S-lemma with equality and its applications

Abstract

Let $f(x)=x^TAx+2a^Tx+c$ and $h(x)=x^TBx+2b^Tx+d$ be two quadratic functions having symmetric matrices $A$ and $B$. The S-lemma with equality asks when the unsolvability of the system $f(x)<0, h(x)=0$ implies the existence of a real number $\mu$ such that $f(x) + \mu h(x)\ge 0, ~\forall x\in \mathbb {R}^n$. The problem is much harder than the inequality version which asserts that, under Slater condition, $f(x)<0, h(x)\le 0$ is unsolvable if and only if $f(x) + \mu h(x)\ge 0, ~\forall x\in \mathbb {R}^n$ for some $\mu \ge 0$. In this paper, we show that the S-lemma with equality does not hold only when the matrix $A$ has exactly one negative eigenvalue and $h(x)$ is a non-constant linear function ($B=0, b\not =0$). As an application, we can globally solve $\inf \{f(x): h(x)=0\}$ as well as the two-sided generalized trust region subproblem $\inf \{f(x): l\le h(x)\le u\}$ without any condition. Moreover, the convexity of the joint numerical range $\{(f(x), h_1(x),\ldots , h_p(x)):x\in \mathbb R^n\}$ where $f$ is a (possibly non-convex) quadratic function and $h_1(x),\ldots ,h_p(x)$ are affine functions can be characterized using the newly developed S-lemma with equality.

Appendix

We discuss the relations among S-Conditions 1,2,3 and 4. First, it is easy to see that the definiteness of $B$ implies $A \succeq \eta B$ for some $\eta .$ Therefore,

$$\begin{aligned} \mathrm{S\text {-}Condition\ 1} ~\Longrightarrow ~\mathrm{S\text {-}Condition\ 2}. \end{aligned}$$

Moreover, when $B$ is definite, $x^TBx=0$ if and only if $x=0$ and thus

$$\begin{aligned} \mathrm{S\text {-}Condition\ 1} ~\Longrightarrow ~\mathrm{S\text {-}Condition\ 4}. \end{aligned}$$

When $h(x)$ is homogeneous, there is $b=d=0$ so that $h(0)=0.$ By choosing $\zeta =0$,

$$\begin{aligned} \mathrm{S\text {-}Condition\ 3} ~\Longrightarrow ~\mathrm{S\text {-}Condition\ 4}. \end{aligned}$$

It is not difficult to verify that neither S-Condition 2 nor S-Condition 4 can imply each other [24]. Consequently, neither S-Condition 2 nor S-Condition 4 is necessary for the S-lemma with equality.

We now show that the statement (4) in S-Condition 4 can be equivalently simplified as

$$\begin{aligned} \left\{ \begin{array}{ll}b\in \mathcal {R}(B),&{}\quad \text {if}~B\succeq 0~ \text {or}~B\preceq 0,\\ B\zeta +b=0,&{}\quad \text {otherwise}. \end{array}\right. \end{aligned}$$

(64)

Notice that (64) trivially implies (4), so it is sufficient to prove the converse.

Suppose $B\succeq 0$ or $B\preceq 0$. Then, $x^TBx=0 ~\Longleftrightarrow ~ Bx=0$ and (4) can be recast as

$$\begin{aligned} Bx=0 ~\Longrightarrow ~ b^Tx=0,~\forall x\in \mathbb {R}^n, \end{aligned}$$

which shows that $b\in \mathcal {R}(B)$.

Now assume that $B$ is indefinite. We first rewrite (4) by

$$\begin{aligned} x^TBx=0 ~\Longrightarrow ~x^T(B\zeta +b)(B\zeta +b)^Tx=0,~\forall x\in \mathbb {R}^n. \end{aligned}$$

Since $(B\zeta +b)(B\zeta +b)^T\succeq 0$, it is further equivalent to

$$\begin{aligned} x^TBx=0 ~\Longrightarrow ~ -x^T(B\zeta +b)(B\zeta +b)^Tx\ge 0,~\forall x\in \mathbb {R}^n. \end{aligned}$$

Since $B$ is indefinite, $h(x)=x^TBx$ takes both positive and negative values. By the S-lemma with equality for homogeneous quadratic forms,

$$\begin{aligned} (\exists \mu \in \mathbb R)~~~-(B\zeta +b)(B\zeta +b)^T+\mu B \succeq 0. \end{aligned}$$

(65)

Since $\mu B \succeq (B\zeta +b)(B\zeta +b)^T \succeq 0$ and $B$ is indefinite, it must be $\mu =0$ and thus $(B\zeta +b)(B\zeta +b)^T=0$. It implies that $B\zeta +b=0$.