1.1.22 Hilbert Spaces

A Hilbert space, named after the math­e­mati­cian David Hilbert, ex­tends the meth­ods of vector al­ge­bra and cal­cu­lus from two and three-di­men­sion­al Eu­clid­ean spaces to spaces $\in \mathbb{R},\mathbb{C}$, of finite and in­fi­nite di­men­sions. A Hilbert space is an ab­stract vector space that is com­plete and has the struc­ture of an inner prod­uct to make pos­si­ble the mea­sure­ment of lengths and angles. In this sec­tion, we use many of the de­f­i­n­i­tions from Ap­plied Analy­sis by Hunter and Nachter­gaele ¹, which is well-writ­ten with a straight­for­ward ap­proach to many of the analy­sis topics con­sid­ered in this book. To start, we must define what the inner prod­uct of a com­plex linear space $X$ is.

De­f­i­n­i­tion (1.1.137) – An inner prod­uct on a com­plex linear space $X$ is a map

such that, for all $x_{j} \in X$ and $a,b\in \mathbb{C}$:

• $\left\langle x_{1},ax_{2} + bx_{3} \right\rangle = a\left\langle x_{1},x_{2} \right\rangle + b\left\langle x_{1},x_{3} \right\rangle$ – Linear in 2^nd ar­gu­ment

• $\left\langle x_{1},x_{2} \right\rangle = \overline{\left\langle x_{1},x_{2} \right\rangle}$ – Her­mit­ian Sym­met­ric

• $\left\langle x_{1},x_{1} \right\rangle \geq 0$ – Non-neg­a­tive

• $\left\langle x_{1},x_{1} \right\rangle = 0$, iff $x_{1} = 0$ – Pos­i­tive-def­i­nite

A linear space with an inner prod­uct space is called a pre-Hilbert space. For $X$, a pre-Hilbert space, we can define a norm on $X$ to be

where $X$ is also a normed linear space, lead­ing us to define a Hilbert space,

De­f­i­n­i­tion (1.1.139) – A Hilbert space is a com­plete inner prod­uct space.

Com­plete, in layman's terms, means no gaps. In math­e­mat­i­cal analy­sis, a metric space $X$ is com­plete if every Cauchy se­quence of points $\in X$ has a limit that is also $\in X$, or if every Cauchy se­quence $\in X$ con­verges $\in X$ , stated as

The stan­dard inner prod­uct on $\mathbb{C}^{n}$, is given by

for $x_{1},x_{2}\in \mathbb{C}$, which rep­re­sents a finite-di­men­sion­al Hilbert space as it is com­plete. An in­fi­nite-di­men­sion­al Hilbert space $\in \mathbb{C}$, where $C\left( \lbrack a,b\rbrack \right)$is the space of all com­plex-valued func­tions, with a de­fined inner prod­uct

for $f,g\lbrack a,b\rbrack: \rightarrow \mathbb{C}$, being con­tin­u­ous func­tions. This space may not be com­plete, as it is bound on the in­ter­val $\lbrack a,b\rbrack$, and the com­ple­tion of the space con­cern­ing its norm

This norm is often de­not­ed as $L^{2}\left( \lbrack a,b\rbrack \right)$, where the 2-su­per­script­ed rep­re­sents the $p$-norm of two for Hilbert spaces. In­fi­nite-di­men­sion­al Hilbert spaces re­quire square-in­te­grable func­tions to be con­ver­gent, and these func­tions would have a finite result for

For the circle space, bound on the in­ter­val $\mathbb{T} =\lbrack 0,2\pi\rbrack$, $L^{2}\left( \mathbb{T} \right)$ is Hilbert space of square-in­te­grable func­tions $f:\mathbb{T} \rightarrow C$, with the inner prod­uct

Going back to the matrix-vector prod­uct ex­am­ple $G\left( X_{1}\left( \frac{i\pi}{8} \right) \right) \bullet \widehat{\epsilon}$, we denote its result $\widehat{v}$, as

and taking the inner prod­uct $\widehat{v} \in \mathbb{C}^{3}$ with itself, this is de­ter­mined by eval­u­at­ing

where $\overline{v_{j}}$ de­notes the com­plex con­ju­gate of $v_{j}$. The result of $\left\langle \widehat{v},\widehat{v} \right\rangle$ is pos­i­tive def­i­nite, and the inner prod­uct would only be equal to zero if $\widehat{v} = 0.$For ex­am­ple, we take a vector $\widehat{u}$ equal to the matrix-vector prod­uct $G\left( X_{1}\left( \frac{i\pi}{3} \right) \right) \bullet \widehat{\epsilon}$,

and con­sid­er the fol­low­ing inner prod­uct $\left\langle \widehat{v},\widehat{u} \right\rangle$

which is equal to its con­ju­gate

Moving from vec­tors to ma­tri­ces $\in \mathbb{C}$, let $\mathbb{C}^{m \times n}$ define the space of all com­plex $m \times n$ ma­tri­ces, and then define the inner prod­uct on $\mathbb{C}^{m \times n}$

