On 14-Aug-03 Feng Zhang wrote: > Thank, Jerome > > The question is if this generalized inverse can make > their product to be identity matrix?
>> On August 14, 2003 09:24 am, Feng Zhang wrote: >> > Dear R-listers, >> > >> > I have a dxr matrix Z, where d > r. >> > And the product Z*Z' is a singular square matrix. >> > The problem is how to get the left inverse U of this >> > singular matrix Z*Z', such that U*(Z*Z') = I? No, not if I is to be a full identity matrix (1s all the way along the diagonal), when Z*Z' is singular. Compare ranks on both sides; or simply observe that det(U*(Z*Z')) = det(U)*det(Z*Z') = 0 while det(I) = 1. However, you could have an incomplete diagonal (r 1s and (n-r) 0s where r is the rank of Z*Z'). Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) <[EMAIL PROTECTED]> Fax-to-email: +44 (0)870 167 1972 Date: 14-Aug-03 Time: 20:41:01 ------------------------------ XFMail ------------------------------ ______________________________________________ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
