j****s
发帖数: 156 | 1 What I have learned as Hadamard's Inequality is that for positive-semi-
definite matrix
det(A)<=a_11a_22.....a_NN
where a_ii is the $i$th diagonal elements of A.
This is different from the definiation of Hadamard's Inequality in Wiki and
Mathworld.
http://en.wikipedia.org/wiki/Hadamard's_inequality
And I cannot find the related matrials from google!!!!!!!!!!!!!!!!
What really is going on?
Thanks | l*********g
发帖数: 9 | 2 The original Hadamard inequality states that
det(G(x_1,...,x_n))<=det(G(x_1,...,x_k))det(G(x_(k+1),...,x_n)) where, G is
the Gramian of vector x_1,...x_n
All other results follow from this. Let A=[x_1,...,x_n]
Then G(x_1,...,x_n)=A(^T)A By Hardamard we have
(detA)^2<=prod(||x_i||^2), which is what you find in wiki.
If H>=0 then H=A^2, where A>=0 A=[x_1,...,x_n] then H=G(x_1,...,x_n). By Had
amard you have det(H)<=prod(H_ii), which is a special case.
Hopefully this will make sense to you.
and
【在 j****s 的大作中提到】
: What I have learned as Hadamard's Inequality is that for positive-semi-
: definite matrix
: det(A)<=a_11a_22.....a_NN
: where a_ii is the $i$th diagonal elements of A.
: This is different from the definiation of Hadamard's Inequality in Wiki and
: Mathworld.
: http://en.wikipedia.org/wiki/Hadamard's_inequality
: And I cannot find the related matrials from google!!!!!!!!!!!!!!!!
: What really is going on?
: Thanks
| j****s
发帖数: 156 | 3 great!!! Thanks a lot
is
Had
【在 l*********g 的大作中提到】
: The original Hadamard inequality states that
: det(G(x_1,...,x_n))<=det(G(x_1,...,x_k))det(G(x_(k+1),...,x_n)) where, G is
: the Gramian of vector x_1,...x_n
: All other results follow from this. Let A=[x_1,...,x_n]
: Then G(x_1,...,x_n)=A(^T)A By Hardamard we have
: (detA)^2<=prod(||x_i||^2), which is what you find in wiki.
: If H>=0 then H=A^2, where A>=0 A=[x_1,...,x_n] then H=G(x_1,...,x_n). By Had
: amard you have det(H)<=prod(H_ii), which is a special case.
: Hopefully this will make sense to you.
:
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