George Hammond <ghammond@mediaone.net> writes:
> Stein A. Strømme wrote:
> >
> > A 1000x1000 real symmetric matrix always admits 1000 linearly
> > independent eigenvectors. This is the well known result that is
> > referred to here.
>
> [Hammond]
> I hope you're not intent on playing mathematical trivia
> with me.. I'm not interested in trivia. In general the
> "rank" of a matrix is not the same thing as the "order"
> of a matrix.
I suppose the explanation is that you are talking about a 1000x1000
real symmetric matrix of rank 13. In that case there are only 13
*non-zero eigenvalues* (counted with the proper multiplicity). If
these are all distinct, then there will be 13 one-dimensional
eigenspaces (and one 987-dimensional eigenspace). Then you need to make
a choice of eigenvector in each of the 13 one-dimensional subspaces in
order to arrive at a set of 13. Is this what you mean?
> An NxN matrix is by definition of order N,
> however the rank of the matrix may be much less than N.
> The rank of the matrix is the number of eigenvectors
> (independent eigenvalues) of the matrix.
No, as explained above. The rank is the number of non-zero
eigenvalues.
Stein
-- http://www.mi.uib.no/~stromme
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