I guess what is meant is the fact that if you compute a normalized PC1 from a correlation matrix based on p variables for data in which PC1 represents size and growth is isometric then the loadings will all be equal to 1/sqrt(p). That is because "normalized" means that the sum of the squared elements adds up to 1 and "isometric" means the elements are all equal. If you then divide each loading by 1/sqrt(p) (that is, if you multiply them by sqrt(p)) the result will be values of 1 for each variable.
If growth is not isometric then some variables will have values less than 1 and some will have values greater than 1. Perhaps the author considers that easier than asking the reader to just compare the PC1 loadings against 1/sqrt(p) directly. -------------------- F. James Rohlf - Dept. Ecology & Evolution SUNY, Stony Brook, NY 11794-5245 > -----Original Message----- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of > [EMAIL PROTECTED] > Sent: Tuesday, March 09, 2004 8:47 AM > To: [EMAIL PROTECTED] > Subject: newbie question > > > Hi there, > > I got a newbie question. I was reading a paper on > interspecific morphological variation, and at some > point the authors say that "each value of the > standardized PC1 for each species was divided by > 1/sqrt(number of characters) to assess morphological > divergence from isometry." > > Could someone please explain to me how that > standardization would allow that assessment? > > morphometrically yours, > > Mark ([EMAIL PROTECTED]) > > == > Replies will be sent to list. > For more information see > http://life.bio.sunysb.edu/morph/morphmet.htm> l. > == Replies will be sent to list. For more information see http://life.bio.sunysb.edu/morph/morphmet.html.
