对称正交反对称矩阵反问题

设P为一给定的对称正交矩阵, 记SAR\+n\-P={A∈R\+\{n×n\}|A\+T=A,(PA)\+T=-PA}. 该文考虑下列问题问题Ⅰ〓给定X∈R\+\{n×m, Λ=diag(λ\-1,λ\-2,…, λ\-m)∈R\+\{m×m\}, 求A∈SAR\+n\-P使AX=XΛ,问题Ⅱ〓给定X,B∈R\+\{n×m, 求A∈SAR\+n\-P使  ‖AX-B‖=min.问题Ⅲ设AKA～]∈R\+\{n×n\},求A\+*∈S\-E使 ‖AKA～]-A\+*‖=infDD(X]A∈S\-EDD)]‖AKA～]-A‖, 其中S\-E为问题Ⅱ的解集合, ‖·‖表示Frobenius范数.该文得到了问题Ⅰ有解的充要条件及解集合的表达式, 给出了解集合S\-E的通式和逼近解A\+*的具体表达式.

The Inverse Problem of Symmetric Ortho-anti-symmetric Matrices

Given P∈OR\+\{n×n satisfying  P\+T=P. Set  SAR\+n\-P={A∈R\+\{n×n|A\+T=A, (PA)\+T=-PA}. This paper discusses the following problems Problem Ⅰ〓Given X∈R\+\{n×m, Λ=diag(λ\-1,λ\-2, …, λ\-m)∈R\+\{m×m. Find A∈SAR\+n\-P such that AX=XΛ.Problem ⅡGiven B, X∈R\+\{n×m.Find A∈SAR\+n\-P such that ‖AX-B‖=min.Problem Ⅲ〓Given AKA～]∈R\+\{n×n. FindA\+*∈S\-E such that‖AKA～]-A\+*‖=infDD(X]A∈S\-EDD)]‖AKA～]-A‖ where S\-E is the solution set of Problem Ⅱ, ‖\5‖ is the Frobenius norm.The necessary and sufficient conditions for the solvability of  Problem Ⅰ have been studied. The general form of the solution set S\-E of Problem Ⅱ has been given. For Problem Ⅲ the expression of the solution A\+* has been provided.