Computing the distance between the linear matrix pencil and the completely positive cone

Abstract

In this paper, we consider the problem of computing the distance between the linear matrix pencil and the completely positive cone. We formulate it as a linear optimization problem with the cone of moments and the second order cone. A semidefinite relaxation algorithm is presented and the convergence is studied. We also propose a new model for checking the membership in the completely positive cone.

Appendix: Proof of Theorem 3.2

Proof

(i) Let $(p^0,z^0)$ be a relative interior point of (D). Then

$$\begin{aligned} c \circ z^0 +p^0=0, \quad \mathrm {vech}(A_i)\bullet (c \circ z^0) = 0, \ i=1,\ldots ,m, \end{aligned}$$

(6.1)

$\Vert z^0\Vert <1$, and $p^0(x) > 0$ on $\Delta$ due to [22, Lemma 3.1]. Since $\Delta$ is compact, there exist $\epsilon _0 > 0$ and $\delta > 0$ such that

$$\begin{aligned} p(x)-\epsilon _0 > \epsilon _0, \; \forall p\in B(p^0,\delta ). \end{aligned}$$

By Nie [20, Theorem 6], there exists $k_0 > 0$ such that

$$\begin{aligned} p(x)-\epsilon _0 \in I_{2k_0}(h)+Q_{k_0} (g),\; \forall p\in B(p^0,\delta ). \end{aligned}$$

Thus, ($D^k$) has a relative interior point for all $k \ge k_0$. So, the strong duality holds between ($P^k$) and ($D^k$). Since (P) is feasible, the relaxation ($P^k$) is also feasible, and it has a minimizer $({\mathbf {x}}^{*,k}, w^{*,k}, y^{*,k}, \gamma ^{*,k},{\tilde{\mathbf {x}}}^{*,k})$ (cf. [1, Theorem 2.4.I]).

(ii) Firstly, we prove that the sequence $\{({\mathbf {x}}^{*,k}, w^{*,k}, y^{*,k}, \gamma ^{*,k})\}$ is bounded. Since ($P^k$) is feasible and $(w^{*,k}, \gamma ^{*,k})\in {\mathcal {L}}_{{\bar{n}}+1}$, we have that $\{(w^{*,k}, \gamma ^{*,k})\}$ is bounded. Note that $A_i (i=1, \ldots , m)$ are linearly independent and $({\mathbf {x}}^{*,k}, w^{*,k}, y^{*,k})$ satisfies

$$\begin{aligned} c \circ \left( {\mathbf {x}}^{*,k} +\sum _{i=1}^m y^{*,k}_i \mathrm {vech}(A_i) \right) =w^{*,k} + c \circ \mathrm {vech}(A_0), \end{aligned}$$

(6.2)

we know that $\{y^{*,k}\}$ is bounded if $\{{\mathbf {x}}^{*,k}\}$ is bounded. Thus, it suffices to prove that $\{{\mathbf {x}}^{*,k}\}$ is bounded.

Let $(p^0,z^0)$ and $\epsilon _0$ be as in the proof of (i). Because $I_{2k_0} (h) + Q_{k_0} (g)$ is dual to $\Gamma _{k_0}$, for all $k \ge k_0$, we have

$$\begin{aligned} 0 \le \langle p^0-\epsilon _0, {\tilde{\mathbf {x}}}^{*,k}\rangle = \langle p^0, {\tilde{\mathbf {x}}}^{*,k}\rangle - \epsilon _0 \langle 1, {\tilde{\mathbf {x}}}^{*,k}\rangle . \end{aligned}$$

(6.3)

It follows from $\langle z^0, w^{*,k}\rangle +\langle 1, \gamma ^{*,k}\rangle \ge 0$, $\gamma ^{*,k}\le \vartheta _P$, (6.1) and (6.2) that

$$\begin{aligned} \langle p^0, {\tilde{\mathbf {x}}}^{*,k}\rangle= & {} \langle -c \circ z^0, {\tilde{\mathbf {x}}}^{*,k}\rangle = \langle -c \circ z^0, {\mathbf {x}}^{*,k}\rangle =\langle z^0, -c \circ {\mathbf {x}}^{*,k}\rangle \nonumber \\= & {} \left\langle z^0, c \circ \sum \limits _{i=1}^m y_i^{*,k} \mathrm {vech}(A_i)-w^{*,k} -c \circ \mathrm {vech}(A_0) \right\rangle \nonumber \\= & {} \sum \limits _{i=1}^m y_i^{*,k} \mathrm {vech}(A_i)\bullet (c \circ z^0)- \langle z^0, w^{*,k}\rangle - \mathrm {vech}(A_0)\bullet (c \circ z^0)\nonumber \\= & {} - \langle z^0, w^{*,k}\rangle - \mathrm {vech}(A_0)\bullet (c \circ z^0)\nonumber \\\le & {} \gamma ^{*,k}- \mathrm {vech}(A_0)\bullet (c \circ z^0)\nonumber \\\le & {} \vartheta _P-\mathrm {vech}(A_0)\bullet (c \circ z^0)\nonumber \\:= & {} K_0. \end{aligned}$$

(6.4)

