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9. Its solution x(u) and the corresponding vector of Lagrange multipliers λ(u) are fully determined by the KKT conditions A BT B O x(u) b = , λ(u) c+u so that x(u) A BT = B O λ(u) −1 b A BT = B O c+u −1 b A BT + B O c −1 o . 4 Equality Constrained Problems 47 ∇f (x)T d(u) = −(BT λ)T d(u) = −λT Bd(u) = −λT u. It follows that −[λ]i can be used to approximate the change of the optimal cost due to the violation of the ith constraint by [u]i . To give more detailed analysis of the sensitivity of the optimal cost with respect to the violation of constraints, let us deﬁne for each u ∈ Rm the primal function p(u) = f (x(u)) .

3. Let A ∈ Rn×n denote a symmetric matrix, let B ∈ Rm×n , and let there be μ > 0 such that xT Ax ≥ μ x 2 , x ∈ KerB. Then A is positive deﬁnite for suﬃciently large . Proof. 17) any x ∈ Rn can be written in the form x = y + z, y ∈ KerB, z ∈ ImBT . 33), we get xT A x = yT Ay + 2yT Az + zT Az + Bz 2 ≥ μ y 2 − 2 A y z + (λmin + σ 2min ) z 2 μ, − A y = y , z . 10 Penalized Matrices 23 We shall complete the proof by showing that the matrix H = μ, − A − A , λmin + σ 2min is positive deﬁnite for suﬃciently large values of .

Alternatively, taking α ∈ (0, 1), we get f (αx + (1 − α)y) < αf (x) + (1 − α)f (y) = f (x), which contradicts the assumption that x is a global minimizer of f on Ω. 3 Existence of Minimizers Since quadratic functions are continuous, existence of at least one minimizer is guaranteed by the Weierstrass theorem provided Ω is compact, that is, closed and bounded. We can also use the following standard results which do not assume that Ω is bounded. 5. Let f be a quadratic function deﬁned on a nonempty closed convex set Ω ⊆ Rn .