 Solution

Stationarity of the equality-constrained quadratic model gives $g_k+H_k{s}_k-A_k^T{y}_{k+1}=0$, while feasibility gives $A_k{s}_k=-c_k$. A symmetric system linking the step and new multiplier isLet $Z$ have columns forming a basis of $\ker A_k$. The saddle-point matrix is nonsingular exactly when $A_k$ has full row rank and $Z^T{H}_{k}Z$ is nonsingular. Full row rank is necessary because a nonzero vector in $\ker A_k^T$ would give a null vector of the block matrix. Under full row rank, solving a homogeneous system forces $s=Zq$ and then $Z^T{H}_{k}Zq=0$; this proves sufficiency and the converse. Positive definiteness of the reduced Hessian is sufficient for a constrained minimum but is stronger than the nonsingularity condition asked here.

A null-space method first finds $s_p$ with $A_k{s}_p=-c_k$, then solvesand recovers $y_{k+1}$ from $A_k^T{y}_{k+1}=g_k+H_k{s}_k$. It needs the stated full row rank and nonsingular reduced matrix; Cholesky decomposition is available when the latter is positive definite.

A range-space method additionally assumes $H_k$ is invertible. It solves the Schur complement equationsIf $H_k$ is positive definite and $A_k$ has full row rank, the Schur matrix is positive definite and both solves can use Cholesky decomposition. An indefinite but invertible $H_k$ instead requires a suitable indefinite factorization; the Schur matrix must remain nonsingular.