线性流形上对称正交反对称矩阵反问题的最小二乘解

One kind of inverse problems for symmetric orthogonal anti-symmetric matrices on the linear manifold

设P是n阶对称正交矩阵,如果n阶矩阵A满足AT=A和(PA)T=-PA,则称A为对称正交反对称矩阵,所有n阶对称正交反对称矩阵的全体记为SARnp.令S={A∈SARnp f(A)=‖AX-B‖=m in,X,B〗∈Rn×m本文讨论了下面两个问题问题Ⅰ给定C∈Rn×p,D∈Rp×p,求A∈S使得CTAC=D问题Ⅱ已知A~∈Rn×n,求A∧∈SE使得‖A^-A∧‖=m inA∈SE‖A^-A‖其中SE是问题Ⅰ的解集合.文中给出了问题Ⅰ有解的充要条件及其通解表达式.进而,指出了集合SE非空时,问题Ⅱ存在唯一解,并给出了解的表达式,从而得到了求解A∧的数值算法.

Let P be an n×n symmetric orthogonal matrix. An n × n matrix A is called a symmetric orthogonal anti-symmetric matrix if A^T = A and （PA）^T =-PA. We denote the set of all n × n symmetric orthogonal anti-symmetric meatrices by SARp^n. Let S={A∈SARp^n f（A）=｜｜AX-B｜｜=min,X,B｜｜∈ER^n×m We discuss the following problems： Problem Ⅰ Given C ∈ Rn×p,D ∈ R^p×p ,find A ∈ S, such that C^TAC = D Problem Ⅱ Given A ∈ R^n×n ,find A ∈ SE ,such that ｜｜A-A｜｜ = min A∈SE｜｜A-｜｜ where SE is the solution set of Problem Ⅰ . The sufficient and necessary condition under which SE is nonempty is obtained. The general form of SE is given. Then the expression of the solution A of Problem Ⅱ is presented and the numerical method is described.