Proof of Finite Direct Sums of a Complex Hilbert Space: the Hilbert Structure, Coordinate Inclusions, Block Entries of Bounded Operators and Commutation with Diagonal Operators

For $\zeta\in H^{m}$ and $k\in[m]$, $\zeta_{k}$ is the $k$-th component of $\zeta$; by definition of the operations, $(\zeta+\zeta')_{k}=\zeta_{k}+\zeta'_{k}$ and $(\lambda\zeta)_{k}=\lambda\zeta_{k}$ for $\zeta,\zeta'\in H^{m}$ and $\lambda\in\mathbb{C}$, and two elements of $H^{m}$ are equal when all their components are equal. For $k\in[m]$ let $\pi_{k}:H^{m}\to H$ be the map $\zeta\mapsto\zeta_{k}$; it is linear by the preceding formulas.

1. (Hilbert structure) Vector space. Each condition of Vector Space over a Field for $H^{m}$, read in the $k$-th component, is the same condition for $H$; the tuple all of whose components are $0$ is a zero vector of $H^{m}$, and $(-\zeta_{1},\dots,-\zeta_{m})$ is an additive inverse of $\zeta$. So $H^{m}$ is a complex vector space.

Inner product. Let $\zeta,\zeta',\zeta''\in H^{m}$ and $\lambda\in\mathbb{C}$. By condition 1 of Complex Inner Product Space for $H$ and claim 4 of Properties of Finite Sums,

By conditions 2 and 3 of Complex Inner Product Space for $H$ and claims 2 and 3 of Properties of Finite Sums,

The summands $\langle\zeta_{k},\zeta_{k}\rangle_{H}=\lVert\zeta_{k}\rVert_{H}^{2}$ of $\langle\zeta,\zeta\rangle_{H^{m}}$ are nonnegative real numbers. Since $\mathbb{R}\subseteq\mathbb{C}$ is closed under addition, the map $\sigma:[m]\to\mathbb{C}$ of Finite Sum Notation in a Field, formed in $\mathbb{C}$ for these summands, takes real values: the set of $j\in\mathbb{N}$ such that $j\notin[m]$ or $\sigma(j)\in\mathbb{R}$ contains $1$ and, with $j$, contains $S(j)$ (if $S(j)\in[m]$ then $j\in[m]$ and $\sigma(S(j))=\sigma(j)+\lVert\zeta_{S(j)}\rVert_{H}^{2}$), so it is $\mathbb{N}$ by Principle of Induction for the Natural Numbers; by the uniqueness in Existence and Uniqueness of Iterates of a Binary Operation it is then also the map formed in $\mathbb{R}$, so $\langle\zeta,\zeta\rangle_{H^{m}}$ is the finite sum of the $\lVert\zeta_{k}\rVert_{H}^{2}$ in $\mathbb{R}$. By claim 5 of Properties of Finite Sums it is nonnegative, and if it is $0$ then $\lVert\zeta_{k}\rVert_{H}^{2}=0$ for every $k\in[m]$, so $\zeta_{k}=0$ by claim 4 of Elementary Properties of a Complex Inner Product, that is, $\zeta=0$. Hence $\langle\cdot,\cdot\rangle_{H^{m}}$ is an inner product, and by Norm Induced by a Complex Inner Product

Consequently, by claim 6 of Properties of Finite Sums and claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field,

Completeness. Let $(\zeta^{(n)})_{n\in\mathbb{N}}$ be a Cauchy sequence in $H^{m}$ and let $k\in[m]$. The $k$-th component of $\zeta^{(n)}-\zeta^{(p)}$ is $\zeta^{(n)}_{k}-\zeta^{(p)}_{k}$, so by (1.1) $\lVert\zeta^{(n)}_{k}-\zeta^{(p)}_{k}\rVert_{H}\le\lVert\zeta^{(n)}-\zeta^{(p)}\rVert_{H^{m}}$ for all $n,p\in\mathbb{N}$, and $(\zeta^{(n)}_{k})_{n\in\mathbb{N}}$ is a Cauchy sequence in $H$. Since $H$ is a complex Hilbert space, it is complete, so this sequence converges to some $\zeta_{k}\in H$, which is unique by Uniqueness of Limits in a Metric Space. Let $\zeta=(\zeta_{1},\dots,\zeta_{m})\in H^{m}$.

