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The Inverse of a Bordered Matrix

  • Hongbo Zhang

摘要

A ring R is said to have stable range one if for any \(a,x\in R\) a , x R , there exists a \(t\in R\) t R such that \(a+(1-ax)t=u\) a + ( 1 - a x ) t = u is a unit of R. A nonsingular matrix \(S_W(X,Y)\) S W ( X , Y ) is constructed over such ring. Consequently for any \(A\in \mathbb {C}^{n\times n}\) A C n × n , \(X\in A\{2\}\) X A { 2 } , it is proved that for any matrix \(B,C^*\) B , C with full column rank satisfying \(R(B)=N(X)\) R ( B ) = N ( X ) and \(N(C)=R(X)\) N ( C ) = R ( X ) , \(\left( {\begin{matrix} A & B \\ C & 0 \end{matrix}}\right) ^{-1}=\left( {\begin{matrix} X & & GB(CB)^{-1}\\ (CB)^{-1}C& & -(CB)^{-1}CAB(CB)^{-1} \end{matrix}}\right) \) A B C 0 - 1 = X G B ( C B ) - 1 ( C B ) - 1 C - ( C B ) - 1 C A B ( C B ) - 1 for some matrix G if and only if \(AX=XA^2X\) A X = X A 2 X and there is a nonsingular matrix W such that \(AXW=AXA\) A X W = A X A and \(W(I_n-XA)=UQ_WY\) W ( I n - X A ) = U Q W Y for some nonsingular matrix \(Q_W\) Q W , where \(I_n-AX=UY\) I n - A X = U Y is a full rank factorization. And G could be any solution of the equation \(G(I_n-AX)=(I_n-XA)\) G ( I n - A X ) = ( I n - X A ) . For \(X\in \{A_d, A^{\tiny \textcircled {\tiny \dag }}, A^{ow}\}\) X { A d , A , A ow } , A and X satisfy this equivalent condition. Besides, \(\left( {\begin{matrix} A & U\\ Y& 0 \end{matrix}}\right) ^{-1}=\left( {\begin{matrix} X & GU\\ Y& -YAU \end{matrix}}\right) .\) A U Y 0 - 1 = X GU Y - Y A U .