} }$$
Then the uniqueness of the solution implies that
$$ \adjt{T}(w_1,\dots,w_n)=(w_2,w_3,\dots,w_n,0) $$
$\blacksquare$
(2) Suppose $T\in\oper{V}$ and $\lambda\in\wF$. Then $\lambda$ is an eigenvalue of $T$ if and only if $\cj\lambda$ is an eigenvalue of $\adjt{T}$.
Proof Equivalence follows from
$$\align{ \lambda\text{ is not an eigenvalue of }T &\iff T-\lambda I\text{ invertible}\tag{by 5.6.a} \\ &\iff S(T-\lambda I)=(T-\lambda I)S=I\text{ for some }S\in\oper{V}\tag{by 3.53} \\ &\iff (T-\lambda I)^*S^*=S^*(T-\lambda I)^*=I\text{ for some }S\in\oper{V}\tag{by 7.6 (d),(e)} \\ &\iff (T-\lambda I)^*\text{ invertible}\tag{by 3.53} \\ &\iff \adjt{T}-\cj{\lambda}I\text{ invertible}\tag{by 7.6 (a),(b),(d)} \\ &\iff \cj{\lambda}\text{ is not an eigenvalue of }\adjt{T}\tag{by 5.6.a} }$$
$\blacksquare$
Alternative Proof Suppose $\lambda$ is an eigenvalue of $T$. Then there exists $v\in V$ with $v\neq0$ and $0=(T-\lambda I)v$. Hence, for any $w\in V$, we have
$$\align{ 0 &= \innprdbg{(T-\lambda I)v}{w}\tag{by 6.7.b} \\ &= \innprdbg{v}{(T-\lambda I)^*w}\tag{by 7.2} \\ &= \innprdbg{v}{(\adjt{T}-\cj{\lambda}I)w}\tag{by 7.6 (a),(b),(d)} \\ }$$
Suppose $\cj{\lambda}$ is not an eigenvalue of $\adjt{T}$. Then $\adjt{T}-\cj{\lambda}I$ is surjective (by 5.6.c). Hence there exists some $u\in V$ such that $(\adjt{T}-\cj{\lambda}I)u=v$. Then the above equation gives
$$ 0=\innprdbg{v}{(\adjt{T}-\cj{\lambda}I)u}=\innprd{v}{v} $$
Then the definiteness of an inner product implies that $v=0$, a contradiction. Hence $\cj{\lambda}$ is an eigenvalue of $\adjt{T}$.
Since $T=(\adjt{T})^*$ (by 7.6.c) and $\lambda=\cj{\cj{\lambda}}$, then, by a symmetrical argument, we’re done. $\blacksquare$
Source of Confusion Suppose $\lambda$ is an eigenvalue of $T$. Then there exists $v\in V$ with $v\neq0$ and $Tv=\lambda v$. Then, for any $w\in V$, we have
$$ \innprd{v}{\adjt{T}w}=\innprd{Tv}{w}=\innprd{\lambda v}{w}=\lambda\innprd{v}{w}=\innprd{v}{\cj{\lambda}w} $$
So every vector in $V$ is an eigenvector of $\adjt{T}$ corresponding to $\cj{\lambda}$?
(3) Suppose $T\in\oper{V}$ and $U$ is a subspace of $V$. Then $U$ is invariant under $T$ if and only if $U^\perp$ is invariant under $\adjt{T}$.
Proof Suppose $U$ is invariant under $T$. Let $w\in U^\perp$. Then $Tu\in U$ for any $u\in U$ and
$$ 0=\innprd{Tu}{w}=\innprd{u}{\adjt{T}w} $$
for any $u\in U$. This implies that $\adjt{T}w\in U^\perp$. Hence $U^\perp$ is invariant under $\adjt{T}$.
Conversely, suppose $U^\perp$ is invariant under $\adjt{T}$. Then, by the first argument, we know that $(U^\perp)^\perp=U$ (by 6.51) is invariant under $(\adjt{T})^*=T$ (by 7.6.c). $\blacksquare$
(4) Suppose $T\in\linmap{V}{W}$. Then
(a) $T$ is injective if and only if $\adjt{T}$ is surjective.
(b) $T$ is surjective if and only if $\adjt{T}$ is injective.
Proof Part (a):
$$\align{ T\text{ is injective } &\iff \nullsp{T}=\set{0}\tag{by 3.16} \\ &\iff \rangsp{\adjt{T}}^\perp = \set{0}\tag{by 7.7.c} \\ &\iff \rangsp{\adjt{T}} = W\tag{by 6.46.b and 6.51} \\ &\iff \adjt{T}\text{ is surjective}\tag{by 3.20} }$$
Then (b) follows by replacing $T$ with $\adjt{T}$ in part (a) and using 7.7.a instead of 7.7.c. $\blacksquare$
(5) For every $T\in\linmap{V}{W}$, we have
$$ \dim{\nullsp{\adjt{T}}}=\dim{\nullsp{T}}+\dim{W}-\dim{V} $$
and
$$ \dim{\rangsp{\adjt{T}}}=\dim{\rangsp{T}} $$
Proof
$$\align{ \dim{\nullsp{\adjt{T}}} &= \dim{\rangsp{T}}^\perp\tag{by 7.7.a} \\ &= \dim{W}-\dim{\rangsp{T}}\tag{by 6.50} \\ &= \dim{W}-\prn{\dim{V}-\dim{\nullsp{T}}}\tag{by 3.22} \\ &= \dim{\nullsp{T}}+\dim{W}-\dim{V} }$$
$$\align{ \dim{\rangsp{\adjt{T}}} &= \dim{(\rangsp{\adjt{T}}^\perp)^\perp}\tag{by 6.51} \\ &= \dim{\nullsp{T}^\perp}\tag{by 7.7.c} \\ &= \dim{V}-\dim{\nullsp{T}}\tag{by 6.50} \\ &= \dim{\rangsp{T}}\tag{by 3.22}\quad\blacksquare }$$
(6) Make $\polyrn{2}$ into an inner product space by defining
$$ \innprd{p}{q}\equiv\int_0^1p(x)q(x)dx $$
Define $T\in\operb{\polyrn{2}}$ by $T(a_0+a_1x+a_2x^2)=a_1x$.
(a) Show that $T$ is not self-adjoint.
(b) The matrix of $T$ with respect to the basis $(1,x,x^2)$ is
$$ \bmtrx{0&0&0\\0&1&0\\0&0&0} $$
This matrix equals its conjugate transpose, even though $T$ is not self-adjoint. Explain why this is not a contradiction.
Solution Note that
$$ \innprd{T1}{x} = \innprd{0}{x} = 0 $$
But
$$ \innprd{1}{Tx} = \innprd{1}{x} = \int_0^1xdx = \frac12\Prn{x^2\eval01}=\frac12 $$
Hence $T$ is not self-adjoint (by W.7.G.2).
For part (b), note that proposition 7.10 specifies that the matrices are relative to orthonormal bases. But $1,x,x^2$ is not an orthonormal basis of $\polyrn{2}$. So the result $\mtrxof{\adjt{T}}=\cj{\mtrxof{T}^t}$ from that proposition doesn’t apply to this problem. $\blacksquare$
(7) Suppose $S,T\in\oper{V}$ are self-adjoint. Then $ST$ is self-adjoint if and only if $ST=TS$.
Proof Suppose $ST$ is self-adjoint. This gives the first equality
$$ ST=(ST)^*=\adjt{T}S^*=TS $$
The second equality follows from 7.6(e). The third equality follows because $\adjt{T}=T$ and $S^*=S$.
Conversely, suppose $ST=TS$. This gives the first equality
$$ ST=TS=\adjt{T}S^*=(ST)^* $$
The second equality follows because $\adjt{T}=T$ and $S^*=S$. The third equality follows from 7.6(e). $\blacksquare$
(8) Suppose $V$ is a real inner product space. Then the set of self-adjoint operators on $V$ is a subspace of $\oper{V}$.
Proof Let $v,w\in V$. Then
$$ \innprd{0v}{w}=\innprd{0}{w}=0=\innprd{v}{0}=\innprd{v}{0w} $$
Hence $0=0^*$ (by W.7.G.2). Hence the set of self-adjoint operators on $V$ contains the zero map.
Let $S$ and $T$ be self-adjoint operators on $V$. Then
$$ (S+T)^*=S^*+\adjt{T}=S+T=(S+T) $$
The first equality is given by 7.6(a). Hence $S+T$ is a self-adjoint operator on $V$. Hence the set of self-adjoint operators on $V$ is closed under addition. Also
$$ (\lambda T)^*=\cj{\lambda}\adjt{T}=\lambda T $$
The first equality is given by 7.6(b). The third equality follows because $\lambda\in\wR$ and $\adjt{T}=T$. Hence $\lambda T$ is self-adjoint. Hence the set of self-adjoint operators on $V$ is closed under scalar multiplication. $\blacksquare$
(9) Suppose $V$ is a complex inner product space with $V\neq\set{0}$. Then the set of self-adjoint operators on $V$ is not a subspace of $\oper{V}$.
Proof Let $v\in V$ with $v\neq0$. Then
$$ (iI)^*v=(\cj{i}I^*)v=(-iI)v=-i(Iv)=-iv\neq iv=i(Iv)=(iI)v $$
$\blacksquare$
(10) Suppose $\dim{V}\geq2$. Then the set of normal operators on $V$ is not a subspace of $\oper{V}$.
Proof Let $e_1,\dots,e_n$ be an orthonormal basis of $V$ and let $w\in V$. Define $S\in\oper{V}$ (a projection onto $\span{e_1,e_2}$) by
$$ Se_1=e_2 \quad\quad\quad Se_2=e_1 \quad\quad\quad Se_k=0\quad\text{for }k=3,\dots,n $$
Note that $\innprd{Se_k}{e_j}=0$ for $k=3,\dots,n$ and for $j=1,\dots,n$. Hence for $k=3,\dots,n$, we get
$$ e_k\sum_{j=1}^n\innprd{w}{e_j}\cj{\innprd{Se_k}{e_j}} = e_k\sum_{j=1}^n\innprd{w}{e_j}\cj{0} = e_k\cdot0 = 0 \\ $$
We also have $\innprd{Se_1}{e_j}=0$ for $j\neq2$ and $\innprd{Se_1}{e_2}=1$. Hence
$$ e_1\sum_{j=1}^n\innprd{w}{e_j}\cj{\innprd{Se_1}{e_j}} = \innprd{w}{e_2}\cj{\innprd{Se_1}{e_2}}e_1=\innprd{w}{e_2}e_1 $$
And we have $\innprd{Te_2}{e_j}=0$ for $j\neq1$ and $\innprd{Se_2}{e_1}=1$. Hence
$$ e_2\sum_{j=1}^n\innprd{w}{e_j}\cj{\innprd{Se_2}{e_j}} = \innprd{w}{e_1}\cj{\innprd{Se_2}{e_1}}e_2=\innprd{w}{e_1}e_2 $$
Hence, formula W.7.CoA.3 gives
$$\align{ S^*w &= \sum_{k=1}^ne_k\sum_{j=1}^n\innprd{w}{e_j}\cj{\innprd{Se_k}{e_j}} \\ &= e_1\sum_{j=1}^n\innprd{w}{e_j}\cj{\innprd{Se_1}{e_j}}+e_2\sum_{j=1}^n\innprd{w}{e_j}\cj{\innprd{Se_2}{e_j}} \\ &= \innprd{w}{e_2}e_1+\innprd{w}{e_1}e_2 }$$
whereas
$$\align{ Sw &= S\Prn{\sum_{k=1}^n\innprd{w}{e_k}e_k} \\ &= \sum_{k=1}^n\innprd{w}{e_k}Se_k \\ &= \innprd{w}{e_1}Se_1+\innprd{w}{e_2}Se_2 \\ &= \innprd{w}{e_1}e_2+\innprd{w}{e_2}e_1 \\ &= \innprd{w}{e_2}e_1+\innprd{w}{e_1}e_2 }$$
Hence $S=S^*$ and $S$ is self-adjoint. Hence $S$ is normal.
Now define $T\in\oper{V}$ to be a projection down to the first two dimensions of $V$ and also make $T$ a clockwise rotation on the first two dimensions. That is, $T$ is defined by
$$ Te_1=-e_2 \quad\quad\quad Te_2=e_1 \quad\quad\quad Te_k=0\quad\text{for }k=3,\dots,n $$
In our ch.7 notes, under the section “Computation of Adjoint”, we showed that $T$ is not self-adjoint but that $T$ is normal. We also showed that $\adjt{T}$ is a projection down to the first two dimensions of $V$ but that it’s a counter-clockwise rotation (whereas $T$ is clockwise):
$$ \adjt{T}e_1=e_2 \quad\quad\quad \adjt{T}e_2=-e_1 \quad\quad\quad \adjt{T}e_k=0\quad\text{ for }k=3,\dots,n $$
Then
$$\align{ &(S+T)e_1=Se_1+Te_1=e_2-e_2=0 \\ &(S+T)e_2=Se_2+Te_2=e_1+e_1=2e_1 \\ &(S+T)e_k=Se_k+Te_k=0+0=0\quad\text{ for }k=3,\dots,n \\ \\ &(S+T)^*e_1=S^*e_1+\adjt{T}e_1=Se_1+\adjt{T}e_1=e_2+e_2=2e_2 \\ &(S+T)^*e_2=S^*e_2+\adjt{T}e_2=Se_2+\adjt{T}e_2=e_1-e_1=0 \\ &(S+T)^*e_k=S^*e_k+\adjt{T}e_k=Se_k+\adjt{T}e_k=0+0=0\quad\text{ for }k=3,\dots,n }$$
Hence
$$\align{ &(S+T)(S+T)^*e_1=(S+T)(2e_2)=2\prn{(S+T)e_2}=2\cdot2e_1=4e_1 \\ &(S+T)^*(S+T)e_1=(S+T)^*(0)=0 \\ \\ &(S+T)(S+T)^*e_2=(S+T)(0)=0 \\ &(S+T)^*(S+T)e_2=(S+T)^*(2e_1)=2\prn{(S+T)^*e_1}=2\cdot2e_2=4e_2 }$$
Hence $(S+T)^*(S+T)\neq(S+T)(S+T)^*$ and $S+T$ is not normal. Hence the set of normal operators on $V$ is not closed under additional and hence this set is not a subspace of $\oper{V}$. $\blacksquare$
(11) Suppose $P\in\oper{V}$ is such that $P^2=P$. Then $P$ is an orthogonal projection if and only if $P$ is self-adjoint.
