Recent content by GeoffO

Hmm... yeah. I'm not sure it's possible. I looked around for a while and found nothing. hyperref does not seem to support it and it's pretty standard. Sorry I can't help more.

I don't know the answer but am curious now. Are you doing this to use with latex2html, or does the pdf format support title text as well?

Good point... It's careful surgery. I'll have to double check, not everything here is symmetrical, so being sure to associate carefully is very important. (I edited the post above lest I confuse any future reader.)

** solved ** I see how you did it, for any three square **SYMMETRIC** (or anti-symmetric) matrices you can reorder at will because of these properties. The first Tr(AB) = Tr(BA) allows us to write Tr(ABC) = Tr(BCA) and the like. The second is Tr(A) = Tr(A^T). With these we can permute any...

Even still, what allows you to go from Tr(M N M^T N^{-1}) to Tr(M M^T N N^{-1})? This is not just a rotation, it's a reordering, right?

Yeah, sorry the inverted terms should have been (XX^T)^{-1} and (YY^T)^{-1}. This would result in all of the X and Y terms canceling out in your reduction. I'm excited to learn what property allows for this step!

Hmm... somewhere I have made an error. (X^TX)^{-1} is a 3x3, so this doesn't make sense... back in a minute after I figure out my error. In the meantime how did you introduce multiplication of traces? What property am I missing? This looks promising!

I see. You mean the following, right? \mathrm{trace}[ (B^{-1}A)(X^TX)^{-1}(B^{-1}A)^T (XX^T) - (B^{-1}A)(Y^TY)^{-1}(B^{-1}A)^T (YY^T) ] I would like to factor out the (B^{-1}A) terms, if possible. (Thanks!)

What property allows me to reorder like that? I only know of the property \mathrm{trace}(AB) = \mathrm{trace}(BA) (Again, thank you.)

Thanks for the reply. I have used this property and end up with \mathrm{trace}[ (B^{-1}A)^T (XX^T(B^{-1}A)(X^TX)^{-1} - YY^T(B^{-1}A)(Y^TY)^{-1} ] Which is okay, but I'd really like to get that common (B^{-1}A) term out of there. We can rewrite this as (Q^T)(R^T Q R^{-1} - S^T...

I am trying to simplify the following, so that I can differentiate it (with respect to X). Ideally I'll have everything in terms of (XX^T - YY^T). \mathrm{trace}[(AX(AX)^T)((BX)(BX)^T)^{-1}] - \mathrm{trace}[(AY(AY)^T)((BY)(BY)^T)^{-1}] Where X and Y are 3 x N and A and B are N x N...