Let $\varepsilon>0$ and put $\delta=\varepsilon/(4m)$. First choose $N\in\mathbb{N}$ with $\lVert\zeta^{(n)}-\zeta^{(p)}\rVert_{H^{m}}<\delta$ for all $n,p\ge N$. Let $n\ge N$ and $k\in[m]$. Next choose $N_{k}\in\mathbb{N}$ with $\lVert\zeta^{(p)}_{k}-\zeta_{k}\rVert_{H}<\delta$ for all $p\ge N_{k}$, and let $p$ be the larger of $N$ and $N_{k}$ (claim 3 of Properties of the Order on the Natural Numbers). By the triangle inequality (claim 2 of The Induced Norm is a Norm, and Induces a Metric) and (1.1),

so $\lVert\zeta^{(n)}_{k}-\zeta_{k}\rVert_{H}^{2}<4\delta^{2}$ by claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. The $k$-th component of $\zeta^{(n)}-\zeta$ is $\zeta^{(n)}_{k}-\zeta_{k}$; by the norm formula, claims 2, 3 and 5 of Properties of Finite Sums (applied to the nonnegative differences $4\delta^{2}-\lVert\zeta^{(n)}_{k}-\zeta_{k}\rVert_{H}^{2}$) and the recursion in claim 1 of that lemma (which gives $\sum_{k=1}^{m}4\delta^{2}=4m\delta^{2}$),

so $\lVert\zeta^{(n)}-\zeta\rVert_{H^{m}}<\varepsilon$ by claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. As $n\ge N$ was arbitrary, $(\zeta^{(n)})$ converges to $\zeta$. Hence $H^{m}$ is complete, and it is a complex Hilbert space.

Two remarks. From now on we use the following facts. (R1) For a map $j\mapsto\xi^{(j)}$ from $[n]$ to $H^{m}$ and $k\in[m]$, the $k$-th component of $\sum_{j=1}^{n}\xi^{(j)}$ is $\sum_{j=1}^{n}\xi^{(j)}_{k}$, by claim 4 of Properties of Finite Sums of Vectors applied to the linear map $\pi_{k}$. (R2) Let $V,W,U$ be among $H$ and $H^{m}$. For fixed $v\in V$ the map $R\mapsto Rv$ from $\mathcal{L}(V,W)$ to $W$ is linear, because sums and scalar multiples in $\mathcal{L}(V,W)$ are formed pointwise; and for fixed $P\in\mathcal{L}(W,U)$ and $P'\in\mathcal{L}(U,V)$ the maps $R\mapsto PR$ and $R\mapsto RP'$ are linear, because composition distributes over sums and commutes with scalar multiples (Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §operations). So, by claim 4 of Properties of Finite Sums of Vectors, finite sums in $\mathcal{L}(V,W)$ may be evaluated termwise at $v$ and composed termwise with $P$ and $P'$.

2. (Coordinate inclusions) Let $k,l\in[m]$. For $v,w\in H$ and $\lambda\in\mathbb{C}$, the tuples $\iota_{k}(v+\lambda w)$ and $\iota_{k}v+\lambda\iota_{k}w$ both have $k$-th component $v+\lambda w$ and all other components $0$, so $\iota_{k}$ is linear. By the norm formula of claim 1 and claim 7 of Properties of Finite Sums (the summands with index $j\ne k$ are $\lVert0\rVert_{H}^{2}=0$), $\lVert\iota_{k}v\rVert_{H^{m}}^{2}=\lVert v\rVert_{H}^{2}$, so $\lVert\iota_{k}v\rVert_{H^{m}}=\lVert v\rVert_{H}$ by claim 3 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field and $1$ is a bound for $\iota_{k}$; thus $\iota_{k}\in\mathcal{L}(H,H^{m})$.