Proof Note that ch.7 assumes that $V$ is finite-dimensional. Hence any subspaces are finite-dimensional. Hence we can invoke proposition 6.47 freely.
First suppose that $P$ is an orthogonal projection. Then there exists a subspace $U$ of $V$ such that $P=P_U$. Let $v_1,v_2\in V$. Then 6.47 gives
$$ v_1=u_1+w_1 \quad\quad\quad\quad v_2=u_2+w_2 $$
where $u_1,u_2\in U$ and $w_1,w_2\in U^\perp$. Hence
$$\align{ \innprd{v_1}{P^*v_2} &= \innprd{Pv_1}{v_2} \\ &= \innprd{u_1}{u_2+w_2} \\ &= \innprd{u_1}{u_2}+\innprd{u_1}{w_2} \\ &= \innprd{u_1}{u_2} \\ &= \innprd{u_1}{u_2}+\innprd{w_1}{u_2} \\ &= \innprd{u_1+w_1}{u_2} \\ &= \innprd{v_1}{Pv_2} }$$
Hence $P=P^*$ and $P$ is self-adjoint.
Note Cross-reference the next part with exercises 6.C.7 and 5.B.4.
Conversely, suppose $P$ is self-adjoint. Let $v\in V$. Because $P(v-Pv)=Pv-P^2v=0$, we get
$$ v-Pv\in\nullsp{P}=\rangsp{P^*}^\perp=\rangsp{P}^\perp $$
The first equality follows from 7.7(c) and the second equality follows because $P$ is self-adjoint. Also note that
$$ v=Pv+(v-Pv) $$
Since $Pv\in\rangsp{P}$, then $v$ can be written as a sum of a vector in $\rangsp{P}$ and a vector in $\rangsp{P}^\perp$. Then proposition 6.47 implies that this sum is uniquely written. Hence $P_{\rangsp{P}}v=Pv$. $\blacksquare$
Alternate Proof to Second Part Conversely, suppose $P$ is self-adjoint. Let $v\in V$. Then proposition 6.47 implies that $V=\rangsp{P}+\rangsp{P}^\perp$. Hence $v$ can be uniquely decomposed as
$$ v=u+w $$
for some $u\in\rangsp{P}$ and for some $w\in\rangsp{P}^\perp$. Hence $P_{\rangsp{P}}v=u$.
Because $P(v-Pv)=Pv-P^2v=0$, we get
$$ v-Pv\in\nullsp{P}=\rangsp{P^*}^\perp=\rangsp{P}^\perp $$
The first equality follows from 7.7(c) and the second equality follows because $P$ is self-adjoint. Also note that
$$ v=Pv+(v-Pv) $$
Since $Pv\in\rangsp{P}$ and $v-Pv\in\rangsp{P}^\perp$ and $v=u+w$ is uniquely decomposed with $u\in\rangsp{P}$ and $w\in\rangsp{P}^\perp$, then it must be that $u=Pv$ and $w=v-Pv$. Hence $P_{\rangsp{P}}v=u=Pv$. $\blacksquare$
(12) Suppose that $T$ is a normal operator on $V$ and that $3$ and $4$ are eigenvalues of $T$. Then there exists a vector $v\in V$ such that $\dnorm{v}=\sqrt2$ and $\dnorm{Tv}=5$.
Proof Let $u$ be a unit eigenvector (i.e. $\dnorm{u}=1$) of $T$ corresponding to eigenvalue $3$. And let $w$ be an unit eigenvector of $T$ corresponding to eigenvalue $4$. Then 7.22 implies that $u$ and $w$ are orthogonal. Hence $u,w$ is an orthonormal list. Then for any $a,b\in\wF$, define $v_{a.b}=au+bw$. Then proposition 6.25 gives
$$ \dnorm{v_{a,b}}^2=\dnorm{au+bw}^2=\norm{a}^2+\norm{b}^2 $$
and
$$ \dnorm{Tv_{a,b}}^2=\dnorm{T(au+bw)}^2=\dnorm{aTu+bTw}^2=\dnorm{a3u+b4w}^2=\norm{3a}^2+\norm{4b}^2=9\norm{a}^2+16\norm{b}^2 $$
So we need to choose $a$ and $b$ such that
$$ \norm{a}^2+\norm{b}^2=2 \quad\quad\quad\quad 9\norm{a}^2+16\norm{b}^2=25 $$
A visible solution is $a\equiv1$ and $b\equiv1$. Hence $v_{1,1}=u+w$ satisfies the requirements of the desired vector. $\blacksquare$
(13) Give an example of an operator $T\in\oper{\wC^4}$ such that $T$ is normal but not self-adjoint.
Solution Let $e_1,e_2,e_3,e_4$ be the standard basis of $\wC^4$. This basis is orthonormal. Define $T\in\oper{\wC^4}$ by
$$ Te_1\equiv-e_2 \quad\quad\quad\quad Te_2\equiv e_1 \quad\quad\quad\quad Te_3\equiv0 \quad\quad\quad\quad Te_4\equiv0 $$
In our ch.7 notes, under the section “Computation of Adjoint”, we showed that $T$ is normal but not self-adjoint. $\blacksquare$
(14) Suppose $T$ is a normal operator on $V$. Suppose also that $v,w\in V$ satisfy the equations
$$ \dnorm{v}=\dnorm{w}=2 \quad\quad\quad\quad Tv=3v \quad\quad\quad\quad Tw=4w $$
Then $\dnorm{T(v+w)}=10$.
Proof Define
$$ v_1\equiv\frac{v}{\dnorm{v}} \quad\quad\quad\quad w_1\equiv\frac{w}{\dnorm{w}} $$
Proposition 7.22 implies that $v$ and $w$ are orthogonal. Since scaling preserves orthogonality (by W.6.CS.2), then $v_1$ and $w_1$ are orthogonal. Hence $v_1,w_1$ is an orthonormal list and
$$\align{ \dnorm{T(v+w)}^2 &= \dnorm{Tv+Tw}^2 \\ &= \dnorm{3v+4w}^2 \\ &= \dnormb{3\dnorm{v}v_1+4\dnorm{w}w_1}^2 \\ &= \dnorm{6v_1+8w_1}^2 \\ &= \norm{6}^2+\norm{8}^2\tag{by 6.25} \\ &= 36+64 \\ &= 100 }$$
Hence $\dnorm{T(v+w)}=10$. $\blacksquare$
(15) Fix $u,x\in V$. Define $T\in\oper{V}$ by
$$ Tv=\innprd{v}{u}x $$
for every $v\in V$.
(a) Suppose $\wF=\wR$. Then $T$ is self-adjoint if and only if $u,x$ is linearly dependent.
(b) $T$ is normal if and only if $u,x$ is linearly dependent.
Proof Whether $\wF=\wC$ or $\wF=\wR$, note that for every $v,w\in V$, we have
$$\align{ \innprd{v}{\adjt{T}w}=\innprd{Tv}{w}=\innprdbg{\innprd{v}{u}x}{w}=\innprd{v}{u}\innprd{x}{w}=\innprdbg{v}{\cj{\innprd{x}{w}}u}=\innprdbg{v}{\innprd{w}{x}u} }$$
Hence $\adjt{T}w=\innprd{w}{x}u$ for every $w\in V$.
Proof of (a) Let $v\in V$ and note that
$$ (T-\adjt{T})v=Tv-\adjt{T}v=\innprd{v}{u}x-\innprd{v}{x}u \tag{7.A.15.1} $$
Suppose $T$ is self-adjoint. Then equation 7.A.15.1 implies that
$$ \innprd{v}{x}u=\innprd{v}{u}x $$
If $u=0$ or $x=0$, then $u,x$ is trivially linearly dependent. Suppose $u\neq0$ and $x\neq0$. Then the above equality implies that $\innprd{v}{x}\neq0$ and $\innprd{v}{u}\neq0$. Hence
$$ u=\frac{\innprd{v}{u}}{\innprd{v}{x}}x $$
Hence $u$ is a scalar multiple of $x$ and hence $u,x$ is linearly dependent.
Conversely, suppose that $u,x$ is linearly dependent. Then $x=cu$ for some $c\in\wR$ and equation 7.A.15.1 becomes
$$\align{ (T-\adjt{T})v &= \innprd{v}{u}x-\innprd{v}{x}u \\ &= \innprd{v}{u}(cu)-\innprd{v}{cu}u \\ &= \innprd{v}{u}(cu)-\innprd{v}{u}(cu) \\ &= 0 }$$
Since $v\in V$ was arbitrarily chosen, then $T-\adjt{T}=0$ and $T$ is self-adjoint. $\blacksquare$
Alternate Proof of (a) Suppose $T$ is self-adjoint. If $u=0$ or $x=0$, then $u.x$ is trivially linearly dependent. Suppose $u\neq0$ and $x\neq0$. Note that proposition 6.47 gives $V=\dirsum{\span{x}}{\span{x}^\perp}$. Hence $u$ can be uniquely decomposed as
$$ u=cx+w $$
where $c\in\wR$ and $w\in\span{x}^\perp$. Since $w$ is orthogonal to $x$, then
$$ 0=\innprd{w}{x}u=\adjt{T}w=Tw=\innprd{w}{u}x $$
Hence $w$ is orthogonal to $u$ as well. Hence
$$ 0=\innprd{u}{w}=\innprd{cx+w}{w}=\innprd{cx}{w}+\innprd{w}{w}=0+\innprd{w}{w} $$
Then the definiteness of an inner product implies that $w=0$. Hence $u=cx$ and $u,x$ is linearly dependent. $\blacksquare$
Proof of (b) Let $v\in V$ and note that
$$\align{ (\tta-\tat)v &= \tta v-\tat v \\ &= T\prn{\innprd{v}{x}u}-\adjt{T}\prn{\innprd{v}{u}x} \\ &= \innprd{v}{x}Tu-\innprd{v}{u}\adjt{T}x \\ &= \innprd{v}{x}\innprd{u}{u}x-\innprd{v}{u}\innprd{x}{x}u \tag{7.A.15.2} }$$
Suppose $T$ is normal. Then equation 7.A.15.2 implies that
$$ \innprd{v}{u}\innprd{x}{x}u=\innprd{v}{x}\innprd{u}{u}x $$
If $u=0$ or $x=0$, then $u,x$ is trivially linearly dependent. Suppose $u\neq0$ and $x\neq0$. Then the above equality implies that $\innprd{v}{u}\innprd{x}{x}\neq0$ and $\innprd{v}{x}\innprd{u}{u}\neq0$. Hence
$$ u=\frac{\innprd{v}{x}\innprd{u}{u}}{\innprd{v}{u}\innprd{x}{x}}x $$
Hence $u$ is a scalar multiple of $x$ and hence $u,x$ is linearly dependent.
Conversely, suppose that $u,x$ is linearly dependent. Then $x=cu$ for some $c\in\wF$ and equation 7.A.15.2 becomes
$$\align{ (\tta-\tat)v &= \innprd{v}{x}\innprd{u}{u}x-\innprd{v}{u}\innprd{x}{x}u \\ &= \innprd{v}{cu}\innprd{u}{u}(cu)-\innprd{v}{u}\innprd{cu}{cu}u \\ &= \cj{c}c\innprd{v}{u}\innprd{u}{u}u-\cj{c}c\innprd{v}{u}\innprd{u}{u}u \\ &= 0 }$$
Since $v$ was arbitrarily chosen, then $\tta-\tat=0$ and $T$ is normal. $\blacksquare$
Alternate Proof of (b) Suppose $T$ is normal. If $u=0$ or $x=0$, then $u.x$ is trivially linearly dependent. Suppose $u\neq0$ and $x\neq0$. Note that proposition 6.47 gives $V=\dirsum{\span{x}}{\span{x}^\perp}$. Hence $u$ can be uniquely decomposed as
$$ u=cx+w $$
where $c\in\wF$ and $w\in\span{x}^\perp$. Since $w$ is orthogonal to $x$, then
$$ 0=\innprd{w}{x}\innprd{u}{u}x=\tta w=\tat w=\innprd{w}{u}\innprd{x}{x}u $$
Hence $w$ is orthogonal to $u$ as well. Hence
$$ 0=\innprd{u}{w}=\innprd{cx+w}{w}=\innprd{cx}{w}+\innprd{w}{w}=0+\innprd{w}{w} $$
Then the definiteness of an inner product implies that $w=0$. Hence $u=cx$ and $u,x$ is linearly dependent. $\blacksquare$
(16) Suppose $T\in\oper{V}$ is normal. Then $\rangsp{T}=\rangsp{\adjt{T}}$.
Proof Proposition 7.7(d) implies that $\rangsp{T}=\nullsp{\adjt{T}}^\perp$. Proposition W.7.G.10 implies that $\nullsp{\adjt{T}}=\nullsp{T}$. And proposition 7.7(b) implies that $\nullsp{T}^\perp=\rangsp{\adjt{T}}$. Hence
$$ \rangsp{T}=\nullsp{\adjt{T}}^\perp=\nullsp{T}^\perp=\rangsp{\adjt{T}} $$
$\blacksquare$
(17) Suppose $T\in\oper{V}$ is normal. Then
$$ \nullsp{T^k}=\nullsp{T} \quad\quad\text{and}\quad\quad \rangsp{T^k}=\rangsp{T} $$
for every positive integer $k$.
Proof Let $v\in\nullsp{T}$. Then
$$ T^kv=T^{k-1}(Tv)=T^{k-1}0=0 $$
The last equation follows because the product (i.e. composition) of linear maps is linear (by W.3.21) and because linear maps take $0$ to $0$ (by 3.11). Hence $v\in\nullsp{T^k}$ and $\nullsp{T}\subset\nullsp{T^k}$.
To prove inclusion in the other direction, suppose now that $v\in\nullsp{T^k}$. Then
$$\align{ \innprd{\tat^{k-1}v}{\tat^{k-1}v} &= \innprd{\tta T^{k-1}v}{T^{k-1}v}\tag{by definition of $\adjt{T}$} \\ &= \innprd{\tat T^{k-1}v}{T^{k-1}v}\tag{because $T$ is normal} \\ &= \innprd{\tat^{k}v}{T^{k-1}v} \\ &= \innprd{\adjt{T}0}{T^{k-1}v}\tag{because $v\in\nullsp{T^k}$} \\ &= \innprd{0}{T^{k-1}v}\tag{because $\adjt{T}$ is linear} \\ &= 0\tag{by 6.7.b} }$$
Then the definiteness of an inner product implies that $0=\tat^{k-1}v$. Hence
$$\align{ 0 &= \innprd{\tat^{k-1}v}{T^{k-2}v} \\ &= \innprd{T^{k-1}v}{TT^{k-2}v}\tag{by W.7.G.11} \\ &= \innprd{T^{k-1}v}{T^{k-1}v} }$$
Hence $0=T^{k-1}v$ and $v\in\nullsp{T^{k-1}}$. The same argument, with $k$ replaced by $k-1$, shows that $v\in\nullsp{T^{k-2}}$. We repeat this process until we reach the conclusion that $v\in\nullsp{T}$. Hence $\nullsp{T^k}\subset\nullsp{T}$ and $\nullsp{T^k}=\nullsp{T}$.
