Re: [eigen] Another LDLt issue

[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