For $\zeta\in H^{m}$ and $v\in H$, claim 3 of Elementary Properties of a Complex Inner Product makes the summands with index $j\ne k$ of $\langle\zeta,\iota_{k}v\rangle_{H^{m}}$ equal to $\langle\zeta_{j},0\rangle_{H}=0$, so by claim 7 of Properties of Finite Sums

Thus the linear map $\pi_{k}$ is an adjoint of $\iota_{k}$, and by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-unique $\iota_{k}^{*}=\pi_{k}$, that is, $\iota_{k}^{*}\zeta=\zeta_{k}$; also $\iota_{k}^{*}\in\mathcal{L}(H^{m},H)$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint, as $H$ is a complex Hilbert space. Hence $\iota_{k}^{*}\iota_{l}v=(\iota_{l}v)_{k}$ is $v$ if $k=l$ and $0$ if $k\ne l$, so $\iota_{k}^{*}\iota_{l}=I_{H}$ if $k=l$ and $\iota_{k}^{*}\iota_{l}=0$ if $k\ne l$. Finally let $\zeta\in H^{m}$ and $j\in[m]$. By (R2) and (R1), the $j$-th component of $\bigl(\sum_{k=1}^{m}\iota_{k}\iota_{k}^{*}\bigr)\zeta=\sum_{k=1}^{m}\iota_{k}\zeta_{k}$ is $\sum_{k=1}^{m}(\iota_{k}\zeta_{k})_{j}$, whose summands with $k\ne j$ are $0$; by claim 7 of Properties of Finite Sums of Vectors it equals $\zeta_{j}$. Hence $\sum_{k=1}^{m}\iota_{k}\iota_{k}^{*}=I_{H^{m}}$.

3. (Block entries) By claim 1, $H^{m}$ is a complex Hilbert space, so every $T\in\mathcal{L}(H^{m})$ has an adjoint $T^{*}\in\mathcal{L}(H^{m})$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint; and $T_{kl}=\iota_{k}^{*}T\iota_{l}\in\mathcal{L}(H)$ by claim 2 and Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §operations. Let $S,T\in\mathcal{L}(H^{m})$, $\zeta,\zeta'\in H^{m}$ and $k,l\in[m]$.

Components. By claim 2 and (R2),

Inner products. By the definition of $\langle\cdot,\cdot\rangle_{H^{m}}$, the formula just proved and claim 5 of Properties of Finite Sums of Vectors,

Adjoints. By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus, $\iota_{l}$ is the adjoint of $\iota_{l}^{*}$, and, applying the rule for composites twice, $T_{lk}=\iota_{l}^{*}(T\iota_{k})$ has the adjoint $(T\iota_{k})^{*}\iota_{l}=\iota_{k}^{*}T^{*}\iota_{l}$. By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-unique, $(T_{lk})^{*}=\iota_{k}^{*}T^{*}\iota_{l}=(T^{*})_{kl}$.

