j******n
发帖数: 271 | 1 给定非负定矩阵:
A = (a_{ij})
问
B = (e^{a_{ij}})
是否非负定？ (note that B != e^A) |
s***e
发帖数: 267 | 2 Consider A = [2, 3.9; 1, 2] which is PD.
However, exp(2)*exp(2)-exp(3.9)*exp(1) < 0.
【在 j******n 的大作中提到】
: 给定非负定矩阵:
: A = (a_{ij})
: 问
: B = (e^{a_{ij}})
: 是否非负定？ (note that B != e^A)
|
h**o
发帖数: 78 | 3 赞
【在 s***e 的大作中提到】
: Consider A = [2, 3.9; 1, 2] which is PD.
: However, exp(2)*exp(2)-exp(3.9)*exp(1) < 0.
|
b**g
发帖数: 949 | 4 平生最佩服构造法解题的高手，最不喜欢反证法。
【在 h**o 的大作中提到】
: 赞
|
D**u
发帖数: 204 | 5 I suppose the question is about symmetric matrices only.
Generally "positive semi-definite" matrices are only defined on symmetric
matrices, and there is no consensus about how to define positive semi-
definite on non-symmetric matrices.
【在 s***e 的大作中提到】
: Consider A = [2, 3.9; 1, 2] which is PD.
: However, exp(2)*exp(2)-exp(3.9)*exp(1) < 0.
|
s***e
发帖数: 267 | 6 Yes you are right. If we look at the quadratic forms, then definition of the
"positive definiteness" of asymmetric
matrices does not make much sense.
If A has to be symmetric, then it seems that the answer is Yes, exp(A) (
elementwise exponential) is also PSD.
One way to see this is to look at expansion of exponential at each element.
Then notice that (1) for PSD, the
diagonal element dominate the corresponding row/column elements; (2) the PSD
property does not change if
we simply rescale the ma
【在 D**u 的大作中提到】
: I suppose the question is about symmetric matrices only.
: Generally "positive semi-definite" matrices are only defined on symmetric
: matrices, and there is no consensus about how to define positive semi-
: definite on non-symmetric matrices.
|
J******d
发帖数: 506 | 7 都不需要是对陈阵么？
【在 s***e 的大作中提到】
: Consider A = [2, 3.9; 1, 2] which is PD.
: However, exp(2)*exp(2)-exp(3.9)*exp(1) < 0.
|
J******d
发帖数: 506 | 8 1. A>=0 推出 (A)^k >= 0. 这里是对A做k次Hadamard product.
2. 因此，(A)^0 + (A)^1 + (A)^2/2 + ... (A)^k/k! >=0.
3. B >=0.
【在 j******n 的大作中提到】
: 给定非负定矩阵:
: A = (a_{ij})
: 问
: B = (e^{a_{ij}})
: 是否非负定？ (note that B != e^A)
|
|
g**********1
发帖数: 1113 | 9 Notice B!=e^A. A^2!=(a_{ij}^2) right? |
j******n
发帖数: 271 | 10 you are right.
【在 J******d 的大作中提到】
: 1. A>=0 推出 (A)^k >= 0. 这里是对A做k次Hadamard product.
: 2. 因此，(A)^0 + (A)^1 + (A)^2/2 + ... (A)^k/k! >=0.
: 3. B >=0.
|
