[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 $a=b^*b$.
[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]
[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*}
[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]
[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]
