Abstract
In this paper, the following two problems are considered: Problem I Given a full column rank matrix X ∈ R n × k , a diagonal matrix Λ ∈ R k × k ( k ≤ n ) and matrices M a ∈ R n × n , C 0 , K 0 ∈ R r × r , find n × n matrices C , K such that M a X Λ 2 + C X Λ + K X = 0 , s. t. C ( [ 1 , r ] ) = C 0 , K ( [ 1 , r ] ) = K 0 , where C ( [ 1 , r ] ) and K ( [ 1 , r ] ) are, respectively, the r × r leading principal submatrices of C and K . Problem II Given n × n matrices C a , K a with C a ( [ 1 , r ] ) = C 0 , K a ( [ 1 , r ] ) = K 0 , find ( C ˆ , K ˆ ) ∈ S E , such that ‖ C a − C ˆ ‖ 2 + ‖ K a − K ˆ ‖ 2 = inf ( C , M ) ∈ S E ( ‖ C a − C ‖ 2 + ‖ K a − K ‖ 2 ) , where S E is the solution set of Problem I. By applying the theory and methods of the algebraic inverse eigenvalue problems, the solvability condition and the general solution to Problem I are derived. The expression of the solution to Problem II is presented.
Published Version
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