Givens reduction terminates with a unitary-equivalent Hessenberg-triangular pair

The triangularisation loops range over finitely many pairs $(i,j)$ with $j=0,\dots ,n-1$ and $i=n-1,\dots ,j+1$. For fixed $j$, descending $i$ zeros $B_{i,j}$ without changing already cleared entries below it. The rotations act on rows with indices at least $j$, whose entries in columns less than $j$ are zero by upper triangularity; hence they do not create below-diagonal entries in earlier columns. After all columns, $B$ is upper triangular.

For $j=0$ there are no earlier Hessenberg columns, so the base condition is vacuous. For general $j$, assume every earlier column already has the Hessenberg pattern. Descend through $i=n-1,\dots ,j+2$; the left factor zeros $A_{i,j}$. Since $B$ was upper triangular, this left factor can create below-diagonal fill only at $B_{i,i-1}$; the paired right factor zeros that entry. It creates no further below-diagonal entries in $B$, because rows below $i$ are zero in columns $i-1,i$. The right factor does not touch column $j$ because $i-1\ge j+1$, and the left factors preserve all earlier Hessenberg columns, whose entries in rows $i-1,i$ are already zero. Thus after the loop, column $j$ of $A$ is zero below row $j+1$.

Induction over $j=0,\dots ,n-3$ proves that the final $A$ is upper Hessenberg and $B$ remains upper triangular. Each left update $A\leftarrow LA$ is matched by $Q\leftarrow Q{L}^{\ast }$, and each right update $A\leftarrow AR$ is matched by $Z\leftarrow ZR$; direct substitution preserves $A_{\mathrm{c}\mathrm{u}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{t}}=Q^{\ast }{A}_{\mathrm{i}\mathrm{n}\mathrm{p}\mathrm{u}\mathrm{t}}Z$ and $B_{\mathrm{c}\mathrm{u}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{t}}=Q^{\ast }{B}_{\mathrm{i}\mathrm{n}\mathrm{p}\mathrm{u}\mathrm{t}}Z$. The finite loops terminate, proving the claim.