# KKT stationarity condition in simple SDP

This question is part of my continuing efforts to answer my original question in A puzzling KKT for LMI vs. scalar constraint.

Consider an SDP \begin{align} &\min - \text{Tr}(P)\\ &\ \ \text{s.t.} \begin{pmatrix} A^TP + PA +C^TPC + Q & PB + C^TPD \\ B^T P + D^TPC& R+D^TPD \end{pmatrix}\succeq0, P\succeq0 \end{align}

In a paper, they define a dual variable as \begin{align} Z= \begin{pmatrix} S&U^T&0\\ U&T&0\\ 0&0&W \end{pmatrix}, \end{align} and claim that \begin{align} AS+SA^T + CSC^T + BU + DUC^T + U^TB^T + CU^TD^T + D^TTD + W + I=0. \end{align} By computing the KKT stationarity condition, we get that the Trace of this equation is zero rather than its argument. What am I missing?

1. How will the equation change if we change to objective to $$-\text{Tr}(R+D^TPD)$$.
2. Is it a standard SDP? I learned that the objective should be expressed as $$c^Tx$$.

You can write the primal as: \begin{align} &\min - \text{Tr}(P)\\ &\ \ \text{s.t.} \begin{pmatrix} A^TP + PA +C^TPC + Q & PB + C^TPD & 0 \\ B^T P + D^TPC& R+D^TPD & 0 \\ 0 & 0 & P \end{pmatrix}\succeq0 \end{align} The Lagrangian is: \begin{align} L(P,Z) = - \text{Tr}(P) - \langle \begin{pmatrix} A^TP + PA +C^TPC + Q & PB + C^TPD & 0 \\ B^T P + D^TPC& R+D^TPD & 0 \\ 0 & 0 & P \end{pmatrix} , Z \rangle \end{align} If you take the derivative with respect to $$P$$ you obtain the stationarity condition: \begin{align} 0&=\frac{\partial L(P,Z)}{\partial P}\\ &= - \frac{\partial}{\partial P} \left\{\text{Tr}(P) + \text{Tr} \left(S^T(A^TP+PA + C^TPC+Q) + U(PB+C^TPD))\right)\right\}\\ &\ \ \ \ \ - \frac{\partial}{\partial P} \left\{ \text{Tr}\left(U^T(B^TP+D^TPC) + T^T(R+D^TPD) + P^TW)\right)\right\}. \end{align} By using the general relation $$\frac{\partial}{\partial P} \text{Tr}(\Gamma P \Lambda) = \frac{\partial}{\partial P} \text{Tr}(\Lambda\Gamma P ) = \Lambda^T\Gamma^T$$ for any $$\Lambda,\Gamma$$ and the cyclic trace property, we get: \begin{align} 0&= I + AS + SA^T + CSC^T + U^TB^T + CU^TD^T + BU + DUC^T + DTD^T+ W. \end{align}
1. the $$I$$ at the end will become $$DD^T$$
2. there is no 'standard' SDP, but an SDP in standard form is: $$\min_{X \succeq O} \{ \text{Tr}(CX) : \text{Tr}(A_iX)=b_i, \; i=1,\ldots,m\}$$ so clearly the given SDP is not in standard form. You can always reformulate it into standard form though (using a different choice for $$X$$).
• @Morad $\langle A,B \rangle = \text{Tr}(A^TB)$, not $\text{Tr}(AB)$ – LinAlg Feb 24 at 22:45
• @Morad I have updated the derivation and found discrepancies in the terms $CSC^T$ and in $DTD^T$ – LinAlg Feb 24 at 22:57
• Correct, there was a typo in the first entry and I corrected it to $C^TPC$. The $DTD^T$ term does not have a typo however. I noticed the inner product typo, but some reviewers did not approve my second edit. – Morad Feb 24 at 23:20
• For the inner product, $A$ is symmetric, no? – Morad Feb 25 at 4:56