# The Schur complement is matrix concave

I am confused regarding the Schur complement. From Boyd and Vandenberghe's Convex Optimization,

Suppose $$X \in S^n_{++}$$ partitioned as $$X = \begin{bmatrix} A & B\\ B^T & C\\ \end{bmatrix}$$ where $$A \in S^k$$. The Schur complement of $$X$$ (with respect to $$A$$) is $$S=C-B^TA^{-1}B$$. Show that the Schur complement, viewed as function from $$S^n$$ into $$S^{n-k}$$, is matrix concave on $$S^n_{++}$$.

I believe the goal is to show that $$y^TSy\leq0$$, where $$y\in\mathbb{R}^{n-k}$$. So, I follow a similar argument as given in the book (i.e., Example 3.4). I define $$h(y,S)=y^TSy$$. Then, I think we should consider the hypograph? That is, if the hypograph of $$h$$ is convex set, then $$h$$ is concave. Following this idea, I have the following:

\begin{align} \mathbf{hypo}\, h &= \left\{ (y,S,t) \mid y^TSy\geq t \right\} \\ &= \left\{ (y,S,t) \mid \begin{bmatrix} S & y \\ y^T & t \end{bmatrix} \leq 0 \right\} \end{align}

Since the last line can be expressed as a linear matrix inequality (LMI) of $$(y,S,t)$$, the hypo of $$h$$ is a convex set. This implies that $$S$$ is matrix concave? But I can also consider the epigraph isn't it? Then, I'll get a similar LMI with a different inequality. So, does that means it is both convex and concave? Moreover, if $$X \in S^n_{++}$$, it also implies that $$A$$ and $$S$$ is positive definite? Do they play role here? I'm very confused.

• Which question is that? Which number? Commented May 21, 2023 at 9:13

You want to show matrix concavity: $$S(\theta X_1 + (1-\theta) X_2) \succeq \theta S(X_1) + (1-\theta) S(X_2), \ \ \forall \theta \in [0,1], X_1, X_2 \in S^n_{++}$$ This is equivalent to showing that the matrix hypograph is a convex set: $$\mathbf{hypo}\ S: =\{ (X,T) \ | \ S(X) \succeq T, \ X \in S^n_{++}, \ T \in S^{n-k} \}$$ But then we can use properties of the Schur Complement to show for positive definite $$X$$ we have: $$\ S(X) = C - B^T A^{-1} B \succeq T \Leftrightarrow \begin{bmatrix} A & B\\ B^T & C- T \end{bmatrix} \succeq 0\Leftrightarrow \ X - \begin{bmatrix} 0 & 0\\ 0 & T \end{bmatrix} \succeq 0$$ So let $$L$$ be the linear map $$S^n \times S^{n-k} \rightarrow S^n$$: $$L(X,T) := X - \begin{bmatrix} 0 & 0\\ 0 & T \end{bmatrix}$$ We conclude: $$\mathbf{hypo}\ S =\{ (X,T) \ | \ L(X,T) \in S^n_{+}, \ X \in S^n_{++}, \ T \in S^{n-k} \}$$ So then we can conclude the hypograph is a convex set because the semidefinite cone is.
• I wish to verify something. So, in the last line, we have $$X\geq\Bigg[\begin{matrix} 0 & 0\\ 0 & T\\ \end{matrix}\Bigg]$$ I suppose the value of T is not important right? Because X is positive definite, so that is suffice to show it is a convex set. Commented Dec 15, 2020 at 14:53