Proof of Square Roots of Positive Bounded Operators on a Complex Hilbert Space, Commuting with Everything that Commutes with the Operator

We use two elementary facts. (Z) If $R\in\mathcal{L}(H)$ and $\lVert R\rVert_{\mathrm{op}}=0$, then $R=0$: by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound $\lVert Rv\rVert\le0$, so $Rv=0$ for every $v\in H$ by positivity of the norm (claim 2 of The Induced Norm is a Norm, and Induces a Metric). (D) For $R,R'\in\mathcal{L}(H)$, $\lVert R-R'\rVert_{\mathrm{op}}=\lVert(-1)(R'-R)\rVert_{\mathrm{op}}=\lVert R'-R\rVert_{\mathrm{op}}$ 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; and $\lVert0\rVert_{\mathrm{op}}=0$, since $0$ is a bound for the zero map and so $0\le\lVert0\rVert_{\mathrm{op}}\le0$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound. By Complex Hilbert Spaces and Bounded Linear Maps: Standing Notation §maps, $T\ge0$ means that $T\in\mathcal{L}(H)$ is self-adjoint and positive semi-definite.

The case $T=0$. Take $S=0\in\mathcal{L}(H)$. By claim 3 of Elementary Properties of a Complex Inner Product, $\langle0u,v\rangle=0=\langle u,0v\rangle$ for all $u,v\in H$, so $S$ is self-adjoint and positive semi-definite, i.e. $S\ge0$; $SS=0=T$; and $B0=0=0B$ for every $B\in\mathcal{L}(H)$. So claims 1 and 2 hold. From now on let $T\ne0$.

Normalisation. Let $c=\lVert T\rVert_{\mathrm{op}}$. By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound, $c$ is a bound for $T$, so $c\ge0$, and $c\ne0$ by (Z); thus $c>0$. Let $Q=c^{-1}T\in\mathcal{L}(H)$. 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-calculus, $T$ is an adjoint of itself, so $Q$ has the adjoint $\overline{c^{-1}}\,T=c^{-1}T=Q$ (claim 1 of Properties of Complex Conjugation and Modulus), and $Q$ is self-adjoint by the same clause. 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, $\lVert Q\rVert_{\mathrm{op}}=c^{-1}c=1$. For $v\in H$, $\langle v,Qv\rangle=c^{-1}\langle v,Tv\rangle$ is a nonnegative real number, so by claim 8 of Properties of Complex Conjugation and Modulus, the Cauchy-Schwarz inequality of Cauchy-Schwarz Inequality in a Complex Inner Product Space in the norm form of claim 1 of The Induced Norm is a Norm, and Induces a Metric and Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound,

Let $A=I-Q\in\mathcal{L}(H)$. Since $I$ and $Q$ are their own adjoints, $A$ is its own 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-calculus, hence self-adjoint; and $\langle v,Av\rangle=\lVert v\rVert^{2}-\langle v,Qv\rangle$ is a real number with $0\le\langle v,Av\rangle\le\lVert v\rVert^{2}$. So $|\langle v,Av\rangle|\le\lVert v\rVert^{2}$ for every $v\in H$ (claim 8 of Properties of Complex Conjugation and Modulus), and Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §quadratic gives $\lVert A\rVert_{\mathrm{op}}\le1$.

Let $\mathcal{C}=\{B\in\mathcal{L}(H):BA=AB\}$; note $A\in\mathcal{C}$.

(1) A map $B\in\mathcal{L}(H)$ commutes with $T$ if and only if $B\in\mathcal{C}$. 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, $BA=B-c^{-1}BT$ and $AB=B-c^{-1}TB$. In the vector space $\mathcal{L}(H)$ these are equal if and only if $c^{-1}BT=c^{-1}TB$, that is (multiplying by $c$), if and only if $BT=TB$.

The iteration. By Definition of Sequences by Recursion on the Natural Numbers §recursion, applied on $\mathcal{L}(H)$ with $a=0$ and $f(n,Y)=\tfrac12(A+YY)$ (which lies in $\mathcal{L}(H)$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §operations), and on $\mathbb{R}$ with $a=0$ and $f(n,y)=\tfrac12(1+y^{2})$, there are sequences $(X_{n})_{n\in\mathbb{N}}$ in $\mathcal{L}(H)$ and $(x_{n})_{n\in\mathbb{N}}$ in $\mathbb{R}$ with

where $n+1=S(n)$ by Natural Numbers.

