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Message |
Country_Chiel
Guest
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 Posted:
Tue Dec 21, 2004 1:37 pm Post subject:
Vector differentiation problem - LMS type |
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This is something I am not clear about
suppose I have the quadratic function (where ' denotes transpose)
J=X'AX
and I wish to differentiate this (as in LMS problems) wrt X
I know dJ/dX = 2AX (I think) assuming A is symmetric.
But how do I diferentiate
J=GAG' ie dJ/dG where G is a matrix and A is symmetric like before?
Thanks |
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Rune Allnor
Guest
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| Posted:
Wed Dec 22, 2004 1:43 pm Post subject:
Re: Vector differentiation problem - LMS type |
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Country_Chiel wrote:
| Quote: | This is something I am not clear about
suppose I have the quadratic function (where ' denotes transpose)
J=X'AX
and I wish to differentiate this (as in LMS problems) wrt X
I know dJ/dX = 2AX (I think) assuming A is symmetric.
But how do I diferentiate
J=GAG' ie dJ/dG where G is a matrix and A is symmetric like before?
Thanks
|
You probably have to write out the differentiation in the nitty
gritty details. For a real-valued vector v and parameter vector a
we have
dv/da = [dv_1/da_1,dv_2/da_2,...,dv_N,da_N]^T. [1]
So
d(v'w)/dw = d(w'v)/dw = v. [2]
So try to write the expression out in terms of the columns of J,
and differentiate each such espression with respect to the
columns of G'. I haven't tried these sorts of computations
myself, but seeing the symmetry in [2] above, one might hope
that the computation ends up with something like
dJ/dG' = 2AG'. [3]
Rune |
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Country_Chiel
Guest
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| Posted:
Thu Dec 23, 2004 1:30 am Post subject:
Re: Vector differentiation problem - LMS type |
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"Rune Allnor" <allnor@tele.ntnu.no> wrote in message
news:1103704995.680277.113000@f14g2000cwb.googlegroups.com...
| Quote: |
Country_Chiel wrote:
This is something I am not clear about
suppose I have the quadratic function (where ' denotes transpose)
J=X'AX
and I wish to differentiate this (as in LMS problems) wrt X
I know dJ/dX = 2AX (I think) assuming A is symmetric.
But how do I diferentiate
J=GAG' ie dJ/dG where G is a matrix and A is symmetric like before?
Thanks
You probably have to write out the differentiation in the nitty
gritty details. For a real-valued vector v and parameter vector a
we have
dv/da = [dv_1/da_1,dv_2/da_2,...,dv_N,da_N]^T. [1]
So
d(v'w)/dw = d(w'v)/dw = v. [2]
So try to write the expression out in terms of the columns of J,
and differentiate each such espression with respect to the
columns of G'. I haven't tried these sorts of computations
myself, but seeing the symmetry in [2] above, one might hope
that the computation ends up with something like
dJ/dG' = 2AG'. [3]
Rune
Yes I think that is a good guess. They appear to call it vec(matrix) for an |
individual vector but it gets complicated with Kroneker products etc.
Country Chiel |
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