Denote by ${\mathbf {0}}$ the zero vector in ${\mathbb {N}}^n$. By (6.3), we obtain

$$\begin{aligned} 0 \le \langle p^0-\epsilon _0, {\tilde{\mathbf {x}}}^{*,k}\rangle \le K_0- \epsilon _0 ({\tilde{\mathbf {x}}}^{*,k})_{{\mathbf {0}}}, \end{aligned}$$

i.e.,

$$\begin{aligned} ({\tilde{\mathbf {x}}}^{*,k})_{{\mathbf {0}}}\le K_1:=K_0/\epsilon _0. \end{aligned}$$

(6.5)

For $\Delta$ given in (2.10), since $I (h) + Q(g)$ is archimedean, there exist $\varrho >0$ and $k_1 \in {\mathbb {N}}$ such that

$$\begin{aligned} \varrho -\Vert x\Vert _2^2 \in I_{2k_1} (h) + Q_{k_1}(g). \end{aligned}$$

So, for all $k \ge k_1$, we have

$$\begin{aligned} 0 \le \langle \varrho -\Vert x\Vert _2^2, {\tilde{\mathbf {x}}}^{*,k}\rangle = \varrho ({\tilde{\mathbf {x}}}^{*,k})_{{\mathbf {0}}}- \sum _{|\alpha |=1} ({\tilde{\mathbf {x}}}^{*,k})_{2\alpha }, \end{aligned}$$

which, together with (6.5), gives

$$\begin{aligned} \sum _{|\alpha |=1} ({\tilde{\mathbf {x}}}^{*,k})_{2\alpha } \le \varrho K_1. \end{aligned}$$

(6.6)

Note that for each $t = 1, \ldots , k -k_1$, we have

$$\begin{aligned} \Vert x\Vert _2^{2t-2}(\varrho -\Vert x\Vert _2^2)\in I_{2k} (h) + Q_{k}(g). \end{aligned}$$

The membership ${\tilde{\mathbf {x}}}^{*,k}\in \Gamma _{k}$ implies that

$$\begin{aligned} \varrho \langle \Vert x\Vert _2^{2t-2}, {\tilde{\mathbf {x}}}^{*,k}\rangle -\langle \Vert x\Vert _2^{2t}, {\tilde{\mathbf {x}}}^{*,k}\rangle \ge 0, \quad t = 1, \ldots , k -k_1. \end{aligned}$$

(6.7)

Combining (6.6) and (6.7), we obtain

$$\begin{aligned} \langle \Vert x\Vert _2^{2t}, {\tilde{\mathbf {x}}}^{*,k}\rangle \le \varrho ^t K_1, \quad t = 1, \ldots , k -k_1. \end{aligned}$$

Let ${\mathbf {x}}^k := {\tilde{\mathbf {x}}}^{*,k}|_{2k-2k_1}$. Then the moment matrix $M_{k-k_1} ({\mathbf {x}}^k) \succeq 0$ and

$$\begin{aligned} \Vert {\mathbf {x}}^k\Vert _2 \le \Vert M_{k-k_1} ({\mathbf {x}}^k)\Vert _F \le \mathrm {trace} (M_{k-k_1} ({\mathbf {x}}^k)) =\sum _{i=0}^{k-k_1} \sum _{|\alpha |=i} ({\tilde{\mathbf {x}}}^{*,k})_{2\alpha }. \end{aligned}$$

Since

$$\begin{aligned} \sum _{|\alpha |=i} ({\tilde{\mathbf {x}}}^{*,k})_{2\alpha }=\left\langle \sum _{|\alpha |=i} x^{2\alpha }, {\mathbf {x}}^k\right\rangle \le \langle \Vert x\Vert _2^{2i},{\mathbf {x}}^k\rangle \le \varrho ^i K_1, \end{aligned}$$

we have

$$\begin{aligned} \Vert {\mathbf {x}}^k\Vert _2\le (1+\varrho +\cdots +\varrho ^{k-k_1})K_1. \end{aligned}$$

(6.8)

Fix $k_2 > k_1$ such that ${\mathbf {x}}^{*,k}$ is a subvector of ${\mathbf {x}}^k|_{k_2-k_1}$. It follows from ${\mathbf {x}}^{*,k} = {\mathbf {x}}^k|_{{\mathcal {E}}}$ that

$$\begin{aligned} \Vert {\mathbf {x}}^{*,k}\Vert _2\le \Vert {\mathbf {x}}^{k_2}\Vert _2\le (1+\varrho +\cdots +\varrho ^{k_2-k_1})K_1. \end{aligned}$$

The above inequality shows that $\{{\mathbf {x}}^{*,k}\}$ is bounded. Therefore, $\{({\mathbf {x}}^{*,k}, w^{*,k}, y^{*,k}, \gamma ^{*,k})\}$ is bounded.

