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Moderator Note: This question is from a contest which ended 1 Dec 2012.

Consider a polynomial $f$ with complex coefficients. Call such $f$ broken if we can find a square matrix $M$ such that $M \neq f(N)$ for any square matrix $N$.

My professor told me it was possible to explicitly characterize all broken polynomials of any degree. What would such a characterization look like?

Thank you!

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    Do you know how the eigenvalues of $f(N)$ relate to those of $N$?2012-11-07
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    no, nothing more about $f$ is given than what was stated. How would the eigenvalues be of use though?2012-11-07
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    Ah I am having trouble finding that initial footing. Perhaps if you showed me how to start with that case, I may be able to proceed?2012-11-07
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    You have missed the intent of my question. Do you know any facts that, given any old matrix $A$, and any old polynomial $p$, relate the eigenvalues of $p(A)$ to those of $A$?2012-11-07
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    Sorry, I am very new to this. My professor simply said this was a challenge problem, but I found it interesting. So if we apply $p$ to a matrix $A$, would that not change the eigenvalue? I apologize for not knowing too much about the subject2012-11-07
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    @Gerry: I fail to see where you comment leads. Since any non-constant polynomial is onto, how can you use eigenvalues to address the surjectivity of the map $N\mapsto f(N)$?2012-11-07
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    @Martin, I guess I was assuming the matrices were supposed to have real entries, but I see there is no justification for that assumption in the statement of the problem.2012-11-08
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    Are all polynomials of degree greater than $1$ broken?2012-11-22

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