(2) For every $n\in\mathbb{N}$, $X_{n}B=BX_{n}$ for every $B\in\mathcal{C}$, and $X_{n}$ is self-adjoint. Let $P$ be the set of $n\in\mathbb{N}$ for which both statements hold; we show $P=\mathbb{N}$ by Principle of Induction for the Natural Numbers. The zero map commutes with every $B$ and is self-adjoint (as in the case $T=0$), so $1\in P$. Let $n\in P$ and $B\in\mathcal{C}$. Since 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),

As $A$ and $X_{n}$ are their own adjoints, 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 shows successively that $X_{n}X_{n}$, $A+X_{n}X_{n}$ and $X_{n+1}$ are their own adjoints, so $X_{n+1}$ is self-adjoint by the same clause. Thus $n+1\in P$, and $P=\mathbb{N}$. Taking $B=A$ shows $X_{p}\in\mathcal{C}$ for every $p\in\mathbb{N}$; hence $X_{n}X_{p}=X_{p}X_{n}$ for all $n,p\in\mathbb{N}$.

(3) For every $n\in\mathbb{N}$, $0\le x_{n}\le1$, $x_{n}\le x_{n+1}$ and $\lVert X_{n}\rVert_{\mathrm{op}}\le x_{n}$. The set of $n$ with $0\le x_{n}\le1$ contains $1$, and if $0\le x_{n}\le1$ then $0\le x_{n}^{2}\le1$ and $\tfrac12\le x_{n+1}\le1$; so by Principle of Induction for the Natural Numbers $0\le x_{n}\le1$ for all $n$. Moreover $x_{n+1}-x_{n}=\tfrac12(1-x_{n})^{2}\ge0$. For the norms, let $P'$ be the set of $n\in\mathbb{N}$ with $\lVert X_{n}\rVert_{\mathrm{op}}\le x_{n}$. By (D), $1\in P'$. If $n\in P'$, then $\lVert X_{n}\rVert_{\mathrm{op}}^{2}\le x_{n}^{2}$ by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, 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 §operations

so $n+1\in P'$, and $P'=\mathbb{N}$ by Principle of Induction for the Natural Numbers.

(4) For every $n\in\mathbb{N}$, $\lVert X_{n+1}-X_{n}\rVert_{\mathrm{op}}\le x_{n+1}-x_{n}$. Let $P''$ be the set of $n\in\mathbb{N}$ for which this holds. Since $X_{2}-X_{1}=\tfrac12A$ and $x_{2}-x_{1}=\tfrac12$, Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §operations gives $1\in P''$. Let $n\in P''$. By (2), $X_{n+1}X_{n}=X_{n}X_{n+1}$, so 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

and, using $n\in P''$, (3) and the nonnegativity of all factors,

So $n+1\in P''$, and $P''=\mathbb{N}$ by Principle of Induction for the Natural Numbers.

(5) For $n,p\in\mathbb{N}$ with $n<p$, $\lVert X_{p}-X_{n}\rVert_{\mathrm{op}}\le x_{p}-x_{n}$. Fix $n$ and let $P_{n}$ be the set of $j\in\mathbb{N}$ with $\lVert X_{n+j}-X_{n}\rVert_{\mathrm{op}}\le x_{n+j}-x_{n}$. By (4), $1\in P_{n}$. If $j\in P_{n}$, then 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 and (4) at $n+j$,

so $j+1\in P_{n}$ and $P_{n}=\mathbb{N}$ by Principle of Induction for the Natural Numbers. Every $p>n$ has the form $n+j$ with $j\in\mathbb{N}$ by claim 7 of Properties of the Order on the Natural Numbers.

Convergence. By (3), $(x_{n})$ is nondecreasing and its set of terms is bounded above by $1$, so by A Bounded Monotone Sequence of Real Numbers Converges §nondecreasing it converges to $s=\sup\{x_{n}:n\in\mathbb{N}\}$, and $x_{n}\le s$ for every $n$. Let $\varepsilon>0$ and choose $N\in\mathbb{N}$ with $|x_{n}-s|<\varepsilon$ for all $n\ge N$. Let $k,l\ge N$. If $k=l$, then $\lVert X_{k}-X_{l}\rVert_{\mathrm{op}}=0<\varepsilon$ by (D). If $l<k$, then by (5)

if $k<l$, the same holds by (D) with $k$ and $l$ exchanged. By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §complete there is $X\in\mathcal{L}(H)$ such that $\lVert X_{n}-X\rVert_{\mathrm{op}}\to0$.

Properties of $X$. In each of (6)–(9) below, let $\varepsilon>0$; first fix the number $\eta>0$ named there, and then choose $N\in\mathbb{N}$ with $\lVert X_{p}-X\rVert_{\mathrm{op}}<\eta$ for all $p\ge N$. Norms and composites are estimated 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.

