Eigenvalues of translates of polynomial

Asked Oct 08 '18 at 05:29

Active Oct 08 '18 at 05:54

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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.

edited Oct 08 '18 at 05:54

asked Oct 08 '18 at 05:29

user511110

I'm not sure if I got your post right, is $p(t)$ the characteristic polynomial of some linear map or matrix? – Diglett Oct 08 '18 at 05:37
The edits might be clear, hopefully. – user511110 Oct 08 '18 at 05:56
I do not know if this useful but if $a$ is an eigenvalue of $A$ then $f(a) $ is an eigenvalue of $ f(A)$ – Shweta Aggrawal Oct 08 '18 at 06:01
If T has eigenvalue $a$ then T+I has eigenvalue 1+a. So p(T+I) will have eigenvalue p(a+1) – Shweta Aggrawal Oct 08 '18 at 06:06
We are not sure whether $p(a+1)=0$ or not. How can you make sure? – user511110 Oct 08 '18 at 06:09
I have not said anything on whether $p(a+1)=0$ or not. And you are right in saying that we can not be sure. – Shweta Aggrawal Oct 08 '18 at 06:11

Eigenvalues of translates of polynomial

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