# SDP reformulation with range space constraint

In page 6 of the paper A semidefinite programming method for integer convex quadratic minimization, it is stated that the following maximization problem

\begin{align} \text{maximize} & \quad -\left(q + \frac{1}{2}\lambda \right)^\top \left(P - \text{diag}(\lambda)\right)^\dagger \left(q + \frac{1}{2}\lambda \right)\\ \text{subject to} & \quad P - \text{diag}(\lambda) \succeq 0 , \\ & \quad q + \frac{1}{2}\lambda \in \mathcal{R}(P - \text{diag}(\lambda)) \\ & \quad \lambda \geq 0 \end{align} is equivalent to \begin{align} \text{maximize} & \quad -\gamma \\ \text{subject to} & \quad \begin{bmatrix} P - \text{diag}(\lambda) & q + \frac{1}{2}\lambda \\ \left(q + \frac{1}{2}\lambda \right)^\top & \gamma \end{bmatrix} \succeq 0 \\ & \quad \lambda \geq 0, \end{align} where $$P \in \mathbb{R}^{n\times n}$$ symmetric, $$q \in \mathbb{R}^n$$, $$\lambda \in \mathbb{R}^n$$, $$\gamma \in \mathbb{R}$$, $$A^\dagger$$ is the Moore-Penrose inverse of the matrix $$A$$, diag($$\lambda$$) is the diagonal matrix with $$\lambda$$ in the main diagonal, $$A\succeq 0$$ means that $$A$$ is positive semi-definite, $$\lambda \geq 0$$ means that all the components of $$\lambda$$ are non-negative and $$b \in \mathcal{R}(A)$$ means that $$b$$ is in the range space of $$A$$.

I understand how we can use Schur complements to reformulate the SDP. However, the point I do not understand is why we can drop the range space constraint $$q + \frac{1}{2}\lambda \in \mathcal{R}(P - \text{diag}(\lambda))$$? I do not see how this constraint is enforced in the reformulation.

• Do you understand how $\gamma$ ties into the second equation, making it equivalent to the original objective? It looks like the (2,2) element should be just (incorrectly) $0$, not $0 + \gamma$, but I don't understand why yet. I've asked a similar question here math.stackexchange.com/questions/4836819/… Jan 2 at 12:17