l******f
发帖数: 32 | 1 假如 A, B are both n-by-n matrix. And B > A > 0. Furthermore, A and B are
both symmetric and non-singular.
inverse( A ) > inverse( B ) is established or not?
非常感谢！ | R*********r
发帖数: 1855 | | p******n
发帖数: 66 | 3 B>A>0 => B-A>0 => I- B^{-0.5}A B^{-0.5}>0=>
==> 0
==> (B^{-0.5}AB^{-0.5})^{-1}-I>0
==> B^{0.5}A^{-1}B^{0.5}-I>0
==> A^{-1}-B^{-1}>0 |
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