for which, this inner prod­uct can be rep­re­sent­ed as

giving us the same inner prod­uct for a matrix in $\mathbb{C}^{m \times n}$, with a vector in $\mathbb{C}^{mn}.$The norm for the cor­re­spond­ing $\mathbb{C}^{m \times n}$ space is called the Hilbert-Schmidt norm, rep­re­sent­ed by

which is just the square root of the Frobe­nius norm

The gen­er­a­tors of $U(3)$ are com­plex square ma­tri­ces of the form $\mathbb{C}^{m \times n}$, where $m = n.$With the inner prod­uct on $\mathbb{C}^{m \times n}$ de­fined, we en­deav­or to show that gen­er­a­tors and their prod­ucts are an in­fi­nite-di­men­sion­al space on the sphere. To do so, we must show that all the gen­er­a­tors and their prod­ucts are square-in­te­grable func­tions over the in­ter­val $\lbrack 0,2\pi\rbrack$, verify that our inner prod­uct space is a pre-Hilbert space, and show that it is com­plete.

De­f­i­n­i­tion (1.1.155) – Let $\mathcal{G}_{\mathbf{m}}\left( \left\lbrack \mathbf{0},\mathbf{2}\mathbf{\pi} \right\rbrack \right)$denote the space of $m \times m$ com­plex matrix func­tion gen­er­a­tors from $G\left( X_{j} \right)$ with an inner prod­uct on $\mathcal{G}_{m}$, as

for $G\left( X_{j} \right),G\left( X_{k} \right): \rightarrow \mathbb{C}^{m}$, as con­tin­u­ous func­tions, and $tr\left( G\left( X_{j} \right)^{*}G\left( X_{k} \right) \right)$ , defin­ing the trace of the prod­uct of the com­plex con­ju­gate trans­form of the ma­tri­ces $G\left( X_{j} \right)$ and $G\left( X_{k} \right)$. The com­ple­tion of $\mathcal{G}_{m}$ con­cern­ing the Hilbert-Schmidt norm

is the Hilbert Space, de­not­ed as $L^{2}\left( \mathbb{T}^{m\mathbb{C}} \right)$, where $\mathbb{T}^{m\mathbb{C}}$ sig­ni­fies these func­tions are on the sphere $\left( S^{m\mathbb{C}} \right)$ of di­men­sion $m\mathbb{C}$.

Lemma (1.1.156) – $\mathcal{G}_{m}$ is a pre-Hilbert space with a de­fined inner prod­uct $\left\langle G\left( X_{j} \right),G\left( X_{k} \right) \right\rangle = tr\left\langle G\left( X_{j} \right)^{*},G\left( X_{k} \right) \right\rangle$.

Proof – Ver­i­fi­ca­tion of the inner prod­uct space on $\mathcal{X}_{m}$ holds for all $x_{j} \in \mathcal{G}_{m}$, and $a,b\in \mathbb{C}$:

• $\left\langle x_{1},x_{2} \right\rangle = \overline{\left\langle x_{1},x_{2} \right\rangle}$ – Her­mit­ian Sym­met­ric

• $\left\langle x_{1},x_{1} \right\rangle \geq 0$ – Non-neg­a­tive

• $\left\langle x_{1},x_{1} \right\rangle = 0$, iff $x_{1} = 0$ – Pos­i­tive-def­i­nite

• $\left\langle x_{1},ax_{2} + bx_{3} \right\rangle = a\left\langle x_{1},x_{2} \right\rangle + b\left\langle x_{1},x_{3} \right\rangle$ – Linear in 2^nd ar­gu­ment

Taking the inner prod­uct of the gen­er­a­tors $G\left( X_{j} \right),G\left( X_{k} \right)$, we con­firm the inner prod­uct space on $\mathcal{G}_{m}$:

• As $X_{j}$ are skew-Her­mit­ian ma­tri­ces, we have that $e^{X_{j}} = G\left( X_{j} \right)$ are uni­tary ma­tri­ces, and as they are uni­tary, $G\left( X_{1} \right)^{*} = G\left( X_{1} \right)^{- 1}.$ As $X_{j}$ are sym­met­ric and uni­tary, we have that $G\overline{\left( X_{1} \right)} = G\left( X_{1} \right)^{- 1}$ is as well,

(Her­mit­ian sym­met­ric.)

• For uni­tary ma­tri­ces, $G\left( X_{1} \right)^{*}G\left( X_{1} \right) = G\left( X_{1} \right)G\left( X_{1} \right)^{*} = I_{m},$

(Non-neg­a­tive and pos­i­tive def­i­nite.)

• For uni­tary ma­tri­ces and scalars, $a,b\in \mathbb{C}$

(Lin­ear­i­ty in 2^nd ar­gu­ment.)

