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This is in continuation with this one but it is more general. I will try to make it self contained.

I have a program and I need to take the Dot product of many matrices. But while taking the MatrixExp, I am facing the numeric problem(which I am familiar with, unfortunately don't know how to go out of it). However, I come up with one technique(not good at all).

MatrixExp::eivn: Incorrect number 0 of eigenvectors for eigenvalue -1. Sqrt[1.405 E^(2 I k)+0.9 E^(I k) Cos[t]+0.9 E^(3 I k) Cos[t]+0.405 E^(2 I k) Cos[2. t]] with multiplicity 1.

This is my program,

a = 1;
tI = 0.0001;
dt = 0.0001;
NStep=30000;
T1[t_] = j1 (Cos[t]); 

T2[t_] = j2 ;

cond = {j1 -> 0.9, j2 -> 1.};
HSm[t_,k_] = ({{0, -(T1[t] + T2[t] Cos[k])}, {-(T1[t] + T2[t] Cos[k]),0}})//. cond;
HStDig[t_, k_Real] = MatrixExp[HSm[t,k]]


us = ParallelTable[(tI + j dt) * HStDig[tI + j dt,k], {j, NStep, 0, -1}];
us1 = Apply[Dot, us];

How to go about this problem? Is there a way to speed up the program?

Note:
Diagonalization can't be taken for the matrix!
k is supposed to be given value from a list of numbers

Improve this question
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  • $\begingroup$ The matrix exponential of your matrix can be computed by hand and gives you a matrix with Cosh and Sinh in its components. This might speed up the computation. $\endgroup$
    – user46676
    Commented Apr 6, 2017 at 19:39
  • $\begingroup$ @marmot I didn't get. $\endgroup$
    – L.K.
    Commented Apr 6, 2017 at 19:42
  • $\begingroup$ Try FullSimplify[MatrixExp[mat]] where mat={{0,a},{a,0}}. Your matrix is of this type. Another problem of your code is that you add the scalar tI to a matrix. $\endgroup$
    – user46676
    Commented Apr 6, 2017 at 19:48
  • $\begingroup$ @marmot I edited my question, thanks for the valuable points. Let me go through them $\endgroup$
    – L.K.
    Commented Apr 6, 2017 at 20:11
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    $\begingroup$ If you use HStDig[t_, k_] = FullSimplify[MatrixExp[HSm[t, k]]] your code works. $\endgroup$
    – user46676
    Commented Apr 6, 2017 at 20:50

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