# Matrix singularity issues with gaussian variogram [closed]

I have defined my gaussian variogram like this

$r(h) = \text{nugget} + \text{partial_sill}\cdot(1 - \exp(-\frac{3h^2}{\text{range}^2}))$

I set nugget = 0.1343 partial_sill = 0.3125 range = 19.8642

I tried to create a variogram for a spatial grid of size 5x5, meaning for each of the location in the 5x5 grid, I calculated the value of the variogram with distance equal to its distance to the other locations in the grid.

So I had a matrix of size 25x25. However, this matrix is not positive definite and is singular. How can I remedy this issue with gaussian variogram?

## closed as off-topic by whuber♦Aug 12 '14 at 19:16

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• What values of $h$ do your 25 distance values take? – AdamO Aug 7 '13 at 19:33
• It's just euclidean distance. I have the grid of size 5x5 a total of 25 locations. So the locations are like (1,1)(1,2)...(1,5)(2,1)... and so on the coordinates, I should say – user34790 Aug 7 '13 at 19:50
• I cannot reproduce this. An SVD of the matrix has singular values ranging from $3.68$ down to $0.1343$ and so obviously is positive definite. Are you sure you have implemented the nugget correctly? By definition it is zero when $h=0$ and otherwise equals $0.1343$ for $h\gt 0$. – whuber Aug 7 '13 at 20:35
• @whuber I am not sure but in that case the diagonal element is zero isn't it. – user34790 Aug 7 '13 at 23:30
• That is correct: instead of "nugget" in your formula you should write nugget$(h)$ to indicate that the value is $0$ when $h=0$ and otherwise equals $0.1343$. Because the Gaussian model is $0$ when $h=0$, that makes all diagonal elements $0$. – whuber Aug 8 '13 at 16:34