Re: [eigen] Another LDLt issue

[Benoit Jacob <jacob.benoit.1@xxxxxxxxx>]

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

0 1
1 0

does not even have a pivoting LDLt decomposition.

Cheers,
Benoit

2009/3/30 Benoit Jacob <jacob.benoit.1@xxxxxxxxx>:
> 2009/3/30 Benoit Jacob <jacob.benoit.1@xxxxxxxxx>:
>>>>Anyway,
>>>> routine dsytrf does an LDLt factorization for a symmetric but not
>>>> necessarily positive definite
>>>> matrix.
>>>
>>> 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...
>>
>> Benoit
>>
>
> 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.
>
> Cheers,
> Benoit
>