Now we will show that $\rangsp{T^k}=\rangsp{T}$. Let $v\in\rangsp{T^k}$. Then there exists $u\in V$ such that $v=T^ku=TT^{k-1}u$. Hence $v\in\rangsp{T}$. Hence $\rangsp{T^k}\subset\rangsp{T}$. Also note that
$$\align{ \dim{\rangsp{T^k}} &= \dim{V}-\dim{\nullsp{T^k}}\tag{by FToLA} \\ &= \dim{V}-\dim{\nullsp{T}}\tag{by $\nullsp{T^k}=\nullsp{T}$} \\ &= \dim{\rangsp{T}}\tag{by FToLA} }$$
Since $\rangsp{T^k}\subset\rangsp{T}$ and $\dim{\rangsp{T^k}}=\dim{\rangsp{T}}$, then proposition W.2.12 gives $\rangsp{T^k}=\rangsp{T}$. $\blacksquare$
(18) Prove or give a counterexample: If $T\in\oper{V}$ and there exists an orthonormal basis $e_1,\dots,e_n$ of $V$ such that $\dnorm{Te_j}=\dnorm{\adjt{T}e_j}$ for each $j$, then $T$ is normal.
Counterexample Define $T\in\oper{V}$ by
$$ Te_1=e_1+e_2 \quad\quad\quad\quad Te_2=-e_1-e_2 $$
Then
$$\align{ &\dnorm{Te_1}^2=\dnorm{e_1+e_2}^2=\dnorm{e_1}^2+\dnorm{e_2}^2=2 \\ &\dnorm{Te_2}^2=\dnorm{-e_1-e_2}^2=\dnorm{-e_1}^2+\dnorm{-e_2}^2=2 }$$
and
$$\align{ \adjt{T}e_1 &= \innprd{e_1}{Te_1}e_1+\innprd{e_1}{Te_2}e_2 \\ &= \innprd{e_1}{e_1+e_2}e_1+\innprd{e_1}{-e_1-e_2}e_2 \\ &= \prn{\innprd{e_1}{e_1}+\innprd{e_1}{e_2}}e_1+\prn{\innprd{e_1}{-e_1}+\innprd{e_1}{-e_2}}e_2 \\ &= \prn{\innprd{e_1}{e_1}+\innprd{e_1}{e_2}}e_1+\prn{-\innprd{e_1}{e_1}-\innprd{e_1}{e_2}}e_2 \\ &= (1+0)e_1+(-1-0)e_2 \\ &= e_1-e_2 }$$
and
$$\align{ \adjt{T}e_2 &= \innprd{e_2}{Te_1}e_1+\innprd{e_2}{Te_2}e_2 \\ &= \innprd{e_2}{e_1+e_2}e_1+\innprd{e_2}{-e_1-e_2}e_2 \\ &= \prn{\innprd{e_2}{e_1}+\innprd{e_2}{e_2}}e_1+\prn{\innprd{e_2}{-e_1}+\innprd{e_2}{-e_2}}e_2 \\ &= \prn{\innprd{e_2}{e_1}+\innprd{e_2}{e_2}}e_1+\prn{-\innprd{e_2}{e_1}-\innprd{e_2}{e_2}}e_2 \\ &= (0+1)e_1+(-0-1)e_2 \\ &= e_1-e_2 }$$
and
$$ \dnorm{\adjt{T}e_2}^2=\dnorm{\adjt{T}e_1}^2=\dnorm{e_1-e_2}^2=\dnorm{e_1}^2+\dnorm{-e_2}^2=2 $$
But
$$\align{ \tat e_1 &= \adjt{T}(e_1+e_2) \\ &= \adjt{T}e_1+\adjt{T}e_2 \\ &= e_1-e_2+e_1-e_2 \\ &= 2(e_1-e_2) }$$
and
$$\align{ \tta e_1 &= T(e_1-e_2) \\ &= Te_1-Te_2 \\ &= e_1+e_2-(-e_1-e_2) \\ &= e_1+e_2+e_1+e_2 \\ &= 2(e_1+e_2) \\ &\neq \tat e_1 }$$
Hence $\tat e_1\neq \tta e_1$ and $T$ is not normal. $\blacksquare$
(19) Suppose $T\in\oper{\wC^3}$ is normal and $T(1,1,1)=(2,2,2)$. Suppose $(z_1,z_2,z_3)\in\nullsp{T}$. Then $z_1+z_2+z_3=0$.
Proof Since $T$ is normal, then $(z_1,z_2,z_3)\in\nullsp{T}=\nullsp{\adjt{T}}$ (by W.7.G.10) and
$$\align{ 0 &= \innprd{\adjt{T}(z_1,z_2,z_3)}{(1,1,1)} \\ &= \innprd{(z_1,z_2,z_3)}{T(1,1,1)} \\ &= \innprd{(z_1,z_2,z_3)}{(2,2,2)} \\ &= 2z_1+2z_2+3z_3 }$$
Dividing by $2$ gives the desired result. $\blacksquare$
(21) Fix a positive integer $n$. In the inner product space of continuous real-valued functions on $[-\pi,\pi]$ with inner product
$$ \innprd{f}{g}\equiv\int_{-\pi}^{\pi}f(x)g(x)dx $$
let
$$ V\equiv\span{1,\cos{x},\cos{2x},\dots,\cos{nx},\sin{x},\sin{2x},\dots,\sin{nx}} $$
(a) Define $D\in\oper{V}$ by $Df=f’$. Then $D^*=-D$ and $D$ is normal but not self-adjoint.
(b) Define $T\in\oper{V}$ by $Tf=f’’$. Then $T$ is self-adjoint.
Note Note that for any $f\in V$, we have
$$\align{ Df &= D\Prn{a_01(x)+\sum_{k=1}^na_k\cos{(kx)}+\sum_{k=1}^nb_k\sin{(kx)}} \\ &= a_0D\prn{1(x)}+\sum_{k=1}^na_kD\prn{\cos{(kx)}}+\sum_{k=1}^nb_kD\prn{\sin{(kx)}} \\ &= 0+\sum_{k=1}^n(-ka_k)\sin{(kx)}+\sum_{k=1}^nkb_k\cos{(kx)} \\ &\in V }$$
Hence $D$ is well-defined as an operator. Similarly, $T$ is well-defined as an operator.
Proof of (a) Note that for any $k\in\mathbb{N}$, we have
$$ 1(\pi)=1=1(-\pi) \quad\quad\quad \sin{(k\pi)}=0=\sin{(-k\pi)} \quad\quad\quad \cos{(k\pi)}=\cos{(-k\pi)} $$
Hence, for any $f\in V$, we have
$$\align{ f(\pi) &= a_01(\pi)+\sum_{k=1}^na_k\cos{(k\pi)}+\sum_{k=1}^nb_k\sin{(k\pi)} \\ &= a_01(-\pi)+\sum_{k=1}^na_k\cos{(-k\pi)}+\sum_{k=1}^nb_k\sin{(-k\pi)} \\ &= f(-\pi) }$$
Now let $f,g\in V$. Then
$$\align{ \innprd{Df}{g} &= \int_{-\pi}^{\pi}f'(x)g(x)dx \\ &= f(x)g(x)\eval{-\pi}{\pi}-\int_{-\pi}^{\pi}f(x)g'(x)dx\tag{integration by parts} \\ &= f(\pi)g(\pi)-f(-\pi)g(-\pi)-\innprd{f}{Dg} \\ &= f(\pi)g(\pi)-f(\pi)g(\pi)+\innprd{f}{-Dg} \\ &= \innprd{f}{-Dg} \\ }$$
Since $f$ and $g$ were chosen arbitrarily, then this is true for every $f,g\in V$. Then, by definition 7.2, $D^*=-D$. Since $D\neq0$, then $D^*=-D\neq D$ and $D$ is not self-adjoint. However, $D$ is normal since
$$ D^*D=-DD=D(-D)=DD^* $$
The second equality follows from W.3.22. $\blacksquare$
Proof of (b) Note that, for any $f\in V$, we have
$$ Tf=f''=(f')'=(Df)'=D(Df)=D^2f $$
Hence $T=D^2$ and
$$\align{ \adjt{T} &= (D^2)^* \\ &= (DD)^* \\ &= D^*D^*\tag{by 7.6.e} \\ &= (-D)(-D) \\ &= -(-D)D\tag{by W.3.22} \\ &= (--D)D\tag{by 3.6.scalar multiplication} \\ &= DD \\ &= T }$$
Hence $T$ is self-adjoint. $\blacksquare$
Exercises 7.B
(1) True or false (and give a proof of your answer): There exists $T\in\oper{\wR^3}$ such that $T$ is not self-adjoint (with respect the usual inner product) and such that there is a basis of $\wR^3$ consisting of eigenvectors of $T$.
Proof True. Let $e_1,e_2,e_3$ be the standard basis of $\wR^3$ and define
$$ Te_1=e_1 \quad\quad\quad Te_2=e_1+2e_2 \quad\quad\quad Te_3=e_3 $$
Then
$$ \innprd{e_1}{Te_2}=\innprd{e_1}{e_2+2e_2}=\innprd{e_1}{e_1}+2\innprd{e_1}{e_2}=1 $$
and
$$ \innprd{Te_1}{e_2}=\innprd{e_1}{e_2}=0\neq1=\innprd{e_1}{Te_2} $$
Hence $T\neq \adjt{T}$ and $T$ is not self-adjoint.
By proposition W.2.18, the list $e_1,e_1+e_2,e_3$ is a basis of $V$. Note that
$$\align{ &Te_1=e_1 \\ &T(e_1+e_2)=Te_1+Te_2=e_1+e_1+2e_2=2(e_1+e_2) \\ &Te_3=e_3 }$$
Hence each basis vector is an eigenvector of $T$. $\blacksquare$
(2) Supose that $T$ is a self-adjoint operator on a finite-dimensional inner product space and that $2$ and $3$ are the only eigenvalues of $T$. Then $T^2-5T+6I=0$.
Proof Before proving this, we note that $b^2=(-5)^2=25\gt24=4\cdot6=4c$. So even though $T$ is self-adjoint, the result – that $T^2-5T+6I$ is not injective (by 3.16) and hence not invertible (by 3.69) – does not conflict with 7.26. This result further strengthens the analogy between the real numbers and self-adjoint operators because $2$ and $3$ are the only roots of the quadractic polynomial $x^2-5x+6=(x-2)(x-3)$.
If $\wF=\wR$, then the self-adjointedness of $T$ implies that $V$ has an orthonormal basis consisting of eigenvectors of $T$ (by the Real Spectral Theorem). Note that self-jointedness implies normality. Hence, if $\wF=\wC$, then again $V$ has an orthonormal basis consisting of eigenvectors of $T$ (by the Complex Spectral Theorem).
Let $e_1,\dots,e_n$ denote an orthonormal basis of $V$ consisting of eigenvectors of $T$. Since $2$ and $3$ are the only eigenvalues of $T$, then, for all $k=1,\dots,n$, it must be that
$$\align{ (T-2I)e_k=0 \quad\quad\text{or}\quad\quad (T-3I)e_k=0 }$$
Proposition W.3.24 or proposition 5.19 give
$$\align{ (T-3I)(T-2I)=T^2-5T+6I=(T-2I)(T-3I) }$$
Hence, for each $k=1,\dots,n$, we have
$$\align{ &(T^2-5T+6I)e_k=(T-3I)(T-2I)e_k=0 }$$
or
$$\align{ &(T^2-5T+6I)e_k=(T-2I)(T-3I)e_k=0 }$$
$\blacksquare$
(3) Give an example of an operator $T\in\oper{\wC^3}$ such that $2$ and $3$ are the only eigenvalues of $T$ and $T^2-5T+6I\neq0$.
Solution Note that we cannot simply find a self-adjoint $T\in\oper{\wC^3}$ and then use 7.26. The reason we cannot do that is
$$ b^2=(-5)^2=25\gt24=4\cdot6=4c $$
Let $e_1,e_2,e_3$ be the standard basis of $\wC^3$. Proposition 5.32 tells us that the diagonal entries of an upper-triangular matrix of $T$ with respect to any basis of $\wC^3$ are precisely the eigenvalues of $T$. So we want to find some $T\in\oper{\wC^3}$ such that
$$ \mtrxof{T,(e_1,e_2,e_3)}=\bmtrx{2&a&b\\0&2&c\\0&0&3} \tag{7.B.3.1} $$
for some complex numbers $a,b,c$. This gives
$$ Te_1=2e_1 \quad\quad\quad Te_2=ae_1+2e_2 \quad\quad\quad Te_3=be_1+ce_2+3e_3 \tag{7.B.3.2} $$
But $T$ must also satisfy $T^2-5T+6I\neq0$. By proposition W.3.12, this is true if $(T^2-5T+6I)e_2\neq0$. Hence
$$\align{ 0 &\neq (T^2-5T+6I)e_2 \\ &= (T-3I)(T-2I)e_2 \\ &= (T-3I)(Te_2-2Ie_2) \\ &= (T-3I)(ae_1+2e_2-2e_2) \\ &= (T-3I)(ae_1) \\ &= a(T-3I)e_1 \\ &= a(Te_1-3Ie_1) \\ &= a(2e_1-3e_1) \\ &= -ae_1 \\ }$$
Any $a\neq0$ will satisfy this equation. Hence, for 7.B.3.2, we choose any $a\neq0$ and any $b,c\in\wC$ and we get that $T^2-5T+6I\neq0$. And 7.B.3.1 implies that the eigenvalues of such a $T$ are precisely $2$ and $3$. $\blacksquare$
(4) Suppose $\wF=\wC$ and $T\in\oper{V}$. Then $T$ is normal if and only if all pairs of eigenvectors corresponding to distinct eigenvalues of $T$ are orthogonal and
$$ V=\eignsp{\lambda_1}{T}\oplus\dotsb\oplus\eignsp{\lambda_m}{T} $$
where $\lambda_1,\dots,\lambda_m$ denote the distinct eigenvalues of $T$.
