Two-sided unitary errors in a matrix pencil accumulate as a backward perturbation pair

If $m=0$, take $Q=Z=I_n$ and ${\mathrm{\Delta }}_A={\mathrm{\Delta }}_B=0$; the equations and bound hold. For $m\ge 1$, form the $3n\times 3n$ block matrix $M_j=\left(\begin{array}{ccc}0& A_j& B_j\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$ and $P_j=\mathrm{diag}(Q_j,Z_j,Z_j)$. Put the local $E_j^A,E_j^B$ in the first block row of ${\mathcal{E}}_j$ and the local $D_j^A,D_j^B$ in the first block row of ${\mathcal{D}}_j$, with all other blocks zero. Then [F2] and the stated updates give $M_j=P_j^{\ast }(M_{j-1}+{\mathcal{E}}_j+{\mathcal{D}}_j)P_j$. Each $P_j$ is unitary/orthogonal because its diagonal blocks are unitary/orthogonal.

For $m\ge 1$, apply [F1] to the augmented sequence. Its accumulated perturbation has only the two first-row, off-diagonal blocks, because conjugation by each block diagonal $P_j$ preserves those block positions. Call them ${\mathrm{\Delta }}_A,{\mathrm{\Delta }}_B$. By [F3], $\mathrm{\parallel }{\mathcal{E}}_j{\mathrm{\parallel }}_F=\sqrt{\mathrm{\parallel }{E}_j^A{\mathrm{\parallel }}_F^2+\mathrm{\parallel }{E}_j^B{\mathrm{\parallel }}_F^2}$ and likewise for ${\mathcal{D}}_j$. Therefore the global bound from [F1] is exactly the displayed pair bound, and the accumulated block products are $Q,Z$.