0
$\begingroup$

I have a linear system of the type

$A x = y$

where A is a dense, square, symmetric, positive definite matrix, $x$ a vector of unknown parameters, and $y$ is a vector of observed quantity. I know that using a Cholesky decomposition based algorithm should be the fastest way to find the values of $x$ that fulfil the equation. The only problem is I would like to do so for a matrix $A$ with dimension 500000x500000, and $x$ and $y$ 500000x1. Is there any scientific computing library to do so? Which kind of hardware should I use?

Improve this question
$\endgroup$
12
  • 7
    $\begingroup$ Well I can tell you it is doable - on 16384 cores of a modern cluster a collaborator and I have demonstrated finding all evals and evecs of a dense, real, symmetric matrix of about that size which is a somewhat tougher calculation than you need. We used ScaLAPACK which also has solvers in, being the distributed memory version of LAPACK. So it is possible, but whether ScaLAPACK is a good way to go for your problem I feel less sure - hence a comment rather than an answer (basically I do diagonlisation, I rarely solve equations) $\endgroup$
    – Ian Bush
    Commented Jan 19, 2022 at 14:05
  • 4
    $\begingroup$ You're going to need a system with a lot of memory to even store the matrix A. Multiply 500000 by 500000 by 8 bytes per entry and you'll have 2000 gigabytes of storage required for A. This isn't something you can do on a desktop machine, but could well be within the capacity of a high performance computing cluster. $\endgroup$
    – Brian Borchers
    Commented Jan 19, 2022 at 15:28
  • 2
    $\begingroup$ But even if possible, it seems questionable whether whatever you are doing is really best described by a dense matrix! $\endgroup$
    – Wolfgang Bangerth
    Commented Jan 19, 2022 at 21:20
  • 1
    $\begingroup$ @WolfgangBangerth why questionable? I can think of numerous problems that are really best described by a dense matrix. $\endgroup$
    – pinpon
    Commented Jan 20, 2022 at 9:10
  • 1
    $\begingroup$ @BrianBorchers You're missing a division by 2 because $A$ is symmetric (but that doesn't change the conclusion). $\endgroup$
    – Federico Poloni
    Commented Jan 20, 2022 at 12:34

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.