Proof First suppose that $T$ is normal. Then 7.22 implies that all pairs of eigenvectors of $T$ corresponding to distinct eigenvalues are orthogonal. Also note that the Complex Spectral Theorem implies that $T$ is diagonalizable. Then 5.41(d) implies that
$$ V=\eignsp{\lambda_1}{T}\oplus\dotsb\oplus\eignsp{\lambda_m}{T} $$
where $\lambda_1,\dots,\lambda_m$ denote the distinct eigenvalues of $T$.
Conversely, suppose that all pairs of eigenvectors corresponding to distinct eigenvalues of $T$ are orthogonal and
$$ V=\eignsp{\lambda_1}{T}\oplus\dotsb\oplus\eignsp{\lambda_m}{T} $$
where $\lambda_1,\dots,\lambda_m$ denote the distinct eigenvalues of $T$. The latter supposition and exercise 2.C.16 imply that
$$ \dim{V}=\dim{\prn{\eignsp{\lambda_1}{T}\oplus\dotsb\oplus\eignsp{\lambda_m}{T}}}=\dim{\eignsp{\lambda_1}{T}}+\dotsb+\dim{\eignsp{\lambda_m}{T}} \tag{7.B.4.1} $$
Assuming $V$ is finite-dimensional (chapter assumption), then each $\eignsp{\lambda_j}{T}$ is finite-dimensional (by 2.38). Choose an orthonormal basis of each $\eignsp{\lambda_j}{T}$ (by 6.34). Now concatenate these bases to form a list $e_1,\dots,e_n$ of eigenvectors of $T$. Then 7.B.4.1 gives $n=\dim{V}$. And our first supposition implies that $e_1,\dots,e_n$ is orthonormal. Hence $e_1,\dots,e_n$ is an orthonormal basis of $V$ (by 6.26) consisting of eigenvectors of $T$. Then the Complex Spectral Theorem implies $T$ is normal. $\blacksquare$
(5) Suppose $\wF=\wR$ and $T\in\oper{V}$. Then $T$ is self-adjoint if and only if all pairs of eigenvectors corresponding to distinct eigenvalues of $T$ are orthogonal and
$$ V=\eignsp{\lambda_1}{T}\oplus\dotsb\oplus\eignsp{\lambda_m}{T} $$
where $\lambda_1,\dots,\lambda_m$ denote the distinct eigenvalues of $T$.
Proof First suppose that $T$ is self-adjoint. Then $T=\adjt{T}$ and $\tta=\tat$ and $T$ is normal. Then 7.22 implies that all pairs of eigenvectors of $T$ corresponding to distinct eigenvalues are orthogonal. Also note that the Real Spectral Theorem implies that $T$ is diagonalizable. Then 5.41(d) implies that
$$ V=\eignsp{\lambda_1}{T}\oplus\dotsb\oplus\eignsp{\lambda_m}{T} $$
where $\lambda_1,\dots,\lambda_m$ denote the distinct eigenvalues of $T$.
Conversely, suppose that all pairs of eigenvectors corresponding to distinct eigenvalues of $T$ are orthogonal and
$$ V=\eignsp{\lambda_1}{T}\oplus\dotsb\oplus\eignsp{\lambda_m}{T} $$
where $\lambda_1,\dots,\lambda_m$ denote the distinct eigenvalues of $T$. The latter supposition and exercise 2.C.16 imply that
$$ \dim{V}=\dim{\prn{\eignsp{\lambda_1}{T}\oplus\dotsb\oplus\eignsp{\lambda_m}{T}}}=\dim{\eignsp{\lambda_1}{T}}+\dotsb+\dim{\eignsp{\lambda_m}{T}} \tag{7.B.4.2} $$
Assuming $V$ is finite-dimensional (chapter assumption), then each $\eignsp{\lambda_j}{T}$ is finite-dimensional (by 2.38). Choose an orthonormal basis of each $\eignsp{\lambda_j}{T}$ (by 6.34). Now concatenate these bases to form a list $e_1,\dots,e_n$ of eigenvectors of $T$. Then 7.B.4.2 gives $n=\dim{V}$. And our first supposition implies that $e_1,\dots,e_n$ is orthonormal. Hence $e_1,\dots,e_n$ is an orthonormal basis of $V$ (by 6.26) consisting of eigenvectors of $T$. Then the Real Spectral Theorem implies $T$ is self-adjoint. $\blacksquare$
(6) A normal operator on a complex inner product space is self-adjoint if and only if all its eigenvalues are real.
Note: This exercise strengthens the analogy (for normal operators) between self-adjoint operators and real numbers.
Proof Let $V$ be a complex inner product space and let $T\in\oper{V}$ be normal.
If $T$ is self-adjoint, then 7.13 implies that all of its eigenvalues are real.
Conversely, suppose that all of $T$’s eigenvalues are real. Since $T$ is normal, then the Complex Spectral Theorem implies that $V$ has an orthonormal basis $e_1,\dots,e_n$ consisting of eigenvectors of $T$. Hence there exist real numbers $\lambda_1,\dots,\lambda_n$ such that
$$ Te_k=\lambda_ke_k\quad\text{for }k=1,\dots,n $$
Hence $\mtrxof{T,(e_1,\dots,e_n)}$ is diagonal and the diagonal entries of $\mtrxof{T,(e_1,\dots,e_n)}$ are precisely $\lambda_1,\dots,\lambda_n$. Recall that a diagonal matrix is symmetric. Hence $\mtrxof{T,(e_1,\dots,e_n)}=\mtrxof{T,(e_1,\dots,e_n)}^t$. And since all of the entries in $\mtrxof{T,(e_1,\dots,e_n)}^t$ are real (zero or a real eigenvalue), then conjugating effects no change. Hence
$$ \mtrxof{T,(e_1,\dots,e_n)}=\cj{\mtrxof{T,(e_1,\dots,e_n)}^t}=\mtrxof{\adjt{T},(e_1,\dots,e_n)} $$
The last equality follows from 7.10. Note that $\mtrxofsb\in\linmapb{\oper{V}}{\wF^{n,n}}$ is an isomorphism (by 3.60). Hence $\mtrxofsb$ is invertible (by 3.58). Hence $\mtrxofsb$ is injective (by 3.56). Hence $T=\adjt{T}$ (by 3.15) and $T$ is self-adjoint. $\blacksquare$
(7) Suppose $V$ is a complex inner product space and $T\in\oper{V}$ is a normal operator such that $T^9=T^8$. Then $T$ is self-adjoint and $T^2=T$.
Proof Since $T$ is normal, then the Complex Spectral Theorem implies that $V$ has an orthonormal basis $e_1,\dots,e_n$ consisting of eigenvectors of $T$. Let $\lambda_1,\dots,\lambda_n$ be the corresponding eigenvalues. Then
$$ Te_k=\lambda_ke_k\quad\text{for }k=1,\dots,n $$
Note that
$$ T^2e_k = T(Te_k) = T(\lambda_ke_k) = \lambda_k Te_k = \lambda_k\lambda_ke_k=\lambda_k^2e_k $$
and
$$ T^3e_k = T(T^2e_k) = T(\lambda_k^2e_k) = \lambda_k^2 Te_k = \lambda_k^2\lambda_ke_k=\lambda_k^3e_k $$
and the induction assumption $T^{n-1}e_k=\lambda^{n-1}e_k$ gives us
$$ T^ne_k = T(T^{n-1}e_k) = T(\lambda_k^{n-1}e_k) = \lambda_k^{n-1} Te_k = \lambda_k^{n-1}\lambda_ke_k=\lambda_k^ne_k $$
Hence
$$ T^8e_k=\lambda_k^8e_k \quad\quad\text{and}\quad\quad T^9e_k=\lambda_k^9e_k $$
Then the hypothesis $T^8=T^9$ implies that
$$ \lambda_k^8e_k=T^8e_k=T^9e_k=\lambda_k^9e_k $$
Hence
$$ \lambda_k^8=\lambda_k^9=\lambda_k^8\lambda_k $$
Certainly $\lambda_k=0$ is a solution to this. Suppose $\lambda_k\neq0$. Dividing by $\lambda_k^8$, we get
$$ 1=\lambda_k $$
Hence $\lambda_k$ equals $0$ or $1$. In particular, all of the eigenvalues of $T$ are real. Then the previous exercise implies that $T$ is self-adjoint. Also note that
$$ T^2e_k=\lambda_k^2e_k=\lambda_ke_k=Te_k $$
The second equality follows because $\lambda_k$ equals $0$ or $1$. Hence $0=(T^2-T)e_k$ for the basis $e_1,\dots,e_n$. Hence $0=T^2-T$ (by W.3.12). $\blacksquare$
(8) Give an example of an operator $T$ on a complex vector space such that $T^9=T^8$ and $T^2\neq T$.
Example Let $V$ be any $2$-dimensional complex vector space and let $e_1,e_2$ be a basis of $V$. Define $T\in\oper{V}$ by
$$ Te_1=0 \quad\quad\quad\quad Te_2=e_1 $$
Then
$$\align{ &T^3e_1=T^2(Te_1)=T^20=0 \\ &T^3e_2=T^2(Te_2)=T^2(e_1)=T(Te_1)=T0=0 }$$
Hence $T^3=0$ (by W.3.12). Hence $T^n=0$ for $n\geq3$. In particular $T^8=0=T^9$. But
$$ T^2e_1=T(Te_1)=T0=0\neq e_1=Te_2 $$
Hence $T^2\neq T$. $\blacksquare$
(9) Suppose $V$ is a complex inner product space. Then every normal operator on $V$ has a square root.
Note: An operator $S\in\oper{V}$ is called a square root of $T\in\oper{V}$ if $S^2=T$.
Proof Let $T\in\oper{V}$ be normal. Then the Complex Spectral Theorem implies that $V$ has an orthonormal basis $e_1,\dots,e_n$ consisting of eigenvectors of $T$. Hence there exist complex numbers $\lambda_1,\dots,\lambda_n$ such that
$$ Te_k=\lambda_ke_k\quad\text{for }k=1,\dots,n $$
Define $S\in\oper{V}$ by
$$ Se_k=\lambda_k^{1/2}e_k\quad\text{for }k=1,\dots,n $$
where $\lambda_k^{1/2}$ denotes a complex square root of $\lambda_k$. Note that every nonzero complex number has two square roots (Baby Rudin, exercise 1.10). We can choose either one. Then
$$ S^2e_k=S(Se_k)=S(\lambda_k^{1/2}e_k)=\lambda_k^{1/2}S(e_k)=\lambda_k^{1/2}\lambda_k^{1/2}e_k=\lambda_k^{1/2+1/2}e_k=\lambda_ke_k=Te_k $$
Hence $S^2=T$ (by W.3.25). $\blacksquare$
(10) Give an example of a real inner product space $V$ and $T\in\oper{V}$ and real numbers $b,c$ with $b^2\lt4c$ such that $T^2+bT+cI$ is not invertible.
Note: This exercise shows that the hypothesis that $T$ is self-adjoint is needed in 7.26, even for real vector spaces.
Example Let’s take a simple example. Let $V=\wR^2$ and let’s consider the operator $T^2+I$. First note that $b=0$ and $c=1$ so that
$$ b^2=0^2=0\lt4=4\cdot1=4c $$
To find $T$ such that $T^2+I$ invertible, we must have
$$\align{ T^2+I\text{ is invertible} &\iff T^2+I\text{ is injective}\tag{by 3.69} \\ &\iff \nullsp{T^2+I}=\set{0}\tag{by 3.16} \\ &\iff \nullsp{T^2-(-1)I}=\set{0} \\ &\iff \eignsp{-1}{T^2}=\set{0}\tag{by 5.36} \\ }$$
So to find $T$ such that $T^2+I$ is not invertible, we must have $\eignsp{-1}{T^2}\neq\set{0}$. So let’s find an operator $T^2\in\oper{\wR^2}$ that has $-1$ as an eigenvalue. That is, we want to find $T^2$ such that
$$ T^2(x,y)=-1\cdot(x,y)=(-x,-y) $$
Geometrically, we can see that $T^2$ must reflect a point $(x,y)$ through the origin. Hence we can use proposition W.5.6 to find $T$. That is, we can rewrite the above equation as
$$ T^2(x,y)=-1\cdot(x,y)=-1\cdot I(x,y)=-I(x,y) $$
so that $T^2=-I$. From the formulas in W.5.6, we define
$$ T(e_1)\equiv\Big(\cos{\frac{\pi}2}\Big)e_1-\Big(\sin{\frac{\pi}2}\Big)e_2=-e_2 $$
and
$$ T(e_2)\equiv\Big(\sin{\frac{\pi}2}\Big)e_1+\Big(\cos{\frac{\pi}2}\Big)e_2=e_1 $$
where $e_1,e_2$ is the standard basis of $\wR^2$. $\blacksquare$
(11) Suppose $V$ is a real or complex inner product space. Then every self-adjoint operator on $V$ has a cube root.
Note: An operator $S\in\oper{V}$ is called a cube root of $T\in\oper{V}$ if $S^3=T$.
Proof Suppose $T\in\oper{V}$ is self-adjoint.
If $V$ is a real inner product space, then the Real Spectral Theorem implies that $V$ has an orthonormal basis $e_1,\dots,e_n$ consisting of eigenvectors of $T$.
Note that $T=\adjt{T}$ hence $\tta=\tat$ and $T$ is normal.
Hence, if $V$ is a complex inner product space, then the Complex Spectral Theorem implies that $V$ has an orthonormal basis $e_1,\dots,e_n$ consisting of eigenvectors of $T$.
Hence there exist scalars $\lambda_1,\dots,\lambda_n\in\wF$ such that
$$ Te_k=\lambda_ke_k\quad\text{for }k=1,\dots,n $$
Then 7.13 implies that the scalars $\lambda_1,\dots,\lambda_n$ are real (whether $V$ is a real or complex space). Define $S\in\oper{V}$ by
$$ Se_k=\lambda_k^{1/3}e_k\quad\text{for }k=1,\dots,n $$
Then
$$ S^3e_k=S^2(Se_k)=S^2(\lambda_k^{1/3}e_k)=\lambda_k^{1/3}S^2(e_k)=\lambda_k^{1/3}S(Se_k)=\lambda_k^{1/3}S(\lambda_k^{1/3}e_k) $$
$$ =\lambda_k^{1/3}\lambda_k^{1/3}Se_k=\lambda_k^{1/3+1/3}\lambda_k^{1/3}e_k=\lambda_ke_k=Te_k $$
Hence $S^3=T$ (by W.3.25). $\blacksquare$
(12) Suppose $T\in\oper{V}$ is self-adjoint, $\lambda\in\wF$, and $\epsilon\gt0$. Suppose there exists $v\in V$ such that $\dnorm{v}=1$ and
$$ \dnorm{Tv-\lambda v}\lt\epsilon $$
Then $T$ has an eigenvalue $\lambda’$ such that $\norm{\lambda-\lambda’}\lt\epsilon$.
