Rounded matrix products give certified linear and normal-equation residual allowances

For $(a+ib)(c+id)$, each rounded real product has error at most $u$ times its absolute exact value. Each final real addition contributes at most $u(1+u)$ times the two corresponding product magnitudes. The Euclidean norm of the resulting real/imaginary error is at most $2(2u+u^2)\mathrm{\mid }a+ib\mathrm{\mid }\mathrm{\mid }c+id\mathrm{\mid }\le 6u\mathrm{\mid }a+ib\mathrm{\mid }\mathrm{\mid }c+id\mathrm{\mid }$, since $u<1$. Conjugation is exact. Real products obey this conservative bound as well.

In a length-$k$ dot product, each product error from step 1.1 is propagated through at most $k-1$ additions. Expansion of each real component and [F2] bound the dot discrepancy by $[6u+{\gamma }_{k-1}(1+6u)]{\sum }_i\mathrm{\mid }{a}_i\mathrm{\mid }\mathrm{\mid }{x}_i\mathrm{\mid }=g_k{\sum }_i\mathrm{\mid }{a}_i\mathrm{\mid }\mathrm{\mid }{x}_i\mathrm{\mid }$, with ${\gamma }_0=0$. Combining rows and applying Cauchy-Schwarz gives $\mathrm{\parallel }\widehat{v}-Ax{\mathrm{\parallel }}_2\le g_{n}FX=V$. Componentwise subtraction has vector error at most $u(\mathrm{\parallel }\widehat{v}{\mathrm{\parallel }}_2+Y)$, so the linear residual allowance is $W$. Zeros cause no difficulty in this expansion.

Apply step 2.1 to $A^{\ast }\widehat{w}$ to get $\mathrm{\parallel }\widehat{z}-A^{\ast }\widehat{w}{\mathrm{\parallel }}_2\le g_{m}F\mathrm{\parallel }\widehat{w}{\mathrm{\parallel }}_2$. Transporting the earlier error through $A^{\ast }$ adds at most $FW$. The scalar product error is at most $u\lambda X$; the last addition adds at most $u(\mathrm{\parallel }\widehat{z}{\mathrm{\parallel }}_2+\mathrm{\parallel }\widehat{l}{\mathrm{\parallel }}_2)$. Their sum proves $\rho$. For basis columns, the triangle inequality combines each computed linear residual with its allowance; summing the squared column bounds gives a Frobenius certificate for $I-AB$, which bounds its induced Euclidean norm by Cauchy-Schwarz.