An implicit linear solve and its adjoint gradient

The system gives $u_2=1$ and $u_1+\theta u_2=2$, hence $u(\theta )=(2-\theta ,1{)}^T$ and $J(\theta )=3-\theta$. Therefore $J^{\mathrm{\prime }}(0)=-1$.

Write $R(u,\theta )=A(\theta )u-b$. At $\theta =0$, $u=(2,1{)}^T$, $R_u=A(0)=I$, $R_{\theta }=A_{\theta }u=(1,0{)}^T$, and $J_u=(1,1)$. The adjoint equation $R_u^T\lambda =J_u^T$ gives $\lambda =(1,1{)}^T$.

The objective has no explicit $\theta$ dependence, so $J_{\theta }=0$; the adjoint formula gives $J_{\theta }-{\lambda }^T{R}_{\theta }=-(1,1)\cdot (1,0)=-1$.

The adjoint result agrees with the direct derivative in step 1.1, verifying this nonsingular linear-solve instance.

In any arithmetic satisfying the operationwise model and representing $0,1,2,3$, with these small operations in range, at $\theta =0$ the primal and basis solves give $\widehat{x}=(2,1{)}^T$ and $B=I$ exactly. Their inverse defect and primal residual are zero. The certified solve-Jacobian result Certified inverse defects and primal residuals bound linear-solve Jacobian and pullback errors therefore gives zero fixed-map discrepancy; the actual dot and product operations here are also exact, and return derivative $-1$.