Timeline for Deriving inverse of Hilbert matrix
Current License: CC BY-SA 3.0
6 events
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| Dec 7, 2011 at 1:52 | history | edited | J. M. isn't a mathematician | CC BY-SA 3.0 |
added Hitotumatu result
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| Nov 28, 2010 at 22:56 | comment | added | J. M. isn't a mathematician | @Gottfried: Yes, the paper also gives explicit expressions for the $\mathbf L\mathbf D\mathbf L^T$ decomposition of the Hilbert matrix. I any event, If you have a Cholesky decomposition, you can easily obtain the $\mathbf L\mathbf D\mathbf L^T$ decomposition, and vice versa. ($\mathbf G=\mathbf L\sqrt{\mathbf D}$) | |
| Nov 28, 2010 at 22:18 | comment | added | Gottfried Helms | I don't have access to the Hitotumato-paper; but the cholesky-decomposition on its own gives triangular matrices with fractional, if not irrational, entries. A slightly different triangular decomposition seem to allow integer arithmetic. If you decompose $ Hilb = M * D * M^t$ having D diagonal and M lower triangular with unit diagonal, then the inverses of M and D are (empirically) integer matrices. I think that should be waterproof-derivable more easily than an argument via cholesky. | |
| Nov 28, 2010 at 16:44 | comment | added | J. M. isn't a mathematician | @Deane: I remember seeing a proof that didn't use Cholesky machinery, but sadly I can't find the reference now. The only reason I remember Hitotumatu's paper is that I was once stress-testing a Cholesky decomposition routine I wrote for a certain obscure language, and was looking for matrices with recognizable Cholesky triangles. | |
| Nov 28, 2010 at 16:31 | comment | added | Deane Yang | I don't consider this a circuitous proof at all. I first saw this question as a starred problem in a textbook (Hoffman-Kunze?) but could not solve it even after asking both professors and, after I entered graduate school, other graduate students. Finally, Takahiro Shiota showed me the calculation done in this paper. | |
| Nov 28, 2010 at 13:00 | history | answered | J. M. isn't a mathematician | CC BY-SA 2.5 |