Talk:Matrix calculus/Archive 1

message: This isn't conventional differentiation. There needs to be a discussion of what differentiation with respect to x means here. Charles Matthews 22:51, 19 Apr 2005 (UTC)
replies: text: Definition of differentiation with regard to a vector is now added.--Fredrik Orderud 14:59, 7 May 2005 (UTC)
thread_title: 1

message: As usual, when I do a heavy rewrite, I will try to give you a point by point listing of structural or conceptual changes.  If I were a better wikipedian, I would probably discuss changes I wanted to make before I start editing, but well, I'm not.  And when the mood hits me to write, I do it.
# I've changed the notation a bit.  A lot of things are not transposed that were, and some things are transposed that weren't
# Emphasis on the directional derivatives, which are important
# Removed the possibility of complex matrices.  Sure, that's possible, but why bother?
# I more or less killed the whole section about Fréchet derivative, mostly for the reason that I decided that this derivative is not a special case of the Fréchet derivative.
# Oh, I also think that the derivatives need to display where they're evaluated at.  Since they have two domains, it doesn't quite parse if you don't.
and that's pretty much it.  What do you old pros think?  And what about you, ashigabou?  This is your baby, afterall.  Oh, also, I got tired before I got to the examples section.  I think some transposes in there still need to be changed. -lethe talk [ +] 17:46, 1 February 2006 (UTC)
replies: text: Oh, I see that this was originally your article, not ashigabou's.  My mistake.  Well anyway. I'm glad you like the changes.  A bit more work to do, but I think it's coming along nicely. -lethe talk [ +] 00:49, 2 February 2006 (UTC)
thread_title:  my heavy rewrite 

message: Does anyone want to try convert the elegant
:d\,\big({X^{-1}}\big) = -X^{-1}\cdot d\, X \cdot X^{-1}
into this notation? Arthur Rubin | (talk) 21:20, 1 February 2006 (UTC)
replies: text: I suppose we needs must do.  It's an important formula. -lethe talk [ +] 00:49, 2 February 2006 (UTC)
replies: text:  there is also the determinant based formula, linked to the inverse formula through the cofactor matrix. I still cannot get my head around all the details for those two cases... 133.186.47.9 05:48, 3 February 2006 (UTC)

text:  Converting \frac{\partial A^{-1}}{\partial A}(X) =  - A^{-1}XA^{-1} looks really ackward to me. From my POV, that's where the notation using partial derivative falls down (at least from a practical point of view). I think maybe we should add the derivative and determinant using this notation instead ? Ashigabou 06:21, 4 February 2006 (UTC)
thread_title:  inverse derivative 

message: There seems to be something wrong with the product-rule equation, since the first term does not differentiate on X:
:
\frac{\partial (\mathbf{YZ})}{\partial \mathbf{X}} = \frac{\partial\mathbf{Y}}{\partial \mathbf{Z}}\mathbf{X} + \mathbf{Y}\frac{\partial\mathbf{Z}}{\partial \mathbf{X}}
Shouldn't the equation be something simmilar to this instead?
:
\frac{\partial (\mathbf{YZ})}{\partial \mathbf{X}} = \frac{\partial\mathbf{Y}}{\partial \mathbf{X}}\mathbf{Z} + \mathbf{Y}\frac{\partial\mathbf{Z}}{\partial \mathbf{X}}
At last  seems to suggest this. --Fredrik Orderud 00:45, 2 February 2006 (UTC)
replies: text: Ooops, of course you're right, that's a mistake.  Lemme fix. -lethe talk [ +] 00:49, 2 February 2006 (UTC)
thread_title:  Typo in the product rule? 

message: The definition using partial derivative can be defined for function which are not differentiable (at least Gateaux or Fréchet differentiability), but I cannot see any use of the definition given at this article on thoses cases. After having checker at several references, I think that this article should really be understood in the Fréchet meaning. According to Universalis Encyclopedia, the article "calcul infinitesimal a plusieurs variables" says that Fréchet derivative is the usual derivative in Banach spaces, and particularly in finite dimension R space vectors with the norm taken from the usual scalar  product. The expression Frechet derivative, still according to Universalis, is not used anymore: it is simply called differential. I read a bit about tensors, and if I understand a tensor can be seen as the representation of a linear map between vectors: for example, a linear map from matrix spaces to matrix space is a 4 rank matrix, a linear map from n-dimension vectors to p-dimension vectors is a 2-rank tensor, equivalent to a matrix. I also found a presentation of Taylor theorem in multi-dimension, which defines mixed-partial derivative as multi-linear maps, and such having the Fréchet definition of derivative: http://gold-saucer.afraid.org/math/taylor/taylor.pdf. If we evoke tensor as linear map, I don't think we need any tensor theory (which should be avoided here, as it is a practical article). Ashigabou 08:56, 6 February 2006 (UTC)
replies: text: As far as I know, the expression "Frechet derivative" is still used in English for derivatives between vector spaces. Whether you call it a tensor or a multi-linear map depends on your background. However, I don't think that a tensor treatment is more theoretical than a "multi-linear map" treatment; actually, my guess would be that physicists would prefer tensors and mathematicians multi-linear maps. -- Jitse Niesen (talk) 12:22, 15 February 2006 (UTC)
thread_title:  Is the differentiation given here really different than Fréchet derivative ?