The inverse of a banded matrix

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Inverting a banded matrix directly can lead to increased complexity, as the resulting inverse will be fully populated rather than maintaining the banded structure. Instead of inverting the matrix, it is more efficient to solve a set of equations or use matrix decomposition methods like LDU, where L and U are lower and upper triangular matrices that retain the bandwidth. This approach avoids unnecessary computations involving zero terms in the decomposed matrices. The discussion emphasizes that matrix inversion is generally not advisable in computational contexts. Understanding these methods can significantly enhance efficiency in handling large matrices.
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Hello all,

I have say 512-by-512 matrix, but based on the structure of this matrix most elements not on the diagonals between -5 to +5 (- stand for diagonal below the main diagonal, and + for diagonal above the main diagonal) are small relative to the elements of the mentioned diagonals. So, I create a 512-by-512 banded matrix, where I null all other elements not on the mentioned diagonals.

Now the question is: will there be a huge complexity saving if I want the inverse of the matrix by inverting its banded version instead of the original matrix?

Thanks
 
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In computer calculations, "inverting a matrix" is almost always the wrong thing to do, even if you have a nice looking math equation with an inverse matrix in it.

In this case there will be a huge "complexity" increase, because the inverse matrix will be fully populated, not banded.

What you really want to do is probably solve a set of equations or something similar. If you decompose your banded matrix as A = LDU or something similar, where L and U are lower and upper triangular and have the same bandwidth as A, you preserve the efficiency by not needing to process all the zero terms in L and U.
 
AlephZero said:
In computer calculations, "inverting a matrix" is almost always the wrong thing to do, even if you have a nice looking math equation with an inverse matrix in it.

In this case there will be a huge "complexity" increase, because the inverse matrix will be fully populated, not banded.

What you really want to do is probably solve a set of equations or something similar. If you decompose your banded matrix as A = LDU or something similar, where L and U are lower and upper triangular and have the same bandwidth as A, you preserve the efficiency by not needing to process all the zero terms in L and U.

Could you please tell me more about this process?
 
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