Why does Jordan function take so long?
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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.
Accepted Answer
John D'Errico
on 11 Dec 2016
0 votes
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.
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on 11 Dec 2016
on 11 Dec 2016
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