Re: [eigen] [patch] LDLt decomposition with rank-deficient matrices |

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*To*: eigen@xxxxxxxxxxxxxxxxxxx*Subject*: Re: [eigen] [patch] LDLt decomposition with rank-deficient matrices*From*: Benoit Jacob <jacob.benoit.1@xxxxxxxxx>*Date*: Wed, 24 Feb 2010 08:26:17 -0500*Dkim-signature*: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=gamma; h=domainkey-signature:mime-version:received:in-reply-to:references :date:message-id:subject:from:to:content-type :content-transfer-encoding; bh=NZvej5Oir8RAzrl+E/rEZ7AAJXqYNfYSdYQ5H1JX6bA=; b=uozmzBhKiHzE8DuyNMFBzLvSUqCisYATUErVUNj6cBuyL/fimPhSWbpOBH4Iv0Vjpo 6cinXlsrt/INbf/Wh3ScP9YxKHv6ZPaApd7EGkvR9iJ9ks/Bhwfiy7Nn/Jy4U7bK3U+c qxHa28uqbyXsBOLf5s02WDs6sM2nb+Co3AlNw=*Domainkey-signature*: a=rsa-sha1; c=nofws; d=gmail.com; s=gamma; h=mime-version:in-reply-to:references:date:message-id:subject:from:to :content-type:content-transfer-encoding; b=V1T6CkdZcm7/0zXw3xD/DaA30CD52VRCDY91kM6aOaLK4QGhApDmsXkwhFsAMnEN// 9vEiGOLfxqtA9kRDfO95ycYVqmxRGYubGqqtRd7ScNnqltxDLaQLG6ys9Z4RBOih6rJ0 OGqyYLwYoZOVWXnDSI6wgCkzWgrm8QYDpR6dU=

2010/2/24 FMDSPAM <fmdspam@xxxxxxxxx>: > Hi Experts, > > please, for my personal clarification, let me ask: > (I'm not asking for a brief lesson in lin alg. but I have to happily live > with those answers :-) ) > > Def.: > D_labil := "how many zeros in vectorD()" > D_instable : "how many entries in vectorD() are smaller or equal 0..0" > > a) LDLt for non singular matrix is rank-revealing (of course, i hope), This is a tautology: if you assume your matrices to be nonsingular, then you know their rank. Nonsingular means rank N. > n_labil = 0 Yes, for (sufficiently) nonsingular matrices, there won't be too small entries in the vectorD. > b) LDLt for positive definite matrix => D_instable = 0 Yes. > > c) LDLt for singular matrix is not rank-revealing, but: n_deficit > 0 What is n_deficit ? Number of zeros in vectorD? I think this kind of works, yes, as you're only interested in knowing if it's 0 or >0. As the LDLt decomposition still gives you the determinant, it in particular tells you very vaguely if the matrix is singular or not, but don't use it for that, it shouldn't be very precise. > d) LDLt for indefinite matrix => D_instable > 0 but counting D(i) <= 0.0 is > meaning less? Yes. The "zeros" in the vectorD will actually be small positive values. There's no magic cutoff where you'd decide that a small value can be considered zero. > > e) Do all this also hold for LLt decompositions? Oops, for LLt decomposition it's even worse. First you're taking sqrt of small values, so you're losing half of the precision. Second, our LLt is non-pivoting so it's totally unstable on singular matrices. Benoit > > Regards > Frank > > > > >

**Follow-Ups**:

**References**:**[eigen] [patch] LDLt decomposition with rank-deficient matrices***From:*Ben Goodrich

**Re: [eigen] [patch] LDLt decomposition with rank-deficient matrices***From:*Benoit Jacob

**Re: [eigen] [patch] LDLt decomposition with rank-deficient matrices***From:*Ben Goodrich

**Re: [eigen] [patch] LDLt decomposition with rank-deficient matrices***From:*Benoit Jacob

**Re: [eigen] [patch] LDLt decomposition with rank-deficient matrices***From:*Ben Goodrich

**Re: [eigen] [patch] LDLt decomposition with rank-deficient matrices***From:*Gael Guennebaud

**Re: [eigen] [patch] LDLt decomposition with rank-deficient matrices***From:*Jitse Niesen

**Re: [eigen] [patch] LDLt decomposition with rank-deficient matrices***From:*Benoit Jacob

**Re: [eigen] [patch] LDLt decomposition with rank-deficient matrices***From:*FMDSPAM

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