Re: Nonorthogonal Eigenvectors

*To*: mathgroup at smc.vnet.net*Subject*: [mg116397] Re: Nonorthogonal Eigenvectors*From*: Leonid Shifrin <lshifr at gmail.com>*Date*: Sun, 13 Feb 2011 03:06:56 -0500 (EST)*References*: <201102121019.FAA20109@smc.vnet.net>

Well, you obviously have chosen a bad example. Your matrix has two degenerate zero eigenvalues, which means that any linear combination of the two corresponding eigenvectors can be chosen as the basis of zero-eigenvalue subspace. The fact that analytical and numerical routines make different choices for that does not make any of the two wrong. What IMO would actually be wrong would be to rely on any particular choice, given the degeneracy. Regards, Leonid On Sat, Feb 12, 2011 at 1:19 PM, Kevin J. McCann <kjm at kevinmccann.com>wrote: > I have seen some threads from the past on this, but never got a > satisfactory answer. > > Suppose I have an exact matrix A: > > A = {{1, 0, 0, 0, 2}, {0, 16, 0, 0, 0}, {0, 0, 9, 0, 0}, {0, 0, 0, 0, > 0}, {2, 0, 0, 0, 4}}; > > P = Eigenvectors[A] > > produces the following > > {{0,1,0,0,0},{0,0,1,0,0},{1,0,0,0,2},{-2,0,0,0,1},{0,0,0,1,0}} > > which is not an unitary matrix, although the vectors are orthogonal, > just not normal, i.e. > > Transpose[P].P > > is not the identity matrix. > > However, if I make A numeric: > > nA = A//N > > then > > nP = Eigenvectors[nA] > > produces > > {{0., 1., 0., 0., 0.}, {0., 0., 1., 0., 0.}, {0.447214, 0., 0., 0., > 0.894427}, {-0.894427, 0., 0., 0., 0.447214}, {0., 0., 0., -1., 0.}} > > and > > Transpose[nP].nP > > is the identity matrix. > > I do not understand why making the matrix inexact produces the result > that I would expect, but when the matrix is exact it doesn't. Also, I > don't think the inconsistency is a useful thing. > > Any ideas why someone decided to do it this way? > > Kevin > >

**References**:**Nonorthogonal Eigenvectors***From:*"Kevin J. McCann" <kjm@KevinMcCann.com>