[proofplan] We prove both implications using the spectral definition of positivity. If $a$ is positive, then continuous functional calculus applied to the square-root function on $\sigma_A(a)\subset[0,\infty)$ produces an element $b=a^{1/2}$ with $b^*b=a$. Conversely, if $a=b^*b$, we faithfully represent $A$ on a [Hilbert space](/page/Hilbert%20Space), observe that the represented operator has nonnegative quadratic form, and use the Hilbert-space positivity criterion together with spectral permanence to transfer nonnegative spectrum back to $A$. [/proofplan]

[step:Construct a square root by functional calculus when $a$ is positive]Assume $a\ge 0$. By the spectral definition of positivity, $a=a^*$ and \begin{align*} \sigma_A(a)\subset [0,\infty). \end{align*} Define the [continuous function](/page/Continuous%20Function) \begin{align*} f:\sigma_A(a)&\to\mathbb C \end{align*} \begin{align*} t&\mapsto t^{1/2}. \end{align*} Since $a$ is self-adjoint, it is normal, so the continuous functional calculus for normal elements applies. Let \begin{align*} b:=f(a)\in A. \end{align*} The function $f$ is real-valued on $\sigma_A(a)$, hence $\overline f=f$. By the $*$-rule in the functional calculus, as in [citetheorem:8553], \begin{align*} b^*=f(a)^*=(\overline f)(a)=f(a)=b. \end{align*} The pointwise identity $f^2(t)=t$ for every $t\in\sigma_A(a)$ gives, by multiplicativity of the functional calculus, \begin{align*} b^*b=b^2=f(a)f(a)=(f^2)(a)=a. \end{align*} Thus $a=b^*b$.[/step]

[guided]Assume $a\ge 0$. In this proof, positivity is being interpreted spectrally: since $a$ is self-adjoint, $a\ge 0$ means \begin{align*} \sigma_A(a)\subset [0,\infty). \end{align*} This inclusion is exactly what makes the square-root function available on the whole spectrum. Define \begin{align*} f:\sigma_A(a)&\to\mathbb C \end{align*} \begin{align*} t&\mapsto t^{1/2}. \end{align*} The map $f$ is continuous because its domain is contained in $[0,\infty)$, where the usual square-root function is continuous. Since $a=a^*$, the element $a$ is normal. Therefore continuous functional calculus for normal elements applies to $a$. Set \begin{align*} b:=f(a)\in A. \end{align*} We now verify that this $b$ has the required algebraic property. First, $f$ is real-valued on $\sigma_A(a)$, so $\overline f=f$. The involution rule for functional calculus, recorded in [citetheorem:8553], gives \begin{align*} b^*=f(a)^*=(\overline f)(a)=f(a)=b. \end{align*} Thus $b$ is self-adjoint. Next, the pointwise identity \begin{align*} f^2(t)=t \end{align*} holds for every $t\in\sigma_A(a)$. The multiplication rule for functional calculus, again as in [citetheorem:8553], gives \begin{align*} b^*b=b^2=f(a)f(a)=(f^2)(a)=a. \end{align*} This constructs an element $b\in A$ satisfying $a=b^*b$.[/guided]

[step:Represent $A$ faithfully on a Hilbert space] Assume conversely that there exists $b\in A$ such that \begin{align*} a=b^*b. \end{align*} By the [citetheorem:8566], there exist a Hilbert space $H$ and a faithful unital $*$-representation \begin{align*} \pi:A\to \mathcal{L}(H). \end{align*} Define \begin{align*} B:=\pi(b)\in \mathcal{L}(H). \end{align*} Since $\pi$ is a $*$-homomorphism, \begin{align*} \pi(a)=\pi(b^*b)=\pi(b)^*\pi(b)=B^*B. \end{align*} Also, \begin{align*} a^*=(b^*b)^*=b^*b=a, \end{align*} so $a$ is self-adjoint, as required for the spectral definition of positivity. [/step]

[step:Show the represented operator has nonnegative spectrum]For every vector $\xi\in H$, \begin{align*} (\pi(a)\xi,\xi)_H=(B^*B\xi,\xi)_H=(B\xi,B\xi)_H=\|B\xi\|_H^2\ge 0. \end{align*} Hence $\pi(a)$ is a positive bounded operator on $H$ in the Hilbert-space quadratic-form sense. By the Hilbert-space positivity criterion for bounded [self-adjoint operators](/page/Self-Adjoint%20Operators), a bounded self-adjoint operator $T\in\mathcal{L}(H)$ with \begin{align*} (T\xi,\xi)_H\ge 0 \end{align*} for every $\xi\in H$ satisfies \begin{align*} \sigma_{\mathcal{L}(H)}(T)\subset [0,\infty). \end{align*} Applying this to $T=\pi(a)$ gives \begin{align*} \sigma_{\mathcal{L}(H)}(\pi(a))\subset [0,\infty). \end{align*}[/step]

[guided]We now translate the algebraic expression $a=b^*b$ into a Hilbert-space positivity statement. The faithful representation has already given us an operator \begin{align*} B:=\pi(b)\in\mathcal{L}(H), \end{align*} and the $*$-homomorphism property gives \begin{align*} \pi(a)=\pi(b^*b)=\pi(b)^*\pi(b)=B^*B. \end{align*} To prove positivity of $B^*B$, we test it on arbitrary vectors. Let $\xi\in H$. Using the defining property of the Hilbert-space adjoint, \begin{align*} (B^*B\xi,\xi)_H=(B\xi,B\xi)_H. \end{align*} The right-hand side is the square of the Hilbert-space norm of $B\xi$, so \begin{align*} (B^*B\xi,\xi)_H=\|B\xi\|_H^2\ge 0. \end{align*} Since $\pi(a)=B^*B$, this says \begin{align*} (\pi(a)\xi,\xi)_H\ge 0 \end{align*} for every $\xi\in H$. We now use the standard Hilbert-space positivity criterion for bounded self-adjoint operators: if $T\in\mathcal{L}(H)$ is self-adjoint and satisfies \begin{align*} (T\xi,\xi)_H\ge 0 \end{align*} for every $\xi\in H$, then \begin{align*} \sigma_{\mathcal{L}(H)}(T)\subset[0,\infty). \end{align*} The operator $\pi(a)$ is self-adjoint because \begin{align*} \pi(a)^*=\pi(a^*)=\pi(a), \end{align*} and the quadratic-form inequality has just been verified. Therefore \begin{align*} \sigma_{\mathcal{L}(H)}(\pi(a))\subset[0,\infty). \end{align*}[/guided]

[step:Transfer the spectral inclusion back to $A$] Because $\pi:A\to\mathcal{L}(H)$ is a faithful unital $*$-homomorphism, it identifies $A$ with the unital $C^*$-subalgebra $\pi(A)\subset\mathcal{L}(H)$. By spectral permanence for unital $C^*$-subalgebras, \begin{align*} \sigma_A(a)=\sigma_{\pi(A)}(\pi(a))=\sigma_{\mathcal{L}(H)}(\pi(a)). \end{align*} The previous step gives \begin{align*} \sigma_{\mathcal{L}(H)}(\pi(a))\subset [0,\infty), \end{align*} and hence \begin{align*} \sigma_A(a)\subset [0,\infty). \end{align*} Since $a=a^*$, the spectral definition of positivity gives $a\ge 0$. This proves the converse implication and completes the proof. [/step]