home *** CD-ROM | disk | FTP | other *** search
- Newsgroups: sci.math
- Path: sparky!uunet!van-bc!balden
- From: balden@wimsey.bc.ca (Bruce Balden)
- Subject: Re: sum of eigenvalues of a covariance matrix = trace ?
- Organization: Wimsey Information Services
- Date: Sat, 9 Jan 1993 23:23:23 GMT
- Message-ID: <C0M0z0.1A6@wimsey.bc.ca>
- References: <1993Jan9.194845.8567@noose.ecn.purdue.edu>
- Lines: 12
-
- In article <1993Jan9.194845.8567@noose.ecn.purdue.edu> kavuri@lips1.ecn.purdue.edu (Surya N Kavuri ) writes:
- > Is the sum of the eigenvalues of a covariance matrix equal to
- > its trace ?
- Properly interpreted, yes.
-
- For any two square matrices A, B of the same size, tr(AB) = tr(BA)
-
- Therefore tr((inv(A)*B)A) = tr(A*inv(A)*B) = tr(B).
-
- Thus, tr(B) is invariant under similarity transformations.
- The diagonalization of B contains the eigenvalues (repeated according
- to multiplicity) and is similar to B. Therefore, it has the same trace.
-