Products. By claim 2 and (R2),

Converse. Let $(A_{kl})_{k,l\in[m]}$ be a family in $\mathcal{L}(H)$ and $T=\sum_{k=1}^{m}\sum_{l=1}^{m}\iota_{k}A_{kl}\iota_{l}^{*}$, an element of the complex vector space $\mathcal{L}(H^{m})$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §operations. Let $p,q\in[m]$. By (R2), $T_{pq}=\sum_{k=1}^{m}\sum_{l=1}^{m}\iota_{p}^{*}\iota_{k}A_{kl}\iota_{l}^{*}\iota_{q}$. For fixed $k$, the summands with $l\ne q$ vanish since $\iota_{l}^{*}\iota_{q}=0$ by claim 2, so by claim 7 of Properties of Finite Sums of Vectors the inner sum is $\iota_{p}^{*}\iota_{k}A_{kq}$; in the same way, since $\iota_{p}^{*}\iota_{k}=0$ for $k\ne p$ and $\iota_{p}^{*}\iota_{p}=I_{H}$, the outer sum is $A_{pq}$. So $T_{pq}=A_{pq}$. If also $T'\in\mathcal{L}(H^{m})$ satisfies $T'_{kl}=A_{kl}$ for all $k,l\in[m]$, then by the component formula $(T'\zeta)_{k}=\sum_{j=1}^{m}A_{kj}\zeta_{j}=(T\zeta)_{k}$ for all $\zeta\in H^{m}$ and $k\in[m]$, so $T'=T$.

4. (Diagonal operators) Let $A\in\mathcal{L}(H)$, $\zeta\in H^{m}$ and $j\in[m]$. By (R2), (R1) and claim 2, the $j$-th component of $\bigl(\sum_{k=1}^{m}\iota_{k}A\iota_{k}^{*}\bigr)\zeta$ is $\sum_{k=1}^{m}(\iota_{k}A\zeta_{k})_{j}$, whose summands with $k\ne j$ are $0$, so by claim 7 of Properties of Finite Sums of Vectors it equals $A\zeta_{j}=(A^{(m)}\zeta)_{j}$. Hence

which lies in $\mathcal{L}(H^{m})$ by claim 2 and Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §operations. Let $D_{kl}=A$ if $k=l$ and $D_{kl}=0$ if $k\ne l$. For fixed $k$ the summands $\iota_{k}D_{kl}\iota_{l}^{*}$ with $l\ne k$ are $0$, so by claim 7 of Properties of Finite Sums of Vectors $\sum_{k=1}^{m}\sum_{l=1}^{m}\iota_{k}D_{kl}\iota_{l}^{*}=A^{(m)}$, and the converse part of claim 3 gives $(A^{(m)})_{kl}=D_{kl}$ for all $k,l\in[m]$.

Adjoint. By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint, $A^{*}\in\mathcal{L}(H)$, so by what was just shown $(A^{*})^{(m)}\in\mathcal{L}(H^{m})$ has the entry $A^{*}$ at $(k,k)$ and $0$ at $(k,l)$ with $k\ne l$. The zero map is its own adjoint, since $\langle0w,v\rangle=0=\langle w,0v\rangle$ by claim 3 of Elementary Properties of a Complex Inner Product, so by the adjoint formula of claim 3 and Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-unique, $((A^{(m)})^{*})_{kl}=((A^{(m)})_{lk})^{*}$ is $A^{*}$ if $k=l$ and $0$ if $k\ne l$. By the uniqueness in the converse part of claim 3, $(A^{(m)})^{*}=(A^{*})^{(m)}$.

Commutation. Let $T\in\mathcal{L}(H^{m})$ and $k,l\in[m]$. By the product formula of claim 3, $(TA^{(m)})_{kl}=\sum_{j=1}^{m}T_{kj}(A^{(m)})_{jl}$ and $(A^{(m)}T)_{kl}=\sum_{j=1}^{m}(A^{(m)})_{kj}T_{jl}$; the summands with $j\ne l$, respectively $j\ne k$, are composites with the zero map and hence $0$, so by claim 7 of Properties of Finite Sums of Vectors

If $TA^{(m)}=A^{(m)}T$, then these entries agree, so $T_{kl}$ commutes with $A$ for all $k,l\in[m]$. Conversely, if $T_{kl}A=AT_{kl}$ for all $k,l\in[m]$, then $TA^{(m)}$ and $A^{(m)}T$ have the same entries and are equal by the uniqueness in the converse part of claim 3.