I am looking for change in eigenvalues as $ p(T)$ is translated as $ p(T+I)$. I thought the following link could be helpful, though I can't figure out any. $T$ being some matrix such that $p(T)=0$

How to prove "eigenvalues of polynomial of matrix $A$ = polynomial of eigenvalues of matrix $A$ "

Can the variation in eigenvalues be seen in terms of the polynomial operator 'p' itself? If so, can this be generalized in any sense? Let me know if the question needs clarification. Possible hints or solution will be a great help. Regards.

  • $\begingroup$ I'm not sure if I got your post right, is $p(t)$ the characteristic polynomial of some linear map or matrix? $\endgroup$ – Diglett Oct 8 '18 at 5:37
  • $\begingroup$ The edits might be clear, hopefully. $\endgroup$ – user511110 Oct 8 '18 at 5:56
  • $\begingroup$ I do not know if this useful but if $a$ is an eigenvalue of $A$ then $f(a) $ is an eigenvalue of $ f(A)$ $\endgroup$ – StammeringMathematician Oct 8 '18 at 6:01
  • $\begingroup$ If T has eigenvalue $a$ then T+I has eigenvalue 1+a. So p(T+I) will have eigenvalue p(a+1) $\endgroup$ – StammeringMathematician Oct 8 '18 at 6:06
  • $\begingroup$ We are not sure whether $p(a+1)=0$ or not. How can you make sure? $\endgroup$ – user511110 Oct 8 '18 at 6:09

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