On matrix characterizations for P-property of the linear transformation in second-order cone linear complementarity problems

A dynamical system based on projection operator for solving absolute value equations associated with second-order cone

“…According to Lemma 3.2, if we set s+t = Ax−b and s−t = x, we can obtain the equivalence between SOCAVEs (1.2) and SOCGLCP (3.1). We should point out that the equivalence between SOCAVEs (1.2) and SOCGLCP (3.1) is implicit in the proof of [35,Theorem 4.1] and the proof of Lemma 3.2 is also inspired by that of [35,Theorem 4.1]. Moreover, SOCGLCP (3.1) is equivalent to the generalized linear variational inequality problem associated with SOC (SOCGLVI) [11]:…”

“…We are interested in SOCAVEs (1.2) and SOCGAVEs (1.3) not only because they are extensions of the standard ones, but also because they are equivalent with some LCPs associated with SOC (SOCLCPs), which have various applications in engineering, control and finance [18,34,35]. Recently, some numerical methods and theoretical results have been developed for SOCAVEs (1.2) and SOCGAVEs (1.3).…”

“…In addition, the unique solvability for SOCAVEs (1.2) and SOCGAVEs (1.3) was also investigated in [36]. Miao and Chen [35] investigated conditions under which the unique solution of SOCAVEs (1.2) is guaranteed, which are different from those in [36]. Hu, Huang and Zhang proved that SOCGAVEs (1.3) is equivalent to the following problem: find x, y ∈ R n such that…”

“…Note that the dimension of the above matrix N is three times the dimension of the matrix A (or B) (i.e., ℓ = 3n). More recently, Miao and Chen [35], under the condition that 1 is not an eigenvalue of A or N , provided the equivalence between SOCAVEs (1.2) and SOCLCP (1.7) without changing the dimension (i.e., ℓ = n).…”

“…Since s = (s 1 , s 2 ) ∈ K n and t = (t 1 , t 2 ) ∈ K n , we haves 1 ≥ s 2 and t 1 ≥ t 2 , which implies that | s 2 , t 2 | ≤ s 2 t 2 ≤ s 1 t 1 .Thus,s, t = s 1 t 1 + s ⊤ 2 t 2 ≥ s 1 t 1 − s 1 t 1 = 0,and the equality is valid if and only if s 2 = kt 2 (k ≥ 0), s 1 = s 2 and t 1 = t 2 . Hence, the vectors s and t in (3.2) share the same Jordan frame[2,35]. Let s = λ 1 e 1…”