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tomcees_math@yahoo.com science forum beginner
Joined: 07 Jun 2006
Posts: 2

Posted: Mon Jun 12, 2006 6:04 pm Post subject:
Elements in Inverse Matrix of A, where a(i.j) = i^j



Hello:
I need to invert the general nxn matrix:
0^0 0^1 0^2 ... 0^(n1)
1^0 1^1 1^2 ... 1^(n1)
2^0 ... ... ... 2^(n1)
....
....
(n1)^0 .... .... (n1)^(n1)
That is, each element in the matrix is i^j where i is the row and j is
the column.
An example when n = 5:
1 0 0 0 0
1 1 1 1 1
1 2 4 8 16
1 3 9 27 81
1 4 16 64 256
[Note that the nature of this problem does require that we accept 0^0
as 1.]
The structure of this matrix does appear to 'ask for' row reduction to
find the inverse. I can do specific cases, but am looking for a
general, symbolic formula for the values of the entries in the inverse
given n.
Can anyone provide the values of the elements in the inverse in terms
of i and j, given n?
Thanks in advance for your help,
TomCee 

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Ronald Bruck science forum Guru
Joined: 05 Jun 2005
Posts: 356

Posted: Mon Jun 12, 2006 11:53 pm Post subject:
Re: Elements in Inverse Matrix of A, where a(i.j) = i^j



In article <1150135463.174896.136770@u72g2000cwu.googlegroups.com>,
<tomcees_math@yahoo.com> wrote:
Quote:  Hello:
I need to invert the general nxn matrix:
0^0 0^1 0^2 ... 0^(n1)
1^0 1^1 1^2 ... 1^(n1)
2^0 ... ... ... 2^(n1)
...
...
(n1)^0 .... .... (n1)^(n1)
That is, each element in the matrix is i^j where i is the row and j is
the column.
An example when n = 5:
1 0 0 0 0
1 1 1 1 1
1 2 4 8 16
1 3 9 27 81
1 4 16 64 256
[Note that the nature of this problem does require that we accept 0^0
as 1.]
The structure of this matrix does appear to 'ask for' row reduction to
find the inverse. I can do specific cases, but am looking for a
general, symbolic formula for the values of the entries in the inverse
given n.
Can anyone provide the values of the elements in the inverse in terms
of i and j, given n?

Look up "Vandermonde matrix". The determinant has a simple form, but I
don't know of a simple form for the inverse. Instead of considering
the special matrix you have, try
1 x1 x1^2 ... x1^(n1)
1 x2 x2^2 ... x2^(n1)
...
1 xn xn^2 ... xn^(n1)
Sometimes generalizing leads to easier solutions.
(Mathworld says the VM can be inverted in O(n^2) operations. So
there's something there. Playing with it in Mathematica, and factoring
the results, there are some patterns.)
Ron Bruck
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