# Representing $\mathbf{Tr}(A + BC^{-1}B^T)< K$ as an LMI

I am trying to implement a convex optimization and I have the constraint $$\mathbf{Tr}(A + BC^{-1}B^T)< K$$ where $$A\succeq0,B,C\succ0$$ are the decision variables. Can it be transformed into a linear constraint or a linear matrix inequality?

• Are there any other constraints on the variables, e.g. positive-semidefinite? – Rammus Nov 21 '20 at 12:31
• As $A + B C^{-1} B^T$ is the Schur complement of the block matrix $\begin{pmatrix} A & B \\ B^T & - C \end{pmatrix}$. Although I don't see immediately how to relate the block matrix to the trace inequality without strengthening it to an operator inequality. – Rammus Nov 21 '20 at 12:36
• Hello @Rammus. Yes, I can assume $A \succeq0$ and $C\succ0$. – Morad Nov 21 '20 at 18:10

$$\mathbf{Tr}(A + B\,C^{-1}B^\top) = \sum_{i=1}^n e_i^\top (A + B\,C^{-1}B^\top)\,e_i, \tag{1}$$
with $$n$$ such that $$A \in \mathbb{R}^{n\times n}$$ and $$e_i$$ the $$i$$th column of an $$n \times n$$ identity matrix. By introducing intermediate scalar variables $$\alpha_i$$ one can write the initial inequality in an indirect way by using
\begin{align} e_i^\top (A + B\,C^{-1}B^\top)\,e_i &< \alpha_i, \forall\,i = 1 \dots n, \tag{2} \\ \sum_{i=1}^n \alpha_i &< K. \tag{3} \end{align}
The inequality from $$(3)$$ is already a linear inequality. The inequality from $$(2)$$ can be formulated as linear (matrix) inequality by using the Schur complement. It can be noted that applying the Schur complement to each inequality from $$(2)$$ does require the additional assumptions that $$C$$ is positive definite.
• Thanks Kwin. I can assume $A\succeq0$ and $C\succ0$ so your answer is perfect. I wanted to ask whether your reply is the easiest way to implement in CVX or that there is another form. Thanks again. – Morad Nov 21 '20 at 18:12