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I am creating with a software a banded matrix, which is also symmetric. In fact, its definition comes from an array, Array[q], whose length is n, where I store my values. Then to define my matrix I do the following:

Matrix[i][j] = Array[abs[i-j]],

where abs indicates the absolute value. The result is, in the case of n=8 for example,

{0.590818, 0.574632, 0.528687, 0.460129, 0.378821, 0.295026, 
   0.21735, 0.151471},
  {0.574632, 0.590818, 0.574632, 0.528687, 0.460129, 0.378821, 
   0.295026, 0.21735},
  {0.528687, 0.574632, 0.590818, 0.574632, 0.528687, 0.460129, 
   0.378821, 0.295026},
  {0.460129, 0.528687, 0.574632, 0.590818, 0.574632, 0.528687, 
   0.460129, 0.378821},
  {0.378821, 0.460129, 0.528687, 0.574632, 0.590818, 0.574632, 
   0.528687, 0.460129},
  {0.295026, 0.378821, 0.460129, 0.528687, 0.574632, 0.590818, 
   0.574632, 0.528687},
  {0.21735, 0.295026, 0.378821, 0.460129, 0.528687, 0.574632, 
   0.590818, 0.574632},
  {0.151471, 0.21735, 0.295026, 0.378821, 0.460129, 0.528687, 
   0.574632, 0.590818}
 }

Now the point is that I need to calculate the inverse of this matrix. As long as n=8, I manage to make the software do that, but as soon as I grow with the dimension, say n=64 for example, I get the error that the matrix is badly conditioned, because the determinant is (almost) 0. In fact, if I calculate the determinant in the case n=64, I get that it is equal to

1.483771072856803*10^-498

My question is: why should such a banded matrix be singular? I cannot see an evident reason.

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    $\begingroup$ It seems to me that Matrix[i][j]=Array[abs[i-j]] defines a symmetric Toeplitz matrix instead of a general symmetric banded matrix. $\endgroup$ – Algebraic Pavel Apr 16 '15 at 10:01
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    $\begingroup$ A small comment: I don't know what software do you use but if it determines bad conditioning by looking at the value of the determinant, I would suggest replacing it by another software since determinant has no connection to the condition number whatsoever. For example, a diagonal $n\times n$ matrix $D:=\mathrm{diag}(0.1,\ldots,0.1)$ has determinant $10^{-n}$ (arbitrarily small for large $n$) but its spectral condition number is 1. $\endgroup$ – Algebraic Pavel Apr 16 '15 at 10:03

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