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Faster inverse of a lower triangular matrix than inv(A)?

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Benson Gou
Benson Gou 2018 年 10 月 19 日
編集済み: Matt J 2018 年 10 月 19 日
Dear All,
I have a sparse lower triangular matrix A. I want to obtain the inverse of A. A has a size of 6000 X 6000. I find inv(A) takes more than 5 seconds. I do not know if there is a faster approach to get the inverse of A?
Thanks. Bei

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Matt J
Matt J 2018 年 10 月 19 日
編集済み: Matt J 2018 年 10 月 19 日
The reason I want to obtain the inverse of A is that I need to select a number of rows from inv(A) for further uses. I do not just solve Ax=b linear equation.
The two are almost the same thing:
B=speye(size(A));
rows = B(rowIndices,:)/A
  1 件のコメント
Benson Gou
Benson Gou 2018 年 10 月 19 日
Thanks a lot, Matt. It works very well. 0.0549 second.
Bei

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その他の回答 (1 件)

Torsten
Torsten 2018 年 10 月 19 日
See here
https://de.mathworks.com/matlabcentral/answers/332233-updating-inverse-of-a-lower-triangular-matrix
why it is a bad idea to invert a lower triangular matrix.
  1 件のコメント
Benson Gou
Benson Gou 2018 年 10 月 19 日
編集済み: Benson Gou 2018 年 10 月 19 日
Thanks, Torsten.
The reason I want to obtain the inverse of A is that I need to select a number of rows from inv(A) for further uses. I do not just solve Ax=b linear equation.
I also used the idea of x=A\b to obtain the rows (800 rows) I want to select from inv(A). But it took more than 9 seconds.
Thank you, anyway.

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