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The inverse of a banded matrix

by S_David
Tags: banded, inverse, matrix
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S_David
#1
Jun12-14, 05:52 PM
P: 597
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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AlephZero
#2
Jun12-14, 07:12 PM
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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.
S_David
#3
Jun21-14, 04:01 PM
P: 597
Quote Quote by AlephZero View Post
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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