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?

I will be happy for help about this step and two more related questions:

  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}

As for your other questions:

  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 $X$).
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    $\begingroup$ @Morad $\langle A,B \rangle = \text{Tr}(A^TB)$, not $\text{Tr}(AB)$ $\endgroup$ – LinAlg Feb 24 at 22:45
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    $\begingroup$ @Morad I have updated the derivation and found discrepancies in the terms $CSC^T$ and in $DTD^T$ $\endgroup$ – LinAlg Feb 24 at 22:57
  • $\begingroup$ 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. $\endgroup$ – Morad Feb 24 at 23:20
  • $\begingroup$ For the inner product, $A$ is symmetric, no? $\endgroup$ – Morad Feb 25 at 4:56
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    $\begingroup$ @Morad not necessarily, it is a valid inner product on the set of all (including asymmetric) matrices $\endgroup$ – LinAlg Feb 25 at 14:14

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