l****y
发帖数: 267 | 1 【 以下文字转载自 Mathematics 讨论区 】
【 原文由 lpuppy 所发表 】
if A, B and B-A are all symmetric non-negative definite,
How to prove that |B|>=|A|?
Thanks! | l****y
发帖数: 267 | 2 还是问问校友吧，xixi
【在 l****y 的大作中提到】
: 【 以下文字转载自 Mathematics 讨论区 】
: 【 原文由 lpuppy 所发表 】
: if A, B and B-A are all symmetric non-negative definite,
: How to prove that |B|>=|A|?
: Thanks!
| a***l
发帖数: 66 | 3 Do you mean |A|=max(|eigenvalue of A|)?
For any |w|=1, w'(B-A)w>=0 => w'Bw>=w'Aw, by letting w to be the
eigenvector corresponding to the largest eigenvalue of A, w'Aw=|A|,
Therefore, w'Bw>=|A|, => |B|>=|A|.
【在 l****y 的大作中提到】
: 还是问问校友吧，xixi
| l****y
发帖数: 267 | 4 thanks, but |A| is the determinant of A
not the largest eigenvalue.
Thanks for your reply!
【在 a***l 的大作中提到】
: Do you mean |A|=max(|eigenvalue of A|)?
: For any |w|=1, w'(B-A)w>=0 => w'Bw>=w'Aw, by letting w to be the
: eigenvector corresponding to the largest eigenvalue of A, w'Aw=|A|,
: Therefore, w'Bw>=|A|, => |B|>=|A|.
| l****y
发帖数: 267 | 5 your proof is to prove the largest eigenvalueof B>= that of A.
Thanks!
【在 a***l 的大作中提到】
: Do you mean |A|=max(|eigenvalue of A|)?
: For any |w|=1, w'(B-A)w>=0 => w'Bw>=w'Aw, by letting w to be the
: eigenvector corresponding to the largest eigenvalue of A, w'Aw=|A|,
: Therefore, w'Bw>=|A|, => |B|>=|A|.
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