In the documentation of the mldivide function, two flow charts report the steps used by MATLAB to decide which method to apply in order to solve the linear system Ax=b with A\b. When A is sparse, if A is not diagonal, the algorithm asks:
"Does A look triangular (Upper or lower bandwidth of 0)?"
If YES, then
"Is A actually triangular (diagonal is structurally nonzero)?"
If NO then
"Is A permuted triangular?"
I am not able to imagine a non diagonal, permuted triangular matrix having 0 upper/lower bandwidth. Any suggestion?

답변 (1개)

Sally Al Khamees
Sally Al Khamees 2016년 12월 22일

1 개 추천

Let matrix A =
1 0 0
2 3 1
4 1 0
Then A looks triangular (upper bandwidth of 0).
A is not actually a triangular matrix because A(2,3) is not 0
A is a permuted triangular. If you switch row 2 with row 3 then A becomes
1 0 0
4 1 0
2 3 1
You can also refer to "LU matrix factorization" documentation page for example of functions that return permuted lower triangular matrix

댓글 수: 1

Thank you Sally for your answer.
However, it seems to me that the matrix A that you propose has an upper bandwidth which is 1 and not 0; MATLAB seems to agree:
[lowband, upband]=bandwidth([1 0 0; 2 3 1; 4 1 0])
returns lowband=2 and upband=1. I would then expect this matrix to fail the first test ("Does A look triangular (Upper or lower bandwidth of 0)?").

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