n****8 发帖数: 37 | 1 我看到:
A,B都是n x n 对称positive definite(pd)。
如果A-B是pd
那么 B逆-A逆一定也是pd. ( B^{-1} - A^{-1} )
请问为什么啊?
谢谢! | h***t 发帖数: 2540 | 2 eigenvalues of A are all greater than B if diagonalized,
so eigenvalues of B inverse are all greater than eigenvalues of A if
diagonalized.
so it is pd
【在 n****8 的大作中提到】 : 我看到: : A,B都是n x n 对称positive definite(pd)。 : 如果A-B是pd : 那么 B逆-A逆一定也是pd. ( B^{-1} - A^{-1} ) : 请问为什么啊? : 谢谢!
| n****8 发帖数: 37 | 3 谢谢这么快答复。
还是不理解你的第一句话。可以具体一点吗?
谢谢! | h***t 发帖数: 2540 | 4 for A & B to be pd, you can always find P such that PD1P'=A, PD2P'=B, D1
and D2 are diagonal.
A>B means D1>D2 for every element | n****8 发帖数: 37 | 5 same P for both A and B? why?
【在 h***t 的大作中提到】 : for A & B to be pd, you can always find P such that PD1P'=A, PD2P'=B, D1 : and D2 are diagonal. : A>B means D1>D2 for every element
| h***t 发帖数: 2540 | 6 I am sorry, seems I misleaded u.
A= PD1P', B=QD2Q'=E1PD2(E1P)'=PE2D2E2P'=PD3P',( P&Q are non-singular, so E1
exist, again E1P is non-singular, there exist E2, such that E1P=PE2)
A>B, then D1>D3
the rest is clear |
|