[Numpy-discussion] Rank deficient matrices
Paul F. Dubois
paul at pfdubois.com
Tue Oct 30 07:21:07 EST 2001
Use LinearAlgebra.singular_value_decomposition
-----Original Message-----
From: numpy-discussion-admin at lists.sourceforge.net
[mailto:numpy-discussion-admin at lists.sourceforge.net] On Behalf Of Nils
Wagner
Sent: Tuesday, October 30, 2001 12:24 AM
To: numpy-discussion at lists.sourceforge.net
Subject: [Numpy-discussion] Rank deficient matrices
Hi,
Let us assume that
r = rank(C) < N (1)
where C is a symmetric NxN matrix. This implies that the number of
non-zero eigenvalues of C is r. Because C is a symmetric matrix there
exists an orthogonal matrix U whose columns are the eigenvectors of C
such that
U^\top C U = [ d , O
O , O]. (2)
In the above equation d \in rxr is a diagonal matrix consisting of only
the non-zero eigenvalues of C. For convenience, partition U as
U = [U_1 | U_2] (3)
where the columns of U_1 (Nxr) are the eigenvectors corresponding to the
non-zero block d and the columns of U_2 are the eigenvectors
corresponding to the rest
(N-r) number
of zero eigenvalues.
Defining a rectangular transformation matrix
R = U_1 (4)
it is easy to show that
R^\top C R = d. (5)
Therefore, the matrix R in equation (4) transforms the originally rank
deficient matrix C to a full-rank (diagonal) matrix of rank r.
I am looking for an efficient Numpy implementation of this
transformation.
Thanks in advance
Nils Wagner
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