(6) $\lVert X\rVert_{\mathrm{op}}\le1$. With $\eta=\varepsilon$, by (D) and (3), $\lVert X\rVert_{\mathrm{op}}\le\lVert X-X_{N}\rVert_{\mathrm{op}}+\lVert X_{N}\rVert_{\mathrm{op}}<\varepsilon+1$. As $\varepsilon>0$ was arbitrary, $\lVert X\rVert_{\mathrm{op}}\le1$ by Comparison of Real Numbers with Arbitrary Positive Slack §slack-above.

(7) $X$ is self-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, $X$ has an adjoint $X^{*}\in\mathcal{L}(H)$ with $\lVert X^{*}-X_{N}\rVert_{\mathrm{op}}=\lVert X-X_{N}\rVert_{\mathrm{op}}$: indeed, as $X_{N}$ is its own adjoint by (2), 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 shows that $X^{*}-X_{N}$ is the adjoint of $X-X_{N}$, so the equality follows from 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 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. With $\eta=\varepsilon/2$ and (D),

By Comparison of Real Numbers with Arbitrary Positive Slack §vanishing, $\lVert X^{*}-X\rVert_{\mathrm{op}}=0$, so $X^{*}=X$ by (Z); thus $X$ is an adjoint of itself and is self-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-calculus.

(8) $XB=BX$ for every $B\in\mathcal{C}$. Let $B\in\mathcal{C}$ and $\eta=\varepsilon/(2\lVert B\rVert_{\mathrm{op}}+1)$. By (2), $BX_{N}=X_{N}B$, so $BX-XB=B(X-X_{N})+(X_{N}-X)B$ and, by (D),

By Comparison of Real Numbers with Arbitrary Positive Slack §vanishing and (Z), $BX=XB$.

(9) $X=\tfrac12(A+XX)$. Let $Y=\tfrac12(A+XX)\in\mathcal{L}(H)$ and $\eta=\varepsilon/2$. Since $X_{N}X_{N}-XX=X_{N}(X_{N}-X)+(X_{N}-X)X$, (3) and (6) give $\lVert X_{N}X_{N}-XX\rVert_{\mathrm{op}}\le(\lVert X_{N}\rVert_{\mathrm{op}}+\lVert X\rVert_{\mathrm{op}})\lVert X_{N}-X\rVert_{\mathrm{op}}\le2\lVert X_{N}-X\rVert_{\mathrm{op}}$. As $X_{N+1}-Y=\tfrac12(X_{N}X_{N}-XX)$ and $N+1\ge N$, by (D)

By Comparison of Real Numbers with Arbitrary Positive Slack §vanishing and (Z), $X=Y$.

The square root of $Q$. Let $S_{0}=I-X\in\mathcal{L}(H)$. As $I$ and $X$ (by (7)) are their own adjoints, so is $S_{0}$ 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; hence $S_{0}$ is self-adjoint. For $v\in H$, self-adjointness of $X$ and condition 1 of Complex Inner Product Space give $\langle v,Xv\rangle=\langle Xv,v\rangle=\overline{\langle v,Xv\rangle}$, so $\langle v,Xv\rangle$ is real by claim 1 of Properties of Complex Conjugation and Modulus. Then, by claim 8 of Properties of Complex Conjugation and Modulus, the Cauchy-Schwarz inequality of Cauchy-Schwarz Inequality in a Complex Inner Product Space in the norm form of claim 1 of The Induced Norm is a Norm, and Induces a Metric, Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound and (6),

so $S_{0}\ge0$. By (9), $XX=2X-A$, so 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

For $B\in\mathcal{C}$, (8) gives $BS_{0}=B-BX=B-XB=S_{0}B$.

1. (Square root) By Existence and Uniqueness of the Nonnegative Square Root there is a real $r\ge0$ with $r^{2}=c$. Let $S=rS_{0}\in\mathcal{L}(H)$. 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-calculus, $S$ has the adjoint $\overline{r}S_{0}=rS_{0}=S$ (claim 1 of Properties of Complex Conjugation and Modulus), so $S$ is self-adjoint; and $\langle v,Sv\rangle=r\langle v,S_{0}v\rangle$ is a nonnegative real number for every $v\in H$. Thus $S\ge0$, 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 §operations

2. (Commutation) Let $B\in\mathcal{L}(H)$ commute with $T$. By (1), $B\in\mathcal{C}$, so $BS_{0}=S_{0}B$, 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 §operations $BS=r\,BS_{0}=r\,S_{0}B=SB$.