[proofplan]
We work in the unital $C^*$-algebra $\mathcal{L}(H)$ of bounded operators on $H$, where the involution is the Hilbert-space adjoint and the positive elements are the Hermitian operators with non-negative spectrum. The candidate for $R$ is the unique positive square root of $TT^*$, which exists by the previous theorem (**Existence of Positive Square Roots**) once we verify $TT^*$ is positive. Invertibility of $T$ forces $R$ to be invertible, so we may set $U := R^{-1}T$. The unitarity of $U$ follows from a direct computation $UU^* = R^{-1}(TT^*)R^{-1} = R^{-1}R^2R^{-1} = I$, using $R = R^*$. Uniqueness: any decomposition $T = RU$ with $R$ positive and $U$ unitary forces $TT^* = R^2$, hence $R = (TT^*)^{1/2}$ uniquely, and then $U = R^{-1}T$.
[/proofplan]
[step:Verify $TT^*$ is positive]
The space $\mathcal{L}(H)$ is a unital $C^*$-algebra: the involution $A \mapsto A^*$ is the Hilbert-space adjoint, the unit is the identity operator $I$, and the $C^*$-identity $\|A^*A\| = \|A\|^2$ holds.
Set $S := TT^*$. Then $S^* = (TT^*)^* = (T^*)^* T^* = TT^* = S$, so $S$ is Hermitian.
For positivity, we use the characterization that an Hermitian operator $S \in \mathcal{L}(H)$ with $(Sx, x)_H \geq 0$ for all $x \in H$ has $\sigma(S) \subseteq [0, \infty)$. (This is the standard characterization of positive operators in a Hilbert space, following from the spectral theory of self-adjoint operators; equivalently, this can be taken as the working definition of a positive operator on $H$.)
Verify the inner-product positivity. For every $x \in H$:
\begin{align*}
(Sx, x)_H = (TT^*x, x)_H = (T^*x, T^*x)_H = \|T^*x\|_H^2 \geq 0,
\end{align*}
using the defining property of the adjoint $(Ay, z)_H = (y, A^*z)_H$ with $A = T$, $y = T^*x$, $z = x$.
Hence $S = TT^*$ is positive in $\mathcal{L}(H)$.
[/step]
[step:Construct $R := (TT^*)^{1/2}$ and verify $R$ is invertible]
By the previous theorem **Existence of Positive Square Roots**, applied to the unital $C^*$-algebra $\mathcal{L}(H)$ and the positive element $S = TT^*$ (verified in Step 1), there exists a unique positive operator $R \in \mathcal{L}(H)$ with $R^2 = S$. We write $R = S^{1/2} = (TT^*)^{1/2}$.
Note that $R$ is Hermitian (positive operators are Hermitian), so $R^* = R$.
*$R$ is invertible.* Since $T$ is invertible (hypothesis), so is its adjoint $T^*$ (in any unital $C^*$-algebra, $(T^{-1})^* = (T^*)^{-1}$, exhibiting an inverse for $T^*$). The product of two invertible operators is invertible, so $S = TT^* \in G(\mathcal{L}(H))$, i.e.\ $0 \notin \sigma(S)$.
We claim $0 \notin \sigma(R)$, hence $R$ is invertible. To see this, observe that $\sigma(R) \subseteq [0, \infty)$ (positivity of $R$). If $0 \in \sigma(R)$, then since the polynomial $p(z) = z^2$ maps $0$ to $0$, by the [Spectral Mapping Theorem for Polynomials](/theorems/2671) applied to the unital Banach algebra $\mathcal{L}(H)$, the element $R \in \mathcal{L}(H)$, and $p(z) = z^2$:
\begin{align*}
\sigma(R^2) = p(\sigma(R)) = \{\lambda^2 : \lambda \in \sigma(R)\}.
\end{align*}
Theorem 2671 requires only a unital Banach algebra and a polynomial; both hypotheses are met. Then $0 = p(0) \in \sigma(R^2) = \sigma(S)$, contradicting $0 \notin \sigma(S)$. Hence $0 \notin \sigma(R)$, i.e.\ $R \in G(\mathcal{L}(H))$.
[/step]
[step:Define $U := R^{-1}T$ and verify $U$ is unitary]
Set
\begin{align*}
U := R^{-1} T \in \mathcal{L}(H).
\end{align*}
This is well-defined because $R$ is invertible (Step 2). By definition, $RU = R \cdot R^{-1}T = T$, so $T = RU$.
We verify $U$ is unitary, i.e.\ $UU^* = U^*U = I$.
Compute $U^*$:
\begin{align*}
U^* = (R^{-1}T)^* = T^* (R^{-1})^* = T^* (R^*)^{-1} = T^* R^{-1},
\end{align*}
using the standard identities $(AB)^* = B^*A^*$ and $(A^{-1})^* = (A^*)^{-1}$ in any unital $C^*$-algebra (the second holds because $(A^*)(A^{-1})^* = (A^{-1}A)^* = I^* = I$ and similarly on the other side), together with $R^* = R$ from Step 2.
