I give a minimalistic example of my problem:

I have a matrix:


I define the eigenvectors as:


and get the result

e[a_,b_]= {{-(Sqrt[b (-a + b)]/b), 1}, {Sqrt[b (-a + b)]/b, 1}}

If I calculate e[a,0] I get

e[a,0]={{1, 0}, {0, 0}}

but if I store the symbolic equation for the eigenvectors, speaking

f[a_,b_]={{-(Sqrt[b (-a + b)]/b), 1}, {Sqrt[b (-a + b)]/b, 1}}

and try to calculate f[a,0] I get the "intermediate" result cause the algorithm tries to divide by 0.

How can I get the symbolic output of eigenvectors to give the correct result?

  • $\begingroup$ Your matrix becomes similar to a Jordan block for $b=0$. Eigenvectors[] does not automatically assume a symbolic matrix is defective. $\endgroup$ – J. M.'s technical difficulties May 27 '16 at 18:42
  • $\begingroup$ How can i add this exeption to get an global solution? $\endgroup$ – MathNut May 27 '16 at 18:49
  • $\begingroup$ A strategy similar to the one done here would work. $\endgroup$ – J. M.'s technical difficulties May 27 '16 at 18:52
  • $\begingroup$ I read the comment but i dont understand how this solves my problem. $\endgroup$ – MathNut May 27 '16 at 19:45
  • $\begingroup$ I have an symbolic expression for the eigenvectors and want to get the right result replacing the variables a and b by the values for every case of a and b $\endgroup$ – MathNut May 27 '16 at 19:46

e[a,0]={{1,0},{0,0}}.But actually {1,0} is not a correct eigenvector to {{0,-a},{0,0}}.(you can try yourself to see). so the symbolic output of e[a_,b_] is not correct when you set b=0. It can not be used to your purpose either.

  • $\begingroup$ Huh? That eigenvector corresponds to an eigenvalue of $0$. What you might have meant is that an arbitrary vector could have been assigned as the eigenvector corresponding to the zero eigenvalue. $\endgroup$ – J. M.'s technical difficulties May 30 '16 at 8:16
  • $\begingroup$ I see. you are right. maybe the eigenvector function doesn't use the symbolic expression to deal with numerical input. if your purpose is to get a symbolic expression for convenience in later use, may be you can first get a symbolic expression, then find its singular points. use the expression while its not singular and use eigenvector function only for those singular points $\endgroup$ – dr.bian May 30 '16 at 11:54
  • $\begingroup$ ok dr.bian, that's what i already feared. There is probably no way to get the eigenvectors of singular points using the symbolic expression of my eigenvectors. In that case it seems to be more easy to me to "not" use the symbolic expression. $\endgroup$ – MathNut May 30 '16 at 17:38

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