|Re: [eigen] Another LDLt issue|
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- Subject: Re: [eigen] Another LDLt issue
- From: Benoit Jacob <jacob.benoit.1@xxxxxxxxx>
- Date: Mon, 30 Mar 2009 20:43:28 +0200
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OK, last email and I don't spam the list anymore....
my previous counterexample was'nt the best because you could still
factor it with pivoting. But the same reasoning shows that the matrix
does not even have a pivoting LDLt decomposition.
2009/3/30 Benoit Jacob <jacob.benoit.1@xxxxxxxxx>:
> 2009/3/30 Benoit Jacob <jacob.benoit.1@xxxxxxxxx>:
>>>> routine dsytrf does an LDLt factorization for a symmetric but not
>>>> necessarily positive definite
>>> Our LDLt does the same,
>> Oops, I misread what you wrote. So, dsytrf doesn't require the matrix
>> to be positive? interesting. I don't have an example of a symmetric
>> matrix that doesn't have an LDLt decomposition, I was just saying that
>> the standard algorithm couldn't handle all of them, but maybe dsytrf
>> uses another algorithm...
> OK, now I have a counter example: the matrix M =
> 0 1
> 1 1
> is symmetric and nonsingular, and does NOT have a LDLt decomposition.
> Proof: if L is
> 1 0
> x 1
> And D is
> a 0
> 0 b
> Then LDLt is
> a ax
> ax (a^2)x+b
> So if LDLt=M then we have a=0 and ax=1, which is impossible.
> So for sure, the DSYTRF function can't find a LDLt for all symmetric
> matrices. I think that what they mean is that they work for all
> positive matrices, not necessarily positive definite. IOW there's a
> little mistake in their docs.