Compute $UU^*$:
\begin{align*}
UU^* = R^{-1}T \cdot T^* R^{-1} = R^{-1} (TT^*) R^{-1} = R^{-1} R^2 R^{-1} = (R^{-1}R)(RR^{-1}) = I \cdot I = I,
\end{align*}
using $TT^* = R^2$ (Step 2) and associativity.
Compute $U^*U$:
\begin{align*}
U^*U = T^* R^{-1} \cdot R^{-1} T = T^* R^{-2} T = T^* (TT^*)^{-1} T.
\end{align*}
Now $(TT^*)^{-1} = (T^*)^{-1} T^{-1}$ (the standard formula $(AB)^{-1} = B^{-1}A^{-1}$ for invertible $A, B$, applied to $A = T$, $B = T^*$, both invertible). Hence
\begin{align*}
U^*U = T^* (T^*)^{-1} T^{-1} T = I \cdot I = I.
\end{align*}
Both $UU^* = I$ and $U^*U = I$ hold, so $U$ is unitary in $\mathcal{L}(H)$. Combined with Step 2 ($R$ positive) and $T = RU$, this completes the existence proof.
[guided]
The candidate for $R$ is forced by the desired equation $T = RU$. *Why?* If $T = RU$ with $R = R^*$ and $U$ unitary, then
\begin{align*}
TT^* = (RU)(RU)^* = RUU^*R^* = RIR = R^2.
\end{align*}
So $R$ must be a square root of $TT^*$. Among square roots in a $C^*$-algebra, *positive* ones are unique (the previous theorem **Existence of Positive Square Roots**), so we are forced to take $R = (TT^*)^{1/2}$.
**Why is $TT^*$ positive?** We need this to invoke the square-root theorem. Self-adjointness $TT^* = (TT^*)^*$ is immediate. Spectral non-negativity follows from the inner-product non-negativity $(TT^*x, x)_H = \|T^*x\|^2 \geq 0$ via the standard equivalence in $\mathcal{L}(H)$ between inner-product positivity and spectral positivity (a Hermitian operator is positive iff $(Ax, x) \geq 0$ for all $x$).
**Why is $R$ invertible?** Because $T$ is invertible by hypothesis. So $T^*$ is invertible (the involution permutes the unit group), so $TT^* = R^2$ is invertible. Now use the Spectral Mapping Theorem: $\sigma(R^2) = \{\lambda^2 : \lambda \in \sigma(R)\}$, so $0 \in \sigma(R)$ would force $0 \in \sigma(R^2)$, contradicting invertibility of $R^2$.
**Why is $U := R^{-1}T$ unitary?** We compute $UU^*$ directly:
\begin{align*}
UU^* = R^{-1}T \cdot T^* R^{-1} = R^{-1} (TT^*) R^{-1} = R^{-1} R^2 R^{-1} = I,
\end{align*}
where we used $R^* = R$ to write $U^* = T^* R^{-1}$, and $TT^* = R^2$ (the defining property of $R$). The same computation in the other order gives $U^*U = I$, but we need to use that $TT^* = T^*T$? No — we use that the inverse of $TT^*$ equals $(T^*)^{-1}T^{-1}$ (general invertibility, no normality required).
**Why this construction has no analogue if $T$ is not invertible.** The whole argument hinges on $R$ being invertible, which in turn hinges on $TT^* \in G(\mathcal{L}(H))$. If $T$ is not invertible, $TT^*$ may have $0$ in its spectrum, so $R = (TT^*)^{1/2}$ has $0$ in its spectrum, and $U := R^{-1}T$ is not defined. (The general polar decomposition of non-invertible $T$ exists with $U$ a partial isometry instead of a unitary, but is a more delicate construction.)
[/guided]
[/step]
[step:Prove uniqueness via the forcing identity $TT^* = R^2$]
Suppose $T = RU = R'U'$ with $R, R'$ positive and $U, U'$ unitary in $\mathcal{L}(H)$.
Compute $TT^*$ in two ways. From $T = RU$:
\begin{align*}
TT^* = RU(RU)^* = RU \cdot U^*R^* = R(UU^*)R = R \cdot I \cdot R = R^2,
\end{align*}
using $U^* = U^{-1}$ (unitarity), $R^* = R$ (positive elements are Hermitian), and associativity. The same calculation gives $TT^* = R'^2$.
Hence $R^2 = R'^2 = TT^*$, with $R$ and $R'$ both positive. By the previous theorem **Existence of Positive Square Roots** (uniqueness clause), the positive square root of $TT^*$ in $\mathcal{L}(H)$ is unique, so $R = R'$.
Now $R$ is invertible (Step 2 applied to the common $R$), so from $T = RU = RU'$ we get $U = R^{-1}T = U'$.
This proves uniqueness. Combined with existence (Steps 1-3), the theorem is proved.
[/step]
