 mersenneforum.org > Math Inverse of a particular matrix
 Register FAQ Search Today's Posts Mark Forums Read  2020-11-10, 22:21 #1 preda   "Mihai Preda" Apr 2015 2×691 Posts Inverse of a particular matrix I have a very special matrix, square n*n, where the lines 0..(n-1) are: line(i)={1/(i+1), 1/(i+2), ... , 1/(i+n)} for example, for n==2: Code: 1 1/2 1/2 1/3 for n==3: Code: 1 1/2 1/3 1/2 1/3 1/4 1/3 1/4 1/5 I would like to find the inverse (as a function of "n"). (I would also be interested in *how* to solve the question (not only in the answer), if it's not too complicated) Thank you! Last fiddled with by preda on 2020-11-10 at 22:22   2020-11-10, 22:44 #2 preda   "Mihai Preda" Apr 2015 2·691 Posts The above matrix originated from attempting to do a least-squares polynomial fit to a function evaluated over a (large) set of equidistant points. (the order of the matrix corresponds to the degree of the polynomial.)   2020-11-10, 23:29 #3 masser   Jul 2003 Behind BB 22×32×53 Posts Have you heard about Hilbert matrices?   2020-11-10, 23:33   #4
preda

"Mihai Preda"
Apr 2015

2·691 Posts Quote:
 Originally Posted by masser Have you heard about Hilbert matrices?
Thanks! I thought that must be some well-known matrix, but I didn't know its name :)   2020-11-10, 23:35 #5 masser   Jul 2003 Behind BB 35648 Posts As the wiki article states, those matrices are useful for testing linear algebra solvers; that's how I first learned about them.   2020-11-11, 00:06 #6 preda   "Mihai Preda" Apr 2015 2·691 Posts In pari/gp: Code: 1 / mathilbert(n) Last fiddled with by preda on 2020-11-11 at 00:07   2020-11-11, 03:27 #7 Dr Sardonicus   Feb 2017 Nowhere 22×1,459 Posts Hilbert matrices are notorious for being "numerically bad." It is however ridiculously easy to prove they're "positive definite," hence invertible. If n is a positive integer, f = f(x) is a polynomial of degree n-1, f = a0 + a1*x + .... + an-1xn-1, we can write [f] as the product of a 1xn and an nx1 matrix (Pari-GP notation), [f] = [a0, a1, ..., an-1]*[1;x;...;xn-1]. Taking the transpose, [f] = [1,x,...,xn-1]*[a0; a1; ...; an-1] Multiplying, [f2] = [a0, a1, ..., an-1]*[1;x;...;xn-1]*[1,x,...,xn-1]*[a0; a1; ...; an-1] Multiplying the middle two matrices gives the nxn matrix whose i, j entry is xi+j-2. Now, integrate from 0 to 1, and we find: The (1x1 matrix whose entry is) the integral from 0 to 1 of f2(x) dx may thus be expressed [a0, a1, ..., an-1]*Hn*[a0; a1; ...; an-1]. The integral is positive unless f is identically zero, so Hn is positive-definite. All you need, then, is a set of polynomials of degree 0, 1, 2, ..., n-1 which are orthogonal WRT integration from 0 to 1, (starting, obviously, with the constant polynomial 1). The "standard" set is the "shifted" Legendre polynomials. (The usual Legendre polynomials are orthogonal WRT integration from -1 to 1.) Last fiddled with by Dr Sardonicus on 2020-11-11 at 03:28 Reason: gixnif ostpy   2020-11-11, 06:15   #8
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

23×32×137 Posts
I want to muck with forum's tex software. (by posting straight from Wiki.)

Quote:
 The inverse of the Hilbert matrix can be expressed in closed form using binomial coefficients; its entries are
$$(H^{-1})_{ij} = (-1)^{i+j}(i + j - 1) \binom {n + i - 1}{n - j} \binom {n + j - 1}{n - i} \binom {i + j - 2}{i - 1}^2$$   2020-11-11, 06:32   #9
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

11001011100112 Posts Quote:
 Originally Posted by Batalov I want to muck with forum's tex software. (by posting straight from Wiki.) $$(H^{-1})_{ij} = (-1)^{i+j}(i + j - 1) \binom {n + i - 1}{n - j} \binom {n + j - 1}{n - i} \binom {i + j - 2}{i - 1}^2$$
That isn't tex, that is relying on some ugly JS interpreter.

This is tex:

At least I can see the tex, although binom isn't defined. Last fiddled with by retina on 2020-11-11 at 06:32   2020-11-11, 07:04   #10
Nick

Dec 2012
The Netherlands

41×43 Posts Quote:
 Originally Posted by retina That isn't tex, that is relying on some ugly JS interpreter. This is tex: At least I can see the tex, although binom isn't defined. Use {n \choose r}.

Last fiddled with by Nick on 2020-11-11 at 07:08 Reason: Typo   2020-11-11, 07:15   #11
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

197316 Posts Thanks

Quote:
 Originally Posted by Nick Use {n \choose r}.    Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post paul0 Math 6 2017-07-25 16:41 wreck NFS@Home 1 2016-05-08 15:44 only_human Miscellaneous Math 26 2012-08-10 02:47 Raman Math 5 2011-04-13 23:29 flouran Math 1 2010-01-18 23:48

All times are UTC. The time now is 14:03.

Thu Jul 7 14:03:39 UTC 2022 up 8:50, 0 users, load averages: 1.38, 1.40, 1.39