v******s
发帖数: 144 | 1 trace(X'AX)
max---------------
trace(X'BX)
s.t. trace(X'X) = 1
A and B are postive-semidefinite
Do we have a close form of X? What is it? Thanks. | D*******a
发帖数: 207 | 2 I have done this type of problem. From my memory, close form solution
depends on how you define close form. Is A^(-1) close form? In reality, when
you implement the algorithm in DSP or computer, you use iteration to
compute A^(-1) any way. So it is really not close form. In this sense, it
does not have close form solution. But, if you think this close form, then
yes. From my memory, not guaranteed to be right, If B is not singular, the
solution is the solution to max(x' AB^(-1) x)
You can google | v******s
发帖数: 144 | 3 Thank you Dannemora.
I know if x is a vector, it is a GSVD problem. But if X is a matrix, can we
generalize that X can be represented by CP, where C is the eigenvectors of B
^(-1)A and P is the waterfilling coefficient? | D*******a
发帖数: 207 | 4 You are welcome. I guess you require the vectors of X to be perpendicular to
each other, and each be normalized. If so, From my memory, and also from
intuition, X should be composed of the eigen vectors of the GSVD associated
with the largest eigen values.
But I have forgotten almost everything related to waterfilling, so no
comments on that. | r*******r
发帖数: 1014 | 5 Fisher's criterion? Looks too expensive/unstable to involve matrix inverse.
【在 v******s 的大作中提到】
: Thank you Dannemora.
: I know if x is a vector, it is a GSVD problem. But if X is a matrix, can we
: generalize that X can be represented by CP, where C is the eigenvectors of B
: ^(-1)A and P is the waterfilling coefficient?
| w*******i
发帖数: 525 | 6 trace(X'X) = 1
这个条件基本是多余的。
因为你随便怎么扩大缩小X，objective function的值都不变。
所以1楼是正解，然后scale X，使得trace是1就好了 |
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