对称正交反对称矩阵反问题

The Inverse Problem of Symmetric Ortho-anti-symmetric Matrices

设 P为一给定的对称正交矩阵 ,记 SARn P={A∈ Rn× n| AT=A,( PA) T=- PA}.该文考虑下列问题问题 给定 X∈Rn× m,Λ=diag( λ1,λ2 ,… ,λm)∈Rm× m,求 A∈ SARn P使AX =XΛ . 问题 给定 X,B∈Rn× m ,求 A∈SARn P使‖ AX - B‖ =min. 问题 设 A∈ Rn× n,求 A* ∈SE使‖ A- A* ‖ =infA∈ SE‖ A- A‖ ,其中 SE为问题 的解集合 ,‖·‖表示 Frobenius范数 .该文得到了问题 有解的充要条件及解集合的表达式 ,给出了解集合 SE的通式和逼近解A*的具体表达式 .

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 Ⅱ Given B, X∈R+{n×m. Find A∈SAR+n-P such that ‖AX-B‖=min. Problem Ⅲ Given ∈R+{n×n. Find A+*∈S-E such that‖-A+*‖= infA∈S-E‖-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.