反对称正交反对称矩阵反问题的最小二乘解

设P为一给定的对称正交矩阵,记AAnp={A∈Rn×n‖AT=-A,(PA)T=-PA}.讨论下列问题问题Ⅰ给定X,B∈Rn×m.求A∈AARnp使‖AX-B‖=min.问题Ⅱ设A∈Rn×n,求A*∈SE使‖A-A*‖=infA∈SE‖A-A‖,其中SE为问题Ⅰ的解集合,‖·‖表示Frobenius范数.研究AARnp中元素的通式,给出问题Ⅰ解的一般表达式,证明了问题Ⅱ存在唯一逼近解A*,且得到了此解的具体表达式.

Least-square solutions of inverse problems for anti-symmetric ortho-antisymmetric matrices

Given P ∈ Orn×n satisfying pT = p. A ∈ Rn×n is called anti-symmetric ortho-antisymmetric matrix if A = -AT, (PA)T = -PA. The set of all n × n anti-symmetric ortho-antisymmetric matrices is denoted by AARp. The following two problems are discussed in this paper:Problem Ⅰ: Given X, B ∈ Rn×m, find A ∈ AARnp such that f(A) = ‖AX - B‖ = min,Problem Ⅱ: Given A ∈ Rn×n, find A* ∈ SE such that ‖A~-A*‖=A∈SE‖A~-A‖ where ‖·‖ is the Frobenius norm, and SE is given. For Problem Ⅱ, the expression of the solution is provided. Futhermore, it is pointed that some results of References 2 and 7 are special cases of this paper.