I noticed that the function "jordan" applied on a square matrix A takes really long to output the canonical Jordan form of A, J, and the change of basis matrix Q, even when A is a small matrix. I am wondering why is it so? Why is "jordan" so slower than function "eig", how is it implemented? Thank you in advance.

 채택된 답변

John D'Errico
John D'Errico 2016년 12월 11일

0 개 추천

Jordan works on the matrix in symbolic form. You cannot possibly expect a symbolic solution to operate as fast as an operation computed using double precision arithmetic.

추가 답변 (0개)

질문:

2016년 12월 11일

답변:

2016년 12월 11일

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by