# A puzzling KKT for LMI vs. scalar constraint

I am trying to understand the KKT conditions for LMI constraints in order to solve my original question in KKT conditions for $\max \log \det(X)$ with LMI constraints.

In the meantime, I found a much simpler problem that does not go through when extending the KKT conditions from scalar case to the vector case. The problem is \begin{align} & \max_{X\succeq0} \log \det(I + B XB^T)\\ \\ & s.t. \begin{pmatrix} AXA^T - X + Q & AXB^T \\ BXA^T& I + BXB^T \end{pmatrix}\succeq0, \end{align} and the goal is to show $$(A - K(X^\ast)B)(A-K(X^\ast)B )^T\prec I$$ where $$K(X)\triangleq AXB^T(I + BXB^T)^{-1}$$. This is an important consequence in control theory since it implies that the optimal solution $$X^*$$ is a stabilizing solution for the corresponding system. From here, I elaborate on my modest progress.

The constraints can be either written as a "big LMI": \begin{align} R(X)&=\begin{pmatrix} AXA^T - X + Q & AXB^T &0\\ BXA^T& I + BXB^T &0 \\ 0&0&X \end{pmatrix}\succeq 0 \end{align} or \begin{align} AXA^T-X+Q - AXB^T(I + BXB^T)^{-1}BXA^T\succeq0\\\ X\succeq0, \end{align} where we used the Schur complement along with $$I+BXB^T\succ0$$.

The scalar case:
By the KKT stationarity condition \begin{align}\label{eq:lagra_1} 0&= -B^2 -\lambda_1+ \lambda_2\{ 1 - A^2 + 2 A^2B^2X(I + BXB^T)^{-1} - A^2B^4X^2(I + BXB^T)^{-2}\} \end{align} for $$\lambda_1,\lambda_2\ge0$$ and correspond to the constraints. If $$B\neq0$$, it follows that $$\lambda_2>0$$ and \begin{align} 0&<1 - A^2 + 2 A^2B^2X^\ast(I + BX^\ast B^T)^{-1} - A^2B^4X^{2\ast}(I + BX^\ast B^T)^{-2}\\ &= 1 - (A - K(X^\ast)B)^2 \end{align} where $$K(X)$$ is defined above. If $$B=0$$, the stability condition holds only if $$A<1$$. The combination of this conditions lead to the following known result: there exists a stabilizing solution $$X$$ iff $$(A,B)$$ is detectable.

The vector case using the Schur complement constraint:
Following a great advice from @LinAlg in the comments, I could make the following progress. The Lagrangian is: \begin{align} L(X,\Lambda_1,\Lambda_2)= - \log \det(I + B XB^T) - \text{Tr}(X\Lambda_1) - \text{Tr}((AXA^T-X+Q - AXB^T(I + BXB^T)^{-1}BXA^T) \Lambda_2). \end{align}

From the stationarity condition, (is the derivative correct?) \begin{align} 0&=\frac{\partial L(X,\Lambda_1,\Lambda_1)}{\partial X}\\ &= - \text{det}(I+BX B^T)\text{Tr}((I+BX B^T)^{-1}BB^T) - \Lambda_1 \\ &\ \ + (I - A^TA + A^TKB + B^TK^TA - B^TK^TKB) \Lambda_2 \end{align} with $$\Lambda_1,\Lambda_2\succeq0$$. By the primal feasibility constraint $$X\succeq0$$, the first summand is positive so that \begin{align} (I - (A-KB)^T(A-KB))\Lambda_2 \succeq 0 \end{align} The complementary slackness conditions read as: \begin{align} 0&= \Lambda_1X\\ 0&= (AXA^T-X+Q - AXB^T(I + BXB^T)^{-1}BXA^T))\Lambda_2\\ &= [(A-KB)X(A-KB)^T - X + Q + KK^T ]\Lambda_2\\ \end{align} I do not know how to proceed from here, but I try to enlighten when I am aiming to arrive. A necessary condition for the existence of stabilizing solution is detectability, i.e., if $$Ax = \lambda x$$ for a vector $$x$$ and $$|\lambda|\ge1$$ then $$Bx\neq0$$.
Let's try to show this fact using contradiction: Assume there exists a vector $$x\neq0$$ such that $$Ax = \lambda x$$ and $$Bx\neq0$$ with $$|\lambda|\ge1$$. We can now pre- and post-multiplying the stationarity condition with $$x^T$$ and $$x$$ and have \begin{align} 0&= - y^Ty -x\Lambda_1 - x^T( )\Lambda_2x \end{align}

The vector case using the big LMI:
Without loss of generality, the dual variable is \begin{align} Z=\begin{pmatrix} S & U&0\\ U^T & T&0\\ 0&0&W \end{pmatrix}. \end{align} The Lagrangian in this case is: \begin{align} L(X,Z)&= - \log \det(I + B XB^T) - \text{Tr}(R(X)Z). \end{align} The KKT stationarity condition gives: \begin{align} 0&= - \text{det}(I+BX B^T)\text{Tr}((I+BX B^T)^{-1}BB^T) \\& \ \ -\text{Tr}( ASA^T - S + B^TU^TA + A^TUB + B^TTB + W ) \end{align} and the complementary slackness condition $$R(X)Z=0$$ simplifies to: \begin{align} 0&= (AXA^T - X + Q)S + AXB^T U^T\\ 0&= (AXA^T - X + Q)U + AXB^T T\\ 0&= BXA^T S + (I+BXB^T)U^T\\ 0&= BXA^T U + (I+BXB^T)T\\ 0&= XW \end{align}

If $$B=I$$, it follows that $$S\succ0$$ (Assume $$Sx=0$$ for some $$x$$ and conclude that $$x=0$$). For the general case, $$Sx=0$$ implies $$Bx=0 \& \& Wx=0$$.

• you just add $-\text{Tr}(\Lambda(AXA^T-X+Q - AXB^T(I + BXB^T)^{-1}BXA^T))$ to the Lagrangian Feb 19, 2021 at 3:33
• That is only true if it holds for all $B \succeq 0$, not just for one specific $B$. You can write $B=UU^T$ and use the cyclic property of the trace to simplify the left hand side to $\sum_{i=1}^p u_i^T (I - (A-BK)(A-BK)^T) u_i$. Feb 19, 2021 at 21:21
• A simpler problem was solved by @LinAlg in <math.stackexchange.com/questions/4034735/…>. I am updating my progress, feel free to contribute. Feb 22, 2021 at 7:00