In system theory, we often encounter Semi-Definite Programs (SDPs) with Linear Matrix Inequality (LMI) constraints, such as those presented in this paper. I have introduced new variables based on the original decision variables and reformulated the LMIs using these new variables. Additionally, I have defined a similar SDP and, through simulation, identified relationships between their optimal solutions. I am now seeking mathematical proofs to substantiate these relationships. Further details will be provided below, and I would greatly appreciate any assistance in addressing my inquiry.
Consider following LMIs, where $ H \in \mathbb{R}^{(n+m) \times (n+m)} $ and $ W \in \mathbb{R}^{n \times n} $ are decision variables ($m \leq n$) and we have: \begin{equation} H = \begin{bmatrix} H_{11} & H_{12} \\ H_{12}^T & H_{22} \end{bmatrix}, H_{11} \in \mathbb{R}^{n \times n}, H_{22} \in \mathbb{R}^{m \times m} \end{equation} Also, $ D \in \mathbb{R}^{n \times (n+m)} $ and $ C \succeq 0 $ are constant matrices. Additionally, in all equations, $ 0 $ represents zero matrices of appropriate dimensions.
\begin{aligned} & \quad H - \begin{bmatrix} W & 0 \\ 0 & 0 \end{bmatrix} \succeq 0, \\ & \quad D^T \left(H_{11} - H_{12} H_{22}^{-1} H_{12}^T \right) D - H + C \succeq 0. \end{aligned}
Suppose $ K \in \mathbb{R}^{n \times j} $ is a constant matrix with rank $ n $ and $ j = n + n \cdot m $. The matrix $ T \in \mathbb{R}^{(n+m) \times (j+m)} $ is constructed as: \begin{equation} T = \begin{bmatrix} K & 0 \\ 0 & I_m \end{bmatrix} \end{equation} Suppose there exists a matrix $ S \in \mathbb{R}^{j \times (j+m)} $ such that $ D \cdot T = K \cdot S $.
The new variable $ X \in \mathbb{R}^{(j+m) \times (j+m)} $ is defined as: \begin{equation} X = T^{T} H T = \begin{bmatrix} X_{11} & X_{12} \\ X_{12}^T & X_{22} \end{bmatrix} \end{equation} Thus, the following relationships are established between the submatrices of $ H $ and $ X $:
\begin{aligned} & X_{11} = K^T H_{11} K, \\ & X_{12} = K^T H_{12}, \\ & X_{22} = H_{22}. \end{aligned}
If we multiply the above LMIs from the left and right by $ T^T $ and $ T $, respectively, we obtain:
\begin{equation} X - \begin{bmatrix} K^T W K & 0 \\ 0 & 0 \end{bmatrix} \succeq 0 \end{equation}
\begin{equation} S^T K^T \left(H_{11} - H_{12} H_{22}^{-1} H_{12}^T \right) K S - T^T H T + T^T C T \succeq 0 \Rightarrow \\ S^T \left(X_{11} - X_{12} X_{22}^{-1} X_{12}^T \right) S - X + T^T C T \succeq 0 \end{equation}
Selecting the new variable $ V $ as $ V = K^T W K $, we construct two SDPs as follows.
First SDP:
\begin{aligned} & \max_{H, W} \operatorname{trace}(W) \\ & \quad \text{subject to:} \\ & \quad H - \begin{bmatrix} W & 0 \\ 0 & 0 \end{bmatrix} \succeq 0, \\ & \quad \begin{bmatrix} D^T H_{11} D - H + C & D^T H_{12}\\ H_{12}^T D & H_{22} \end{bmatrix} \succeq 0. \end{aligned}
Second SDP:
\begin{aligned} & \max_{X, V} \operatorname{trace}(V) \\ & \quad \text{subject to:} \\ & \quad X - \begin{bmatrix} V & 0 \\ 0 & 0 \end{bmatrix} \succeq 0, \\ & \quad \begin{bmatrix} S^T X_{11} S - X + T^T C T & S^T X_{12}\\ X_{12}^T S & X_{22} \end{bmatrix} \succeq 0. \end{aligned}
My Question:
Let $ H^* $ and $W^*$ be the optimal solutions of the first SDP, and $ X^* $ and $V^*$ the optimal solution of the second SDP. We define $Z^*$ as follows: \begin{equation} {Z^*} = T^{T} H^* T = \begin{bmatrix} {Z}_{11}^* & {Z}_{12}^* \\ ({Z}_{12}^*)^T & {Z}_{22}^* \end{bmatrix} \end{equation}
I seek to prove the following equalities:
- $V^*= K^T W^* K$
- $ ({X}_{22}^*)^{-1} \cdot ({X}_{12}^*)^T = ({Z}_{22}^*)^{-1} \cdot ({Z}_{12}^*)^T $
My Attempt:
I have implemented both SDPs for an example with $ n = 2 $ and $ m = 1 $, using different $ K $ matrices in MATLAB. In all instances, the aforementioned equalities hold. However, I am uncertain how to establish a mathematical proof for these results.
Edits:
- I have tested numerous examples with $ n \leq 6 $ and $ m \leq n $ using randomly generated $K$ and $D$ matrices, and in all cases, the equalities hold. If necessary, I can provide my MATLAB code for review.
- Using Schur complement property, the equivalent of second LMIs substituted