$\square$

Corol­lary (1.1.157) – $\mathcal{X}_{m}$ is a pre-Hilbert space with a de­fined inner prod­uct $\left\langle X_{j},X_{k} \right\rangle = tr\left\langle X_{j}^{*},X_{k} \right\rangle$.

Proof – Ver­i­fi­ca­tion of the inner prod­uct space on $\mathcal{X}_{m}$ holds for all $x_{j} \in \mathcal{X}_{m}$, and $a,b\in \mathbb{C}$:

• As $X_{j}$ are skew-Her­mit­ian ma­tri­ces and square-sym­met­ric about the main di­ag­o­nal, we have that $X_{j}^{*} = - X_{j},$ $X_{j}^{*} = \left( \overline{X_{j}} \right)^{T} = \overline{X_{j}^{T}},$and $X_{j}^{T} = X_{j},$where­as for the trace we know that $tr\left( X_{j}X_{k} \right) = tr\left( X_{k}X_{j} \right)$.

(Her­mit­ian sym­met­ric.)

• For the skew-Her­mit­ian ma­tri­ces, $X_{j},$

and$\ tr\left( X_{j}^{*}X_{j} \right) = 0$, only for $x_{j} = 0.$

(Non-neg­a­tive and pos­i­tive def­i­nite.)

• For skew-Her­mit­ian ma­tri­ces and scalars $a,b\in \mathbb{C}$

(Lin­ear­i­ty in 2^nd ar­gu­ment.)

$\square$

The com­ple­tion of $\mathcal{G}_{m}\left( \lbrack 0,2\pi\rbrack \right)$ as­so­ci­at­ed with the Hilbert-Schmidt norm

can also be as­so­ci­at­ed with the norm

for the matrix gen­er­a­tor $G\left( X_{j} \right)$, and for each gen­er­a­tor $G\left( X_{j} \right) \in U(3)$, we find

Lemma (1.1.161) – The matrix gen­er­a­tors $G\left( X_{j} \right)$ are a com­plete metric space on the in­ter­val $\lbrack 0,2\pi\rbrack$, since these matrix func­tions are square-in­te­grable on this in­ter­val

Proof – The gen­er­a­tors$\ G\left( X_{j} \right) \in U(3)$ are uni­tary, there­fore $\left| G\left( X_{j} \right) \right|^{2} = I_{3}$, and when in­te­grat­ed over the in­ter­val $\lbrack 0,2\pi\rbrack$, con­cern­ing $dx_{j}$, will be finite for all angles of $X_{j}.$

We have now shown the math­e­mat­ics to sup­port De­f­i­n­i­tion (1.1.139), where the com­ple­tion of $\mathcal{G}_{m}\left( \lbrack 0,2\pi\rbrack \right)$ con­cern­ing the Hilbert-Schmidt norm, de­not­ed by $L^{2}\left( \mathbb{T}^{m\mathbb{C}} \right)$, is a Hilbert space. To con­sid­er the in­ter­val of our space $\mathcal{G}_{m}\left( \lbrack 0,2\pi\rbrack \right)$, a de­f­i­n­i­tion of the inner prod­uct of com­plex matrix func­tions on $\mathcal{G}_{m}\left( \lbrack 0,2\pi\rbrack \right)$, that con­sis­tent with the or­thog­o­nal­i­ty re­la­tions of the gen­er­a­tor, is ex­pressed as

We in­te­grate the prod­uct $G\left( X_{j} \right)^{*}G\left( X_{k} \right)$ over both $x_{k}$ and$\ x_{j}$, on the or­thog­o­nal­ly sit­u­at­ed cir­cles of the $X_{k}/X_{j}$-planes, having in­ter­vals $\lbrack 0,2\pi\rbrack.$We would expect $\left\langle G\left( X_{j} \right),G\left( X_{j} \right) \right\rangle = 1$ for our unit gen­er­a­tors, and $j \neq k$ $\left\langle G\left( X_{j} \right),G\left( X_{k} \right) \right\rangle = 0$. The fol­low­ing two ex­am­ples nicely demon­strate this, with the first show­ing the inner prod­uct of $G\left( X_{1} \right)$ with itself, yields the iden­ti­ty

and for the second, we demon­strate the or­thog­o­nal re­la­tion be­tween the func­tions, $G\left( X_{1} \right),G\left( X_{2} \right)$, where we in­te­grate over both $x_{1}$ and $x_{2}$

The inner prod­uct on $\mathcal{G}_{m}\left( \lbrack 0,2\pi\rbrack \right)$ from De­f­i­n­i­tion (1.1.155), can also be ap­plied to vec­tors in the space, seen by set­ting ${\widehat{v}}_{1},{\widehat{v}}_{2}$ to be equal to

we can verify the or­thog­o­nal­i­ty con­di­tion be­tween any vector $\in X_{1}$, with an­oth­er $\in X_{2}$

1. Hunter and Nachter­gaele, "Ap­plied Analy­sis", World Sci­en­tif­ic, 2001