Proof Since $T$ is self-adjoint, then it’s normal. Hence we can invoke the appropriate Spectral Theorem (Complex if $V$ is complex and Real if $V$ is real) to imply that $V$ has an orthonormal basis $e_1,\dots,e_n$ consisting of eigenvectors of $T$. Let $\lambda_1,\dots,\lambda_n$ be the corresponding eigenvalues so that
$$ Te_k=\lambda_ke_k\quad\text{for }k=1,\dots,n $$
Since $e_1,\dots,e_n$ is orthonormal basis of $V$ and $v\in V$, then proposition 6.30 gives
$$ v=\sum_{k=1}^n\innprd{v}{e_k}e_k $$
Hence
$$ Tv=T\Prn{\sum_{k=1}^n\innprd{v}{e_k}e_k}=\sum_{k=1}^n\innprd{v}{e_k}Te_k=\sum_{k=1}^n\innprd{v}{e_k}\lambda_ke_k $$
Hence
$$\align{ \epsilon^2 &\gt \dnorm{Tv-\lambda v}^2 \\ &= \dnormBgg{\sum_{k=1}^n\innprd{v}{e_k}\lambda_ke_k-\lambda\sum_{k=1}^n\innprd{v}{e_k}e_k}^2 \\ &= \dnormBgg{\sum_{k=1}^n\lambda_k\innprd{v}{e_k}e_k-\sum_{k=1}^n\lambda\innprd{v}{e_k}e_k}^2 \\ &= \dnormBgg{\sum_{k=1}^n\prn{\lambda_k\innprd{v}{e_k}e_k-\lambda\innprd{v}{e_k}e_k}}^2 \\ &= \dnormBgg{\sum_{k=1}^n(\lambda_k-\lambda)\innprd{v}{e_k}e_k}^2 \\ &= \sum_{k=1}^n\normb{(\lambda_k-\lambda)\innprd{v}{e_k}}^2\tag{by 6.25} \\ &= \sum_{k=1}^n\normw{\lambda_k-\lambda}^2\normb{\innprd{v}{e_k}}^2 \\ &\geq \sum_{k=1}^n\Prn{\min\setb{\normw{\lambda_1-\lambda}^2,\dots,\normw{\lambda_n-\lambda}^2}}\normb{\innprd{v}{e_k}}^2 \\ &= \min\setb{\normw{\lambda_1-\lambda}^2,\dots,\normw{\lambda_n-\lambda}^2}\sum_{k=1}^n\normb{\innprd{v}{e_k}}^2 \\ &= \min\setb{\normw{\lambda_1-\lambda}^2,\dots,\normw{\lambda_n-\lambda}^2}\dnorm{v}^2\tag{by 6.30} \\ &= \min\setb{\normw{\lambda_1-\lambda}^2,\dots,\normw{\lambda_n-\lambda}^2} \\ }$$
Let $j\equiv\text{argmin}\setb{\normw{\lambda_1-\lambda}^2,\dots,\normw{\lambda_n-\lambda}^2}$. Then
$$ \normw{\lambda_j-\lambda}^2\leq\dnorm{Tv-\lambda v}^2\lt\epsilon^2 $$
and
$$ \normw{\lambda_j-\lambda}\lt\epsilon $$
$\blacksquare$
(13) Give an alternative proof of the Complex Spectral Theorem that avoids Schur’s Theorem and instead follows the pattern of the proof of the Real Spectral Theorem.
Proof First suppose (c) holds, so $T$ has a diagonal matrix with respect to some orthonormal basis of $V$. The matrix of $\adjt{T}$ (with respect to the same basis) is obtained by taking the conjugate transpose of the matrix of $T$; hence $\adjt{T}$ also has a diagonal matrix. Any two diagonal matrices commute; thus $T$ commutes with $\adjt{T}$ which means that $T$ is normal. In other words, (a) holds.
Next we will show that (a) implies (b) by induction on $\dim{V}$. To get started, suppose $\dim{V}=1$. Then $V\neq\set{0}$. Hence there exists a nonzero vector $u\in V$. Put $v\equiv\frac{u}{\dnorm{u}}$ so that
$$ \dnorm{v}=\dnorm{\frac{u}{\dnorm{u}}}=\normw{\frac1{\dnorm{u}}}\dnorm{u}=\frac1{\dnorm{u}}\dnorm{u}=1 \tag{7.B.13.1} $$
Hence $v$ is an orthonormal basis of $V$. Hence every vector in $V$ is a scalar multiple of $v$. Since $Tv\in V$, then there exists a scalar $\lambda\in\wR$ such that $Tv=\lambda v$. That is, $v$ is a eigenvector of $T$ corresponding to the eigenvalue $\lambda$. Hence $V$ has an orthonormal basis $v$ consisting of eigenvectors of $T$. Hence (a) implies (b) when $\dim{V}=1$.
Now assume that $\dim{V}\gt1$ and that (a) implies (b) for all real inner product spaces of smaller dimension than $V$.
Suppose (a) holds, so that $T\in\oper{V}$ is normal. Hence $\adjt{T}$ is normal too (by W.7.G.12). Proposition 5.21 gives the existence of an eigenvalue $\lambda$ of $T$. Definition 5.5 gives the existence of a nonzero eigenvector $u$ corresponding to $\lambda$. Put $v\equiv\frac{u}{\dnorm{u}}$. Equation 7.B.13.1 shows $\dnorm{v}=1$. And $v$ is also an eigenvector of $T$:
$$ Tv=T\Prn{\frac{u}{\dnorm{u}}}=\frac1{\dnorm{u}}Tu=\frac1{\dnorm{u}}\lambda u=\lambda\frac1{\dnorm{u}}u=\lambda\Prn{\frac{u}{\dnorm{u}}}=\lambda v $$
Then $\span{v}$ is a $1$-dimensional subspace that is invariant under $T$ (by W.5.7). Hence, by exercise 7.A.3, $\span{v}^\perp$ is also invariant under $\adjt{T}$. Then by definition 5.14 and the discussion on p.137, the operator $(\adjt{T})\bar_{\span{v}^\perp}$ is well-defined. By proposition 9.30, the operator $T\bar_{\span{v}^\perp}$ is well-defined (by part (a)) and normal (by part (d)). And proposition 9.30(d) also gives that $(\adjt{T})\bar_{\span{v}^\perp}$ is normal.
Proposition 6.50 gives
$$ \dim{\span{v}^\perp}=\dim{V}-\dim{\span{v}}=\dim{V}-1\lt\dim{V} $$
Then the induction assumption implies that there exists an orthonormal basis $v_2,\dots,v_{\dim{V}}$ of $\span{v}^\perp$ consisting of eigenvectors of $(\adjt{T})\bar_{\span{v}^\perp}$. By proposition 7.21, $v_2,\dots,v_{\dim{V}}$ is also an orthonormal list of eigenvectors of $T\bar_{\span{v}^\perp}$. By proposition W.7.G.13, $v_2,\dots,v_{\dim{V}}$ is also an orthonormal list of eigenvectors of $T$.
Adjoining $v_2,\dots,v_{\dim{V}}$ to $v$, we get an orthonormal basis of $V$ consisting of eigenvectors of $T$. Hence (a) implies (b).
We have proved that (c) implies (a) and that (a) implies (b). Clearly (b) implies (c) (by 5.41(b) and (c)), completing the proof. $\blacksquare$
(14) Suppose $U$ is a finite-dimensional real vector space and $T\in\oper{U}$. Then $U$ has a basis consisting of eigenvectors of $T$ if and only if there is an inner product on $U$ that makes $T$ into a self-adjoint operator.
Proof First suppose that $U$ has a basis $e_1,\dots,e_n$ consisting of eigenvectors of $T$. Then every vector in $U$ can be uniquely written as a linear combination of $e_1,\dots,e_n$ (by 2.29 Criterion for a basis). Let $u,v\in U$. Then
$$ u=\sum_{k=1}^na_ke_k \quad\quad\quad\quad v=\sum_{k=1}^nb_ke_k $$
for some scalars $a_k,b_k\in\wR$. Define the inner product $\innprd{\cdot}{\cdot}_U$ by
$$ \innprd{u}{v}_U\equiv\innprdBgg{\sum_{k=1}^na_ke_k}{\sum_{k=1}^nb_ke_k}_U\equiv\sum_{k=1}^na_kb_k $$
The proof that $\innprd{\cdot}{\cdot}_U$ is a valid inner product is similar to that of proposition W.6.G.9. Note that, for $i\neq j$, we have
$$ \innprd{e_i}{e_j}_U=\innprdBgg{\sum_{k=1}^n\mathbb{1}_{k=i}e_k}{\sum_{k=1}^n\mathbb{1}_{k=j}e_k}_U=\sum_{k=1}^n\mathbb{1}_{k=i}\mathbb{1}_{k=j}=\sum_{k=1}^n\mathbb{1}_{k=i,k=j}=0 $$
The next-to-last equality follows because
$$ \mathbb{1}_{k=i}\mathbb{1}_{k=j}=\cases{1&\text{if and only if }k=i\text{ and }k=j\\0&\text{otherwise}} =\mathbb{1}_{k=i,k=j} $$
Also note that
$$ \innprd{e_j}{e_j}_U=\innprdBgg{\sum_{k=1}^n\mathbb{1}_{k=j}e_k}{\sum_{k=1}^n\mathbb{1}_{k=j}e_k}_U=\sum_{k=1}^n\mathbb{1}_{k=j}\mathbb{1}_{k=j}=\sum_{k=1}^n\mathbb{1}_{k=j}=1 $$
Hence, under the inner product $\innprd{\cdot}{\cdot}_U$, the real vector space $U$ has an orthonormal basis consisting of eigenvectors of $T$. Hence the Real Spectral Theorem implies that $T$ is self-adjoint under the inner product $\innprd{\cdot}{\cdot}_U$.
Conversely, suppose that there is an inner product on $U$ that makes $T$ into a self-adjoint operator. Then the Real Spectral Theorem implies that $U$ has a basis consisting of eigenvectors of $T$. $\blacksquare$
(15) Find the matrix entry below that is covered up.
Solution Let $V$ be a complex inner product space with orthonormal basis $e_1,e_2,e_3$. Let $T\in\oper{V}$ be an operator such that
$$ \mtrxof{T,(e_1,e_2,e_3)}=\bmtrx{1&1&0\\0&1&1\\1&0&z} $$
We want to find the value of $z$ that makes $T$ normal. For $T$ to be normal, we must find $z$ such that $\tat=\tta$. Since $\mtrxof{\cdot}$ is an isomorphism (by 3.60), this is equivalent to $\mtrxof{\tat}=\mtrxof{\tta}$. By 3.43, this is equivalent to
$$ \mtrxof{\adjt{T}}\mtrxof{T}=\mtrxof{\tat}=\mtrxof{\tta}=\mtrxof{T}\mtrxof{\adjt{T}} \tag{7.B.15.1} $$
By 7.10, we have
$$ \mtrxof{\adjt{T}}=\cj{\bmtrx{1&1&0\\0&1&1\\1&0&z}^t}=\cj{\bmtrx{1&0&1\\1&1&0\\0&1&z}}=\bmtrx{1&0&1\\1&1&0\\0&1&\cj{z}} $$
So that
$$\align{ \mtrxof{\adjt{T}}\mtrxof{T} &= \bmtrx{1&0&1\\1&1&0\\0&1&\cj{z}}\bmtrx{1&1&0\\0&1&1\\1&0&z} \\\\ &= \bmtrx{1+0+1&1+0+0&0+0+z\\1+0+0&1+1+0&0+1+0\\0+0+\cj{z}&0+1+0&0+1+z\cj{z}} \\\\ &= \bmtrx{2&1&z\\1&2&1\\\cj{z}&1&1+z\cj{z}} }$$
and
$$\align{ \mtrxof{T}\mtrxof{\adjt{T}} &= \bmtrx{1&1&0\\0&1&1\\1&0&z}\bmtrx{1&0&1\\1&1&0\\0&1&\cj{z}} \\\\ &= \bmtrx{1+1+0&0+1+0&1+0+0\\0+1+0&0+1+1&0+0+\cj{z}\\1+0+0&0+0+z&1+0+z\cj{z}} \\\\ &= \bmtrx{2&1&1\\1&2&\cj{z}\\1&z&1+z\cj{z}} }$$
Then 7.B.15.1 becomes
$$ \bmtrx{2&1&z\\1&2&1\\\cj{z}&1&1+z\cj{z}}=\mtrxof{\adjt{T}}\mtrxof{T}=\mtrxof{T}\mtrxof{\adjt{T}}=\bmtrx{2&1&1\\1&2&\cj{z}\\1&z&1+z\cj{z}} $$
Hence we must have $z=1$ and $\cj{z}=1$. Hence $z=1$. $\blacksquare$
Exercises 7.C
(1) Prove or give a counterexample: If $T\in\oper{V}$ is self-adjoint and there exists an orthonormal basis of $e_1,\dots,e_n$ of $V$ such that $\innprd{Te_j}{e_j}\geq0$ for each $j$, then $T$ is a positive operator.