Secondly, we prove that every accumulation point of $\{({\mathbf {x}}^{*,k}, w^{*,k}, y^{*,k}, \gamma ^{*,k})\}$ is a minimizer of (P). Without loss of generality, we assume $({\mathbf {x}}^{*,k}, w^{*,k}, y^{*,k}, \gamma ^{*,k})\rightarrow ({\mathbf {x}}^{*}, w^{*}, y^{*}, \gamma ^{*})$ as $k\rightarrow +\infty$. We first show that ${\mathbf {x}}^*\in {\mathcal {R}}$. Note that $\Delta$ is compact. Up to a scaling, we can assume $\Delta \subseteq B(0, \varrho )$ with $\varrho < 1$. By (6.8), we have

$$\begin{aligned} \Vert {\mathbf {x}}^{k}\Vert _2\le K_1/(1-\varrho ), \end{aligned}$$

which implies that $\{{\mathbf {x}}^{k}\}$ is bounded. By adding zero entries to the tailing, each tms ${\mathbf {x}}^k$ can be extended to a vector in ${\mathbb {R}}^{{\mathbb {N}}^n_{\infty }}$, which is a Hilbert space equipped with the inner product

$$\begin{aligned} \langle u, v\rangle =\sum _{\alpha \in {\mathbb {N}}^{n}} u_{\alpha } v_{\alpha }, \; \forall u, v \in {\mathbb {R}}^{{\mathbb {N}}^n_{\infty }}. \end{aligned}$$

So, $\{{\mathbf {x}}^{k}\}$ is also bounded in ${\mathbb {R}}^{{\mathbb {N}}^n_{\infty }}$. By Alaoglu’s Theorem (cf. [16, Theorem C.18]), there exist a subsequence $\{{\mathbf {x}}^{k_j}\}$ that is convergent in the weak-$*$ topology, i.e., there exists ${\bar{\mathbf {x}}}^* \in {\mathbb {R}}^{{\mathbb {N}}^n_{\infty }}$ such that

$$\begin{aligned} \langle f, {\mathbf {x}}^{k_j}\rangle \rightarrow \langle f, {\bar{\mathbf {x}}}^*\rangle , \; \text {as}\; j\rightarrow \infty \end{aligned}$$

for all $f\in {\mathbb {R}}^{{\mathbb {N}}^n_{\infty }}$. So, for each $\alpha \in {\mathbb {N}}^n$,

$$\begin{aligned} ({\mathbf {x}}^{k_j})_{\alpha } \rightarrow ({\bar{\mathbf {x}}}^*)_{\alpha }. \end{aligned}$$

(6.9)

Since ${\mathbf {x}}^k|_{{\mathcal {E}}} = {\mathbf {x}}^{*,k} \rightarrow {\mathbf {x}}^{*}$, we have ${\bar{\mathbf {x}}}^{*}|_{{\mathcal {E}}} = {\mathbf {x}}^*$. By the feasibility of ${\tilde{\mathbf {x}}}^{*,k_j}$ and the definition of ${\mathbf {x}}^{k_j}$, it is easy to check that ${\bar{\mathbf {x}}}^{*}\in {\mathbb {R}}^{{\mathbb {N}}^n_{\infty }}$ is a full moment sequence whose localizing matrices of all orders are positive semidefinite. Thus, ${\bar{\mathbf {x}}}^*$ admits a $\Delta$-measure (cf. [28, Lemma 3.2]). This implies that ${\mathbf {x}}^*={\bar{\mathbf {x}}}^{*}|_{{\mathcal {E}}}\in {\mathcal {R}}$.

Since $({\mathbf {x}}^{*,k}, w^{*,k}, y^{*,k}, \gamma ^{*,k})$ satisfies (6.2) and $(w^{*,k},\gamma ^{*,k}) \in {\mathcal {L}}_{{\bar{n}}+1}$, we know that $({\mathbf {x}}^{*}, w^{*}, y^{*}, \gamma ^{*})$ satisfies

$$\begin{aligned} c \circ \left( {\mathbf {x}}^{*} +\sum _{i=1}^m y^{*}_i \mathrm {vech}(A_i)\right) -w^{*} =c \circ \mathrm {vech}(A_0), \quad (w^{*},\gamma ^{*}) \in {\mathcal {L}}_{{\bar{n}}+1}. \end{aligned}$$

In view of ${\mathbf {x}}^*\in {\mathcal {R}}$, we see that $({\mathbf {x}}^{*}, w^{*}, y^{*}, \gamma ^{*})$ is feasible for (P) and $\vartheta _P \le \gamma ^{*}$. Because ($P^k$) is a relaxation of (P) and $({\mathbf {x}}^{*,k}, w^{*,k}, y^{*,k}, \gamma ^{*,k})$ is a minimizer of ($P^k$), it holds that

$$\begin{aligned} \vartheta _P \ge \gamma ^{*,k}, \quad k=2,3,\ldots . \end{aligned}$$

Hence, we get

$$\begin{aligned} \vartheta _P \ge \lim _{k\rightarrow \infty }\gamma ^{*,k}= \gamma ^{*}. \end{aligned}$$

Therefore, $\vartheta _P= \gamma ^{*}$ and $({\mathbf {x}}^{*}, w^{*}, y^{*}, \gamma ^{*})$ is a minimizer of (P). Since $\{\gamma ^{*,k}\}$ is monotonically increasing, it converges to the minimum of (1.4). $\square$