Complementarity problems with respect to Loewnerian cones

Abstract

This work deals with the analysis and numerical resolution of a broad class of complementarity problems on spaces of symmetric matrices. The complementarity conditions are expressed in terms of the Loewner ordering or, more generally, with respect to a dual pair of Loewnerian cones.

Appendix

Let $\sigma (\mathfrak {L}, \mathcal P_n)$ denote the set of Loewner-eigenvalues of a linear map $\mathfrak {L}:\mathcal S_n\rightarrow \mathcal S_n$. Such set is called the Loewner-spectrum of $\mathfrak {L}$. The following proposition displays a map $\mathfrak {L}$ whose Loewner-spectrum is a set of infinite cardinality.

Proposition 5.1

Let $\mathfrak {L}:\mathcal S_n\rightarrow \mathcal S_n$ be given by $\mathfrak {L}(X)=\langle C,X\rangle I_n$, where $C\succeq \mathbf{0}$. Then

$$\begin{aligned} \left[ \lambda _\mathrm{min}(C), \lambda _\mathrm{max}(C)\right] \;\subseteq \; \sigma (\mathfrak {L}, \mathcal P_n)\;=\; \bigcup _{r=1}^n \left[ f_r(C), g_r(C)\right] \,, \end{aligned}$$

(47)

where $f_r(C)$ and $g_r(C)$ indicate respectively the sum of the $r$ smallest and the sum of the $r$ largest eigenvalues of $C$.

Proof

Assume that $C\not =\mathbf{0}$, otherwise (47) holds trivially. A scalar $\lambda$ belongs to $\sigma (\mathfrak {L}, \mathcal P_n)$ if and only if there exists $X\in \mathcal S_n$ such that

$$\begin{aligned} \left\{ \begin{array}{ll} X\succeq \mathbf{0}, \;\mathrm{tr}(X)=1,\\ \langle C,X\rangle I_n\succeq \lambda X,\\ \langle C,X\rangle = \lambda \Vert X\Vert ^2. \end{array} \right. \end{aligned}$$

Hence,

$$\begin{aligned} \sigma (\mathfrak {L}, \mathcal P_n)= \left\{ \Vert X\Vert ^{-2}\langle C,X\rangle : X\in \Omega \right\} , \end{aligned}$$

(48)

where $\Omega$ stands for the set of matrices $X\in \mathcal S_n$ satisfying

$$\begin{aligned}&\displaystyle X\succeq \mathbf{0}, \;\mathrm{tr}(X)=1,\end{aligned}$$

(49)

$$\begin{aligned}&\displaystyle \langle C,X\rangle (\Vert X\Vert ^2- \lambda _\mathrm{max} (X))\ge 0. \end{aligned}$$

(50)

Under (49), the inequality (50) can be written as an equality. The set on the right-hand side of (48) remains unchanged if one uses

$$\begin{aligned} \Omega _0:= \left\{ X\in \mathcal S_n: X\succeq \mathbf{0}, \;\mathrm{tr}(X)=1, \; \lambda _\mathrm{max} (X)= \Vert X\Vert ^2 \right\} \end{aligned}$$

instead of $\Omega$. On the other hand, one can check that

$$\begin{aligned} \Omega _0= \bigcup _{r=1}^n\left\{ r^{-1} QQ^T: Q\in \mathcal O (n,r) \right\} , \end{aligned}$$

where $Q\in \mathcal O (n,r)$ indicates that $Q$ is matrix of size $n\times r$ such that $Q^TQ=I_r$. Hence,

$$\begin{aligned} \sigma (\mathfrak {L}, \mathcal P_n)= \bigcup _{r=1}^n\left\{ \langle C,QQ^T\rangle : Q\in \mathcal O (n,r) \right\} . \end{aligned}$$

Note that $\langle C,QQ^T\rangle$ ranges over the interval $\left[ f_r(C), g_r(C)\right]$ as the variable $Q$ ranges over $\mathcal O (n,r)$. $\square$