Proof We give a counterexample. Define $T\wiorn{2}$ by
$$\align{ Te_1 = e_1 \dq\dq Te_2 = -e_2 }$$
where $e_1, e_2$ is the standard basis of $\mathbb{R}^2$. Then $\mtrxof{T,(e_1,e_2))}$ is diagonal and hence symmetric:
$$ \mtrxof{T,(e_1,e_2)}=\bmtrx{1 & 0\\0 & -1}=\mtrxof{\adjt{T},(e_1,e_2)} $$
Hence $T$ is self-adjoint. Note that the basis $f_1\equiv\frac{1}{\sqrt{2}}(e_1 + e_2)$, $f_2\equiv\frac{1}{\sqrt{2}}(e_1 - e_2)$ is orthonormal:
$$\align{ \innprd{f_1}{f_2} &= \innprdBg{\frac1{\sqrt2}(e_1+e_2)}{\frac1{\sqrt2}(e_1-e_2)} \\ &= \frac1{\sqrt2}\frac1{\sqrt2}\innprd{e_1+e_2}{e_1-e_2} \\ &= \frac12\prn{\innprd{e_1}{e_1-e_2}+\innprd{e_2}{e_1-e_2}} \\ &= \frac12\prn{\innprd{e_1}{e_1}+\innprd{e_1}{-e_2}+\innprd{e_2}{e_1}+\innprd{e_2}{-e_2}} \\ &= \frac12\prn{\innprd{e_1}{e_1}-\innprd{e_1}{e_2}+\innprd{e_2}{e_1}-\innprd{e_2}{e_2}} \\ &= \frac12(1-0+0-1) \\ &= 0 }$$
$$\align{ \innprd{f_1}{f_1} &= \innprdBg{\frac1{\sqrt2}(e_1+e_2)}{\frac1{\sqrt2}(e_1+e_2)} \\ &= \frac1{\sqrt2}\frac1{\sqrt2}\innprd{e_1+e_2}{e_1+e_2} \\ &= \frac12\prn{\innprd{e_1}{e_1+e_2}+\innprd{e_2}{e_1+e_2}} \\ &= \frac12\prn{\innprd{e_1}{e_1}+\innprd{e_1}{e_2}+\innprd{e_2}{e_1}+\innprd{e_2}{e_2}} \\ &= \frac12(1+0+0+1) \\ &= 1 }$$
$$\align{ \innprd{f_2}{f_2} &= \innprdBg{\frac1{\sqrt2}(e_1-e_2)}{\frac1{\sqrt2}(e_1-e_2)} \\ &= \frac1{\sqrt2}\frac1{\sqrt2}\innprd{e_1-e_2}{e_1-e_2} \\ &= \frac12\prn{\innprd{e_1}{e_1-e_2}+\innprd{-e_2}{e_1-e_2}} \\ &= \frac12\prn{\innprd{e_1}{e_1}+\innprd{e_1}{-e_2}+\innprd{-e_2}{e_1}+\innprd{-e_2}{-e_2}} \\ &= \frac12\prn{\innprd{e_1}{e_1}-\innprd{e_1}{e_2}-\innprd{e_2}{e_1}+\innprd{e_2}{e_2}} \\ &= \frac12(1-0-0+1) \\ &= 1 }$$
Also note that
$$ Tf_1 = T\Prn{\frac{1}{\sqrt{2}}(e_1 + e_2)} = \frac1{\sqrt2}Te_1+\frac1{\sqrt2}Te_2 = \frac1{\sqrt2}e_1-\frac1{\sqrt2}e_2 = \frac1{\sqrt2}(e_1-e_2) = f_2 $$
and
$$ Tf_2 = T\Prn{\frac{1}{\sqrt{2}}(e_1 - e_2)} = \frac1{\sqrt2}Te_1-\frac1{\sqrt2}Te_2 = \frac1{\sqrt2}e_1+\frac1{\sqrt2}e_2 = \frac1{\sqrt2}(e_1+e_2) = f_1 $$
Hence
$$ 0=\innprd{f_2}{f_1}=\innprd{Tf_1}{f_1} \dq\text{and}\dq 0=\innprd{f_1}{f_2}=\innprd{Tf_2}{f_2} $$
But $T$ is not positive:
$$ \innprd{Te_2}{e_2} = \innprd{-e_2}{e_2}= -\innprd{e_2}{e_2} = -1 $$
$\wes$
(2) Suppose $T$ is a positive operator on $V$. Suppose $v,w\in V$ are such that
$$ Tv=w \dq\dq Tw=v $$
Then $v=w$.
Proof Since $T$ is a square root of $T^2$ and $T$ is positive, then $T^2$ is positive (by 7.35(c) or (d) or (e)). Hence $T$ is the unique positive square root of $T^2$ (by 7.36). Note that
$$ T^2v=TTv=Tw=v $$
Hence $v$ is an eigenvector of $T^2$ corresponding to the eigenvalue $1$. From the proof of 7.36, we see that $Tv = \sqrt{1}v=v$. Hence $v=Tv=w$. $\wes$
Alternate Proof Since $T$ is a positive operator, then
$$\align{ 0 &\leq \innprd{T(v-w)}{v-w} \\ &= \innprd{Tv-Tw}{v-w} \\ &= \innprd{w-v}{v-w} \\ &= \innprd{-(v-w)}{v-w} \\ &= -\innprd{v-w}{v-w} }$$
But the positivity of an inner product implies that $-\innprd{v-w}{v-w}\leq0$. Hence $\innprd{v-w}{v-w}=0$. And the definiteness of an inner product implies that $v-w=0$. $\wes$
(3) Suppose $T$ is a positive operator on $V$ and $U$ is a subspace of $V$ invariant under $T$. Then $T\bar_U\in\oper{U}$ is a positive operator on $U$.
Proof By 7.28(b), $T\bar_U\in\oper{U}$ is self-adjoint. And for any $u\in U$, we have $\innprd{T\bar_Uu}{u}=\innprd{Tu}{u}\geq0$. $\wes$
(4) Suppose $T\in\linmap{V}{W}$. Then $\tat$ is a positive operator on $V$ and $\tta$ is a positive operator on $W$.
Proof By 7.35(e), $\tat$ is positive. Or we can prove this directly: $(\tat)^*=\adjt{T}(\adjt{T})^*=\tat$ hence $\tat$ is self-adjoint. And for any $v\in V$, we have $\innprd{\tat v}{v}=\innprd{Tv}{Tv}\geq0$.
Similarly $(\tta)^*=(\adjt{T})^*\adjt{T}=\tta$ hence $\tta$ is self-adjoint. And for any $w\in W$, we have
$$ \innprd{\tta w}{w}=\innprd{(\adjt{T})^*\adjt{T}w}{w}=\innprd{\adjt{T}w}{\adjt{T}w}\geq0 $$
$\wes$
(5) The sum of two positive operators on $V$ is positive.
Proof Let $S,T\wiov$ be positive operators. Then 7.6(a) gives the first equality and the self-adjointedness of $S$ and $T$ gives the second equality:
$$ (S+T)^*=S^*+\adjt{T}=S+T $$
Hence $S+T$ is self-adjoint. Let $v\in V$. Then
$$ \innprd{(S+T)v}{v}=\innprd{Sv+Tv}{v}=\innprd{Sv}{v}+\innprd{Tv}{v}\geq0 $$
since $\innprd{Sv}{v}\geq0$ and $\innprd{Tv}{v}\geq0$. $\wes$
(6) Suppose $T\wiov$ is positive. Then $T^k$ is positive for every positive integer $k$.
Proof Since $T$ is positive, then there exists a self-adjoint operator $S\wiov$ such that $S^2=T$ (by 7.35(d)). For any positive integer $k$, we have $(S^k)^*=(S^*)^k=S^k$ by W.7.G.16 and $S=S^*$. Hence $S^k$ is self-adjoint.
Note that $(S^k)^2=S^{2k}=(S^2)^k=T^k$. Hence $T^k$ has a self-adjoint square root. Hence, by 7.35(d) and 7.35(a), $T^k$ is positive. $\wes$
(7) Suppose $T$ is a positive operator on $V$. Then $T$ is invertible if and only if
$$ \innprd{Tv}{v}\gt0 $$
for every $v\in V\setminus\set{0}$.
Proof First suppose that $T$ is invertible. Since $T$ is positive, then there exists an operator $R\wiov$ such that $T=R^*R$ (by 7.35(a)). Let $v\in V\setminus\set{0}$. If $Rv=0$, then $Tv=R^*Rv=R^*0=0$ (by 3.11). But this implies $T$ is not injective (by 3.16) and hence not invertible (by 3.69). This is a contradiction so $Rv\neq0$. Hence
$$ \innprd{Tv}{v}=\innprd{R^*Rv}{v}=\innprd{Rv}{Rv}\gt0 $$
Conversely suppose that $\innprd{Tv}{v}\gt0$ for every $v\in V\setminus\set{0}$. If it was that case that $Tv’=0$ for some $v’\in V\setminus\set{0}$, then $\innprd{Tv’}{v’}=0$ for this $v’\in V\setminus\set{0}$. This is a contradiction so $Tv\neq0$ for every $v\in V\setminus\set{0}$. Hence $\nullsp{T}=\set{0}$ and hence $T$ is injective. Hence $T$ is invertible. $\wes$
(8) Suppose $T\wiov$. For $u,v\in V$, define $\innprd{u}{v}_T$ by
$$ \innprd{u}{v}_T\equiv\innprd{Tu}{v} $$
Then $\innprd{\cdot}{\cdot}_T$ is an inner product on $V$ if and only if $T$ is an invertible positive operator with respect to the original inner product $\innprd{\cdot}{\cdot}$.
Proof First suppose that $\innprd{\cdot}{\cdot}_T$ is an inner product on $V$. Then, for any $v\in V$, we have
$$ \innprd{Tv}{v}=\innprd{v}{v}_T\geq0 $$
Hence $T$ is positive. Also, for any nonzero $v\in V$, we have
$$ \innprd{Tv}{v}=\innprd{v}{v}_T\gt0 $$
If it was the case that $Tv’=0$ for some nonzero $v’\in V$, then $\innprd{Tv’}{v’}=0$ for this $v’$, contradicting the strict inequality above. Hence $Tv\neq0$ for every nonzero $v\in V$. Hence $\nullsp{T}=\set{0}$ and hence $T$ is injective. Hence $T$ is invertible.
Conversely, suppose $T$ is an invertible positive operator. Then we check the definition 6.3:
Positivity: For any $v\in V$, we have
$$ \innprd{v}{v}_T=\innprd{Tv}{v}\geq0 $$
Definiteness: If $v=0$, then
$$ \innprd{v}{v}_T=\innprd{Tv}{v}=0 $$
Conversely, suppose $0=\innprd{v}{v}_T=\innprd{Tv}{v}$ and $v\neq0$. Since $T$ is an invertible, positive operator, this contradicts the previous exercise. Hence $0=\innprd{v}{v}_T$ implies $v=0$.
Additivity in the first slot:
$$ \innprd{u+v}{w}_T=\innprd{T(u+v)}{w}=\innprd{Tu+Tv}{w}=\innprd{Tu}{w}+\innprd{Tv}{w}=\innprd{u}{w}_T+\innprd{v}{w}_T $$
Homogeneity in the first slot:
$$ \innprd{\lambda u}{v}_T=\innprd{T(\lambda u)}{v}=\innprd{\lambda Tu}{v}=\lambda\innprd{Tu}{v}=\lambda\innprd{u}{v}_T $$
Conjugate Symmetry:
$$ \innprd{u}{v}_T=\innprd{Tu}{v}=\innprd{u}{\adjt{T}v}=\innprd{u}{Tv}=\cj{\innprd{Tv}{u}}=\cj{\innprd{v}{u}_T} $$
The third equality equality follows because $T$ is positive and hence self-adjoint. $\wes$
(9) The identity operator on $\wF^2$ has infinitely many self-adjoint square roots.
Proof For $r\in[-1,1]$, let $T_r$ denote the operator whose matrix (with respect to the standard basis) is
$$ \mtrxof{T_r}=\bmtrx{r&\sqrt{1-r^2}\\\sqrt{1-r^2}&-r} $$
Since $\mtrxof{T_r}$ is symmetric and real, then $\mtrxof{T_r}=\cj{\mtrxof{T_r}^t}=\mtrxof{T_r^*}$. Since $\mtrxof{\cdot}$ is an isomorphism (by 3.60), then it’s invertible and hence injective (by 3.56). Hence $T_r=T_r^*$ and $T_r$ is self-adjoint. $T_r$ is also a square root of the identity operator since
$$\align{ \mtrxof{T_r^2} &= \mtrxof{T_r}^2\tag{by W.5.4} \\\\ &= \bmtrx{r&\sqrt{1-r^2}\\\sqrt{1-r^2}&-r}^2 \\\\ &= \bmtrx{r^2+\prn{\sqrt{1-r^2}}^2&r\sqrt{1-r^2}-r\sqrt{1-r^2}\\t\sqrt{1-r^2}-t\sqrt{1-r^2}&\prn{\sqrt{1-r^2}}^2+(-r)^2} \\\\ &= \bmtrx{r^2+1-r^2&0\\0&1-r^2+t^2} \\\\ &= \bmtrx{1&0\\0&1} \\\\ &= \mtrxof{I} }$$
Hence there are uncountably many self-adjoint square roots of the identity operator on $\wF^2$. $\wes$
(10) Suppose $S\wiov$. Then the following are equivalent:
(a) $S$ is an isometry;
(b) $\innprd{S^*u}{S^*v}=\innprd{u}{v}$ for all $u,v\in V$;
(c) $S^*e_1,\dots,S^*e_m$ is an orthonormal list for every orthonormal list of vectors $e_1,\dots,e_m$ in $V$;
(d) $S^*e_1,\dots,S^*e_n$ is an orthonormal basis for some orthonormal basis $e_1,\dots,e_n$ of $V$.
Proof Note that 7.42(a) and 7.42(g) give (a)$\iff$(e) where
(e) $S^*$ is an isometry
Then (a),(b),(c), and (d) from 7.42 give the equivalence of (e),(b),(c), and (d) from above. $\wes$
(11) Suppose $T_1,T_2$ are normal operators on $\oper{\wF^3}$ and both operators have $2,5,7$ as eigenvalues. Then there exists an isometry $S\in\oper{\wF^3}$ such that $T_1=S^*T_2S$.
Cross-reference with Exercise 5.C.12
Proof Since $T_j$ [$j=1,2$] has $3$ distinct eigenvalues and $\dim{\wF^3}=3$, then $T_j$ is diagonalizable (by 5.44). Hence $\wF^3$ has an eigenbasis $e_1,e_2,e_3$ of $T_1$ and an eigenbasis $f_1,f_2,f_3$ of $T_2$ corresponding to eigenvalues $2,5,7$, respectively. Since scaling preserves eigenpairs, we can assume that $e_j$ and $f_j$ are unit vectors. Since $T_j$ is normal, then $e_1,e_2,e_3$ and $f_1,f_2,f_3$ are orthogonal lists (by 7.22). Hence they’re orthonormal lists. Hence they’re orthonormal eigenbases of $\wF^3$. Hence …
Alternate Beginning to Proof Proposition W.7.G.17 or the Complex Spectral Theorem imply that $\wF^3$ has orthonormal eigenbases $e_1,e_2,e_3$ of $T_1$ and $f_1,f_2,f_3$ of $T_2$ corresponding to the eigenvalues $2,5,7$, respectively. Hence …
Yet Another Alternate Beginning to Proof If $\wF=\wR$, then exercise 7.B.5, proposition 7.22, and the equality (by 5.38)
$$ \wR^3=\eignsp{2}{T_j}\oplus\eignsp{5}{T_j}\oplus\eignsp{7}{T_j} $$
imply that $T_1$ and $T_2$ are self-adjoint. Hence the Real Spectral Theorem or the Complex Spectral Theorem imply that $\wF^3$ has orthonormal eigenbases $e_1,e_2,e_3$ of $T_1$ and $f_1,f_2,f_3$ of $T_2$ corresponding to the eigenvalues $2,5,7$, respectively. Hence
$$\nbmtrx{ \nbmtrx{ T_1e_1 = 2e_1\\ T_1e_2 = 5e_2\\ T_1e_3 = 7e_3} &&&& \nbmtrx{ T_2f_1 = 2f_1\\ T_2f_2 = 5f_2\\ T_2f_3 = 7f_3} }$$
Define $S\in\oper{\wF^3}$ by
$$ Se_j=f_j\quad\text{for }j=1,2,3 \tag{7.C.11.1} $$
The equivalence of 7.42(d) and 7.42(a) implies that $S$ is an isometry. We can also show this directly. Let $v\in V$. Then 6.30 gives $v=\sum_{k=1}^n\innprd{v}{e_k}e_k$. Hence
$$ \dnorm{Sv}^2=\dnorm{S\Prn{\sum_{k=1}^n\innprd{v}{e_k}e_k}}^2=\dnorm{\sum_{k=1}^n\innprd{v}{e_k}Se_k}^2=\dnorm{\sum_{k=1}^n\innprd{v}{e_k}f_k}^2=\sum_{k=1}^n\normb{\innprd{v}{e_k}}^2=\dnorm{v}^2 $$
The next-to-last equality follows from 6.25 (not 6.30) and the last equality follows from 6.30. Hence $S$ is an isometry.
