Why is jordan() much slower than eig()? Both in symbolic forms
조회 수: 17 (최근 30일)
이전 댓글 표시
I have a sparse matrix (attached), and I want to obtain its Jordan canonical form. I used [V,D]=jordan(sym(A)), and found that this matrix is diagonalizable. So I also tried [V,D]=eig(sym(A)), and found eig() is much faster than jordan(). As they are both in symbolic form, I wonder why this happens.
A further question is that, I want to use the obtained decompostion matrices to conduct stablity analysis, which of the following choices is safer? Or, they are all not safe due to numerical problems?
- [V,D]=jordan(sym(A)); V=eval(V); D=eval(D);
- [V,D]=eig(sym(A)); V=eval(V); D=eval(D);
- [V,D]=jordan(A);
- [V,D]=eig(A);
Thanks!
댓글 수: 0
채택된 답변
Sourav Bairagya
2019년 8월 20일
Difference between speed of two functions came from the difference between the algorithms used in the implementation of the two functions. Speed also depends on the type of the problem that is being solved.
The difficulty with the Jordan Canonical Form is that it is extremely sensitive to any kind of perturbation or error in data, arithmetic roundoff error or linearization of nonlinear functions. If any of those occurs, then the JCF as well as the Transformation it generates will get changed. Hence, it is always necessary to convert numeric inputs to exact symbolic form before using “jordan” function to compute JCF of a matrix. By converting your data into its exact symbolic form, your data will remain protected from the above-mentioned errors.
For “eig” there is no such issues. But, again for safety, you can convert your data into exact symbolic form before using that function. Hence, it is safer to use 1 and 2 from the choices you have given.
추가 답변 (0개)
참고 항목
카테고리
Help Center 및 File Exchange에서 Linear Algebra에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!