Exact single-qubit Clifford+T ring characterization

If $U$ is a finite Clifford+T circuit matrix, [F1] puts all its entries in $D$.

Conversely, let $U\in U(2)$ have entries in $D$. By [F2] its first column $(u,v{)}^T$ is a normalized column. Apply the descent in [F3] repeatedly while $s=\mathrm{sde}(\mathrm{\mid }u{\mathrm{\mid }}^2)>3$. Each step replaces the current matrix $M$ by $H{T}^{-k}M$ and decreases the nonnegative integer $s$ by exactly one; hence after finitely many steps the first column lies in the base regime.

By [F3], the base column has a word $W_0$ with $W_0\mathrm{\mid }0⟩=(u_0,v_0{)}^T$. Reversing each descent step by prepending its recorded $T^{k}H$ gives a word $W$ with first column equal to the original $(u,v{)}^T$. The unitary $W^{\ast }U$ has first column $(1,0{)}^T$, so orthogonality of its columns makes it $\mathrm{diag}(1,\lambda )$ for a unit scalar $\lambda$. Its entries lie in $D$, since $W,U$ have entries in that ring. Therefore [F4] gives $\lambda ={\omega }^j$, and $U=W{T}^j$ is a finite Clifford+T word.

Both implications are proved. The proof constructs the converse word with only exact coefficient operations and the finite base lookup; no approximation, ancilla, or multi-qubit inference is used.