Note that $S$ is invertible (by W.3.17 or 7.42(h)). Hence, if we multiply both sides of 7.C.11.1 on the left by $S^{-1}$, then we get
$$ e_j=S^{-1}f_j\quad\text{for }j=1,2,3 $$
Then, because $S^{-1}=S^*$ (by 7.42(h)), we have $S^*f_j=S^{-1}f_j=e_j$. Hence
$$ T_1e_1=2e_1=2S^*f_1=S^*(2f_1)=S^*(T_2f_1)=S^*(T_2Se_1)=(S^*T_2S)e_1 \\ T_1e_2=5e_2=5S^*f_2=S^*(5f_2)=S^*(T_2f_2)=S^*(T_2Se_2)=(S^*T_2S)e_2 \\ T_1e_3=7e_3=7S^*f_3=S^*(7f_3)=S^*(T_2f_3)=S^*(T_2Se_3)=(S^*T_2S)e_3 $$
Hence $T_1=S^*T_2S$. $\wes$
(12) Give an example of two self-adjoint operators $T_1,T_2\in\oper{\wF^4}$ such that the eigenvalues of both operators are $2,5,7$ but there does not exist an isometry $S\in\oper{\wF^4}$ such that $T_1=S^*T_2S$. Be sure to explain why there is no isometry with the required property.
Cross-reference with Exercise 5.C.13
Example Let $e_1,e_2,e_3,e_4$ denote an orthonormal basis of $\wF^4$. Define $T_1,T_2\in\oper{\wF^4}$ by
$$\nbmtrx{ \nbmtrx{ T_1e_1 = 2e_1\\ T_1e_2 = 2e_2\\ T_1e_3 = 5e_3\\ T_1e_4 = 7e_4} &&&& \nbmtrx{ T_2e_1 = 2e_1\\ T_2e_2 = 5e_2\\ T_2e_3 = 5e_3\\ T_2e_4 = 7e_4} }$$
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 $T_1=S^*T_2S$. Then $S$ is invertible (by 7.42) and hence surjective and injective (by 3.69). Let $v\in V$ be the vector that $S$ maps to $e_2$. Then
$$ T_1v = S^*T_2Sv = S^*T_2e_2 = 5S^*e_2 = 5S^{-1}e_2 = 5v $$
Hence $v\in\eignsp{5}{T_1}=\span{e_3}$. Let $w \in V$ be the vector that $S$ maps to $e_3$. Then
$$ T_1w = S^*T_2Sw = S^*T_2e_3 = 5S^*e_3 = 5S^{-1}e_3 = 5w. $$
Hence $w\in\eignsp{5}{T_1}=\span{e_3}$. But this is a contradiction: Note that the linear independence of $Sv=e_2,Sw=e_3$ implies the linear independence of $v,w$ (by exercise 3.A.4). But we can’t have a linearly independent list of length $2$ 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 an isometry. The key idea here is that the eigenspaces of $T_1$ and $T_2$ don’t fit.
(13) Prove or give a counterexample: if $S\in\oper{V}$ and there exists an orthonormal basis $e_1,\dots,e_n$ of $V$ such that $\dnorm{Se_j}=1$ for each $e_j$, then $S$ is an isometry.
Counterexample Let $e_1,e_2$ be the standard basis of $\wF^2$. It’s orthonormal under the typical dot product. Define $S\in\oper{\wF^2}$ by
$$ Se_1\equiv e_1 \dq\dq Se_2=e_1 $$
Then $\dnorm{Se_1}=\dnorm{e_1}=1$ and $\dnorm{Se_2}=\dnorm{e_1}=1$. But
$$ \dnorm{S(e_1-e_2)}=\dnorm{Se_1-Se_2}=\dnorm{e_1-e_1}=0 $$
while
$$\align{ \dnorm{e_1-e_2} &= \sqrt{\innprd{e_1-e_2}{e_1-e_2}} \\ &= \sqrt{\innprd{e_1}{e_1-e_2}+\innprd{-e_2}{e_1-e_2}} \\ &= \sqrt{\innprd{e_1}{e_1}+\innprd{e_1}{-e_2}+\innprd{-e_2}{e_1}+\innprd{-e_2}{-e_2}} \\ &= \sqrt{\innprd{e_1}{e_1}-\innprd{e_1}{e_2}-\innprd{e_2}{e_1}+\innprd{e_2}{e_2}} \\ &= \sqrt{1-0-0+1} \\ &= \sqrt2 }$$
$\wes$
(14) Let $T$ be the second derivative operator in Exercise 21 in Section 7.A. Then $-T$ is a positive operator.
Proof By 7.35(e), it suffices to show that there exists an operator $D\wiov$ such that $-T=D^*D$. In that exercise, we defined $D\wiov$ to be the first derivative operator $Df\equiv f’$ and we showed that $D^*=-D$. We also showed that $T=D^2$. Hence
$$ -T=-(DD)=(-D)D=D^*D $$
$\wes$
Exercises 7.D
(1) Fix $u,x\in V$ with $u\neq0$. Define $T\wiov$ by
$$ Tv\equiv\innprd{v}{u}x $$
for every $v\in V$. Then
$$ \sqrt{\tat}v=\frac{\dnorm{x}}{\dnorm{u}}\innprd{v}{u}u $$
for every $v\in V$.
Proof Note that
$$\align{ \innprd{w}{\adjt{T}v} &= \innprd{Tw}{v} \\ &= \innprdbg{\innprd{w}{u}x}{v} \\ &= \innprd{w}{u}\innprd{x}{v} \\ &= \innprdbg{w}{\cj{\innprd{x}{v}}u} \\ &= \innprdbg{w}{\innprd{v}{x}u} \\ }$$
Hence $\adjt{T}v=\innprd{v}{x}u$ and
$$ (\tat)v=\adjt{T}\prn{\innprd{v}{u}x}=\innprd{v}{u}\adjt{T}x=\innprd{v}{u}\innprd{x}{x}u $$
Define
$$ Rv\equiv\frac{\dnorm{x}}{\dnorm{u}}\innprd{v}{u}u $$
Note that, for any $v\in V$, we have
$$\align{ R^2v &= R(Rv) \\ &= R\Prn{\frac{\dnorm{x}}{\dnorm{u}}\innprd{v}{u}u} \\ &= \Prn{\frac{\dnorm{x}}{\dnorm{u}}\innprd{v}{u}}Ru \\ &= \Prn{\frac{\dnorm{x}}{\dnorm{u}}\innprd{v}{u}}\Prn{\frac{\dnorm{x}}{\dnorm{u}}\innprd{u}{u}u} \\ &= \innprd{v}{u}\dnorm{x}^2\frac{\innprd{u}{u}}{\dnorm{u}^2}u \\ &= \innprd{v}{u}\innprd{x}{x}u \\ &= (\tat)v }$$
Hence $R$ is a square root of $\tat$. To show that it’s the unique (by 7.36) positive square root $\sqrt{\tat}$ (7.44), we must show that $R$ is self-adjoint and that $\innprd{Rv}{v}\geq0$. The latter proof is
$$ \innprd{Rv}{v}=\innprdBg{\frac{\dnorm{x}}{\dnorm{u}}\innprd{v}{u}u}{v}=\frac{\dnorm{x}}{\dnorm{u}}\innprd{v}{u}\innprd{u}{v}=\frac{\dnorm{x}}{\dnorm{u}}\innprd{v}{u}\cj{\innprd{v}{u}}=\frac{\dnorm{x}}{\dnorm{u}}\normb{\innprd{v}{u}}^2\geq0 $$
Let $e_1,\dots,e_n$ be an orthonormal basis of $V$ (by 6.34). Then, for any $v\in V$, we have
$$\align{ \adjt{R}v &= \sum_{j=1}^n\innprd{v}{Re_j}e_j\tag{by W.7.CoA.1} \\ &= \sum_{j=1}^n\innprdBg{v}{\frac{\dnorm{x}}{\dnorm{u}}\innprd{e_j}{u}u}e_j \\ &= \sum_{j=1}^n\cj{\frac{\dnorm{x}}{\dnorm{u}}\innprd{e_j}{u}}\innprd{v}{u}e_j \\ &= \sum_{j=1}^n\frac{\dnorm{x}}{\dnorm{u}}\cj{\innprd{e_j}{u}}\innprd{v}{u}e_j \\ &= \frac{\dnorm{x}}{\dnorm{u}}\innprd{v}{u}\sum_{j=1}^n\cj{\innprd{e_j}{u}}e_j \\ &= \frac{\dnorm{x}}{\dnorm{u}}\innprd{v}{u}\sum_{j=1}^n\innprd{u}{e_j}e_j \\ &= \frac{\dnorm{x}}{\dnorm{u}}\innprd{v}{u}u\tag{by 6.30} \\ &= Rv }$$
Hence $R$ is self-adjoint. $\wes$
(2) Give an example of $T\in\oper{\wC^2}$ such that $0$ is the only eigenvalue of $T$ and the singular values of $T$ are $5,0$.
Example Let’s try
$$ \mtrxof{T}=\bmtrx{0&5\\0&0} $$
so that $Te_1=0e_1$ and $Te_2=5e_1$. Since this is an upper-triangular matrix, then 5.32 gives that $0$ is the only eigenvalue of $T$. And W.7.CoA.1 gives
$$\align{ \adjt{T}e_1=\innprd{e_1}{Te_1}e_1+\innprd{e_1}{Te_2}e_2=\innprd{e_1}{0}e_1+\innprd{e_1}{5e_1}e_2=5e_2 \\ \adjt{T}e_2=\innprd{e_2}{Te_1}e_1+\innprd{e_2}{Te_2}e_2=\innprd{e_2}{0}e_1+\innprd{e_2}{5e_1}e_2=0e_2 }$$
Hence
$$ \mtrxof{\adjt{T}}=\bmtrx{0&0\\5&0} $$
and
$$ \mtrxof{\tat}=\mtrxof{\adjt{T}}\mtrxof{T}=\bmtrx{0&0\\5&0}\bmtrx{0&5\\0&0}=\bmtrx{0&0\\0&25} $$
so that $\tat e_1=0e_1$ and $\tat e_2=25e_2$. Since this is an upper-triangular matrix, then 5.32 gives that the eigenvalues of $\tat$ are $25,0$. And we can see that $1=\dim{\eignsp{25}{\tat}}=\dim{\eignsp{0}{\tat}}$. Hence W.7.G.22 or 7.52 give that the singular values of $T$ are $5,0$. $\wes$
(3) Suppose $T\wiov$. Then there exists an isometry $S\wiov$ such that
$$ T=\sqrt{\tta}S $$
Proof By the Polar Decomposition (7.45), there exists an isometry $S \in \mathcal{L}(V)$ such that
$$ \adjt{T} = S\sqrt{\adjtdbl{T}\adjt{T}} = S\sqrt{\tta} $$
Taking the adjoint of each side, we get
$$ T = \adjt{\prn{S\sqrt{\tta}}} = \adjt{\prn{\sqrt{\tta}}}\adjt{S} = \sqrt{\tta}\adjt{S} $$
where the last equality follows because $\sqrt{\tta}$ is positive (7.44) and hence self-adjoint. Note that $\adjt{S}$ is an isometry (by 7.42(g)). $\wes$
(4) Suppose $T\wiov$ and $s$ is a singular value of $T$. Then there exists a vector $v\in V$ such that $\dnorm{v}=1$ and $\dnorm{Tv}=s$.
Proof Since $s$ is a singular value of $T$, then $s$ is an eigenvalue of $\sqrt{\tat}$ (by 7.49). Let $v \in V$ be an eigenvector of $\sqrt{\tat}$ corresponding to $s$ with $\dnorm{v}=1$. And let $S\wiov$ be an isometry such that $T=S\sqrt{\tat}$. Then
$$ \dnorm{Tv} = \dnorm{S\sqrt{\tat}v} = \dnorm{\sqrt{\tat}v} = \dnorm{sv} = \norm{s}\dnorm{v} = \norm{s} = s $$
where the last equality follows from 7.35(b) because $\sqrt{\tat}$ is positive. $\wes$
(5) Suppose $T\in\oper{\wC^2}$ is defined by $T(x,y)=(-4y,x)$. Find the singular values of $T$.
Solution Note that $T(x,y)=(-4y,x)=(-4y,0)+(0,x)=-4ye_1+xe_2$. Hence
$$\align{ &Te_1=T(1,0)=-4\cdot0e_1+1e_2=e_2 \\ &Te_2=T(0,1)=-4\cdot1e_1+0e_2=-4e_1 }$$
and
$$ \mtrxof{\tat} = \mtrxof{\adjt{T}}\mtrxof{T} = \cj{\bmtrx{0&-4\\1&0}^t}\bmtrx{0&-4\\1&0} = \bmtrx{0&1\\-4&0}\bmtrx{0&-4\\1&0} = \bmtrx{1&0\\0&16} $$
This is an upper-triangular matrix. Hence 5.32 gives that the eigenvalues of $\tat$ are $1,16$. Hence W.7.G.22 or 7.52 give that the singular values of $T$ are $1,4$. $\wes$
(6) Find the singular values of the differentiation operator $D\in\operb{\polyrn{2}}$ defined by $Dp=p’$, where the inner product on $\polyrn{2}$ is as in Example 6.33.
Solution From Example 6.33, we have the following orthonormal basis of $\polyrn{2}$
$$ e_1=\sqrt{\frac12} \dq\dq e_2=\sqrt{\frac32}x \dq\dq e_3=\sqrt{\frac{45}8}\prn{x^2 - \frac13} $$
The action of $D$ on this basis is
$$\align{ &De_1=D\sqrt{\frac{1}{2}} = 0 \\ &De_2=D\sqrt{\frac{3}{2}}x = \sqrt{3}\Prn{\sqrt{\frac{1}{2}}} = \sqrt{3}e_1 \\ &De_3=D\sqrt{\frac{45}{8}}\Prn{x^2 - \frac{1}{3}} = \sqrt{15}\Prn{\sqrt{\frac{3}{2}}x} = \sqrt{15}e_2 }$$
Hence
$$ \mtrxof{D} = \bmtrx{0&\sqrt{3}&0\\0&0&\sqrt{15}\\0&0&0} $$
and
$$ \mtrxof{\adjt{D}D} = \mtrxof{\adjt{D}}\mtrxof{D} = \bmtrx{0&0&0\\\sqrt{3}&0&0\\0&\sqrt{15}&0}\bmtrx{0&\sqrt{3}&0\\0&0&\sqrt{15}\\0&0&0} = \bmtrx{0&0&0\\0&3&0\\0&0&15} $$
This is an upper-triangular matrix. Hence 5.32 gives that the eigenvalues of $\adjt{D}D$ are $0,3,15$. Hence W.7.G.22 or 7.52 give that the singular values of $D$ are $0,\sqrt3,\sqrt{15}$. $\wes$
(7) Define $T\in\oper{\wF^3}$ by
$$ T(z_1,z_2,z_3) \equiv (z_3,2z_1,3z_2) $$
Find (explicitly) an isometry $S\in\oper{\wF^3}$ such that $T=S\sqrt{\tat}$.
Solution Note that
$$\align{ \innprdbg{(w_1,w_2,w_3)}{\adjt{T}(z_1,z_2,z_3)} &= \innprdbg{T(w_1,w_2,w_3)}{(z_1,z_2,z_3)} \\ &= \innprdbg{(w_3,2w_1,3w_2)}{(z_1,z_2,z_3)} \\ &= w_3z_1+2w_1z_2+3w_2z_3 \\ &= 2w_1z_2+3w_2z_3+w_3z_1 \\ &= \innprdbg{(w_1,w_2,w_3)}{(2z_2,3z_3,z_1)} }$$
Hence $\adjt{T}(z_1,z_2,z_3)=(2z_2,3z_3,z_1)$ and
$$ \tat(z_1,z_2,z_3) = \adjt{T}(z_3,2z_1,3z_2) = (2\cdot2z_1,3\cdot3z_2,z_3) = (4z_1,9z_2,z_3) $$
and
$$\align{ \tat e_1&=(4,0,0)=4e_1\\ \tat e_2&=(0,9,0)=9e_2\\ \tat e_3&=(0,0,1)=1e_3 }$$
where $e_1,e_2,e_3$ is the standard basis of $\wF^3$. Hence
$$\align{ &\sqrt{\tat}e_1=2e_1\\ &\sqrt{\tat}e_2=3e_2\\ &\sqrt{\tat}e_3=1e_3 }$$
or
$$ \sqrt{\tat}(z_1,z_2,z_3) = (2z_1,3z_2,z_3) $$
Hence we want to find $S$ such that
$$ S(2z_1,3z_2,z_3) = S\sqrt{\tat}(z_1,z_2,z_3) = T(z_1,z_2,z_3) = (z_3,2z_1,3z_2) $$
So we define $S\in\oper{\wF^3}$ by
$$ S(z_1,z_2,z_3) \equiv (z_3,z_1,z_2) $$
or
$$\align{ &Se_1\equiv e_3\dq\qd &Se_2\equiv e_1\dq\qd &Se_3\equiv e_2 }$$
Then $S$ maps an orthonormal basis to an orthonormal basis. Hence $S$ is an isometry (by 7.42(d)). And we can verify the desired result:
$$ S\sqrt{\tat}(z_1,z_2,z_3) = S(2z_1,3z_2,z_3) = (z_3,2z_1,3z_2) = T(z_1,z_2,z_3) $$
$\wes$
(8) Suppose $T\wiov$, $S\wiov$ is an isometry, and $R\wiov$ is a positive operator such that $T=SR$. Then $R=\sqrt{\tat}$.
This exercise shows that if we write $T$ as the product of an isometry and a positive operator (as in the Polar Decomposition 7.45), then the positive operator equals $\sqrt{\tat}$.
Proof Taking adjoints of both sides of the equality $T=SR$, we get
$$ \adjt{T}=\adjt{(SR)}=\adjt{R}\adjt{S}=R\adjt{S} $$
where the last equality holds because $R$ is positive, and hence self-adjoint. Multiplying together our formulas $\adjtT$ and $T$, we get
$$ \adjt{T}T=R\adjt{S}SR=R\inv{S}SR=RR=R^2 $$
The second equality follows from 7.42(h). Hence $R$ is a square root of $\tat$ (by 7.33). Since $R$ is positive, this implies that $R=\sqrt{\tat}$ (by 7.36 and 7.44). $\wes$
Alternate Proof Proposition 7.42(h) gives that $S$ is invertible and $\inv{S}=\adjt{S}$. Multiplying both sides of the equality $T=SR$ on the left by $\inv{S}=\adjt{S}$, we get
$$ \adjt{S}T=\inv{S}T=\inv{S}SR=R \tag{7.D.8.1} $$
Let $X\wiov$ be an isometry such that $T=X\sqrt{\tat}$ (existence by Polar Decomposition). Multiplying both sides of this equality on the left by $\inv{X}=\adjt{X}$ (by 7.42(h)), we get
$$ \adjt{X}T=\inv{X}T=\inv{X}X\sqrt{\tat}=\sqrt{\tat} \tag{7.D.8.2} $$
By 7.42(g), the operators $\adjt{S}$ and $\adjt{X}$ are isometries. Then
$$\align{ 0 &= \dnorm{Tv}^2 - \dnorm{Tv}^2 \\ &= \dnorm{\adjt{S}(Tv)}^2 - \dnorm{\adjt{X}(Tv)}^2\tag{by 7.42(g) and 7.37} \\ &= \dnorm{(\adjt{S}T)v}^2 - \dnorm{(\adjt{X}T)v}^2 \\ &= \dnorm{Rv}^2 - \dnorm{\sqrt{\tat}v}^2\tag{by 7.D.8.1 and 7.D.8.2} \\ &= \innprdbg{Rv}{Rv}-\innprdbg{\sqrt{\tat}v}{\sqrt{\tat}v} \\ &= \innprdbg{\adjt{R}Rv}{v}-\innprdbg{\adjt{\prn{\sqrt{\tat}}}\sqrt{\tat}v}{v} \\ &= \innprdbg{RRv}{v}-\innprdbg{\sqrt{\tat}\sqrt{\tat}v}{v}\tag{7.D.8.3} \\ &= \innprdbg{R^2v}{v}-\innprdbg{\prn{\sqrt{\tat}}^2v}{v} \\ &= \innprdbg{R^2v}{v}-\innprdbg{\tat v}{v} \\ &= \innprdbg{R^2v-\tat v}{v} \\ &= \innprdbg{(R^2-\tat)v}{v} }$$
7.D.8.3 follows because $R$ and $\sqrt{\tat}$ are both positive and hence self-adjoint. The positivity of $R$ and $\sqrt{\tat}$ also implies that $R^2$ and $\tat=\prn{\sqrt{\tat}}^2$ are positive (by 7.35(c)) and hence self-adjoint. Hence $R^2-\tat$ is self-adjoint (by exercise 7.A.8.content). Then 7.16 implies that $R^2=\tat$. Hence $R$ is a square root of $\tat$ (by 7.33). Since $R$ is positive, this implies that $R=\sqrt{\tat}$ (by 7.36 and 7.44). $\wes$
(9) Suppose $T\wiov$. Then $T$ is invertible if and only if there exists a unique isometry $S\wiov$ such that $T=S\sqrt{\tat}$.
Proof First suppose that $T$ is invertible. The Polar Decomposition (7.45) gives an isometry $S\wiov$ such that
$$ T=S\sqrt{\tat} \tag{7.D.9.1} $$
Since $T$ is invertible, then applying exercise 3.D.9 to 7.D.9.1 implies that $\sqrt{\tat}$ is invertible. Multiplying both sides of 7.D.9.1 on the right by $\inv{\prn{\sqrt{\tat}}}$, we get $S=T\inv{\prn{\sqrt{\tat}}}$. Because $S$ must be given by this formula and because $T$ and $\inv{\prn{\sqrt{\tat}}}$ are unique (by 3.5, 7.36, and 3.54), we wee that there is a unique isometry $S\wiov$ such that $T=S\sqrt{\tat}$.
Conversely, suppose that there exists a unique isometry $S\wiov$ such that $T=S\sqrt{\tat}$. Then the map $S_2$ from the proof of Polar Decomposition (7.45) must be $0$. To show this, suppose $S_2\neq0$. Let’s change the definition of $S\wiov$ from the book’s proof. For $v=u+w$ with $u\in\rangspb{\sqrt{\tat}}$ and $w\in\rangspb{\sqrt{\tat}}^\perp$, define:
$$ Sv\equiv S_1u-S_2w $$
Since $S_2\neq0$, then this definition produces a different $S$ than that from the book’s proof. And we can verify that this $S$ is an isometry and that it satisfies the Polar Decomposition equation:
$$ \dnorm{Sv}^2=\dnorm{S_1u+(-S_2w)}^2=\dnorm{S_1u}^2+\dnorm{-S_2w}^2=\dnorm{S_1u}^2+\norm{-1}^2\dnorm{S_2w}^2=\dnorm{u}^2+\dnorm{w}^2=\dnorm{v}^2 $$
where the second equality holds because $S_1u\in\rangsp{T}$ and $-S_2w\in\rangsp{T}^\perp$. And the Polar Decomposition is
$$ S\prn{\sqrt{\tat}v}=S_1\prn{\sqrt{\tat}v}-S_2(0)=S_1\prn{\sqrt{\tat}v}=Tv $$
Hence, the assumption that $S_2\neq0$ contradicts our hypothesis that a unique isometry $S$ satisfies $T=S\sqrt{\tat}$. Hence $S_2=0$.
Note the definition of the orthonormal bases $e_1,\dots,e_m$ and $f_1,\dots,f_m$ in the proof. And note that the definition of $S_2$ in the proof gives $S_2e_j=f_j$ for $j=1,\dots,m$. Since $e_1,\dots,e_m$ is a basis of $\rangspb{\sqrt{\tat}}^\perp$ and $f_1,\dots,m$ is a basis of $\rangsp{T}^\perp$, then $S_2\in\linmapb{\rangspb{\sqrt{\tat}}^\perp}{\rangsp{T}^\perp}$ is surjective (by W.3.11). Hence $\rangsp{S_2}=\rangsp{T}^\perp$. We showed above that $S_2=0$ hence
$$ \set{0}=\rangsp{S_2}=\rangsp{T}^\perp $$
Then 6.47 gives
$$ V=\dirsum{\rangsp{T}}{\rangsp{T}^\perp}=\dirsum{\rangsp{T}}{\set{0}}=\rangsp{T} $$
Hence $T$ is surjective (by 3.20) and invertible (by 3.69). $\wes$
(10) Suppose $T\wiov$ is self-adjoint. Then the singular values of $T$ equal the absolute values of the eigenvalues of $T$, repeated appropriately.
Proof The Spectral Theorem implies that $V$ has an orthonormal eigenbasis $e_1,\dots,e_n$ corresponding to the eigenvalues $\lambda_1,\dots,\lambda_n$ of $T$. Hence, for $j=1,\dots,n$, we have
$$ Te_j=\lambda e_j $$
and
$$ \tat e_j=T^2e_j=T(Te_j)=T(\lambda_je_j)=\lambda_jTe_j=\lambda_j^2e_j $$
Then W.7.G.22 or 7.52 imply that the singular values of $T$ are
$$ s_j=\sqrt{\lambda_j^2} $$
for $j=1,\dots,n$. Note that $\lambda_1,\dots,\lambda_n\in\wR$ (by 7.13). Hence $s_j=\norm{\lambda_j}$ for $j=1,\dots,n$. $\wes$
(11) Suppose $T\wiov$. Then $T$ and $\adjtT$ have the same singular values.
Proof By exercise 5.A.23, $\tat$ and $\tta$ have the same eigenvalues. By W.7.G.22, the singular values of $T$ are the square roots of the eigenvalues of $\tat$. And by W.7.G.22, the singular values of $\adjtT$ are the square roots of the eigenvalues of $\adjt{(\adjtT)}\adjt{T}=T\adjtT$. $\wes$
(12) Prove or give a counterexample: if $T\wiov$, then the singular values of $T^2$ equal the squares of the singular values of $T$.
Counterexample 1 Take the operator $T\in\oper{\wC^2}$ from exercise 7.D.5. We have
$$\align{ \mtrxofb{\adjt{(T^2)}T^2} &= \mtrxofb{\adjt{(T^2)}}\mtrxof{T^2}\tag{by 3.43} \\ &= \mtrxofb{(\adjtT)^2}\mtrxof{T}^2\tag{by W.7.G.16 and W.5.4} \\ &= \mtrxof{\adjtT}^2\mtrxof{T}^2\tag{by W.5.4} \\ &= \bmtrx{0&1\\-4&0}^2\bmtrx{0&-4\\1&0}^2 \\ &= \bmtrx{-4&0\\0&-4}\bmtrx{-4&0\\0&-4} \\ &= \bmtrx{16&0\\0&16} }$$
Hence the eigenvalues of $\adjt{(T^2)}T^2$ are $16,16$ (by 5.32). Hence the singular values of $T^2$ are $4,4$ (by W.7.G.22) whereas the singular values of $T$ are $1, 4$. $\wes$
(13) Suppose $T\wiov$. Then $T$ is invertible if and only if $0$ is not a singular of $T$.
Proof We have
$$\align{ T\text{ is invertible} &\iff \sqrt{\tat}\text{ is invertible}\tag{by W.7.G.24} \\ &\iff 0\text{ is not an eigenvalue of }\sqrt{\tat}\tag{by W.5.17} \\ &\iff 0\text{ is not a singular value of }T\tag{by 7.49} \\ }$$
$\wes$