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$A \in M_{n}(\mathbb{Q})$ and I have to find $\det A$ and $\operatorname{Tr} A$ if $\det(A+\sqrt[n]{3}I_n)=0$. I observed that $\sqrt[n]{3}$ is an eigenvalue of $A$,but I don't know how to continue.
EDIT : My bad,the matrix has rational entries.

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    $\begingroup$ I don't think this is possible to answer if we only know that $a \in M_n(\Bbb C)$. Do we have some other information? Does $A$ have integer or rational entries? $\endgroup$ Dec 22, 2018 at 20:52
  • $\begingroup$ Yes,I made an edit,sorry for the typo $\endgroup$
    – Math Guy
    Dec 22, 2018 at 20:56

1 Answer 1

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$\det (A+\lambda I_n)$ is a monic polynomial of degree $n$ in $\lambda$. Note that if $n$-th root of $3$ is a root of a degree $n$-polynomial, the polynomial must be a multiple of $\lambda^n-3$, hence $\det$ and Tr must be $(-1)^n\cdot -3$ and $0$ respectively (by using the fact that Tr is the coefficient of $\lambda^{n-1}$ in the characteristic polynomial and $\det$ is the constant term times $(-1)^n$ in the characteristic polynomial).

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  • $\begingroup$ Why must the polynomial be a multiple of $\lambda ^n -3$? $\endgroup$
    – Math Guy
    Dec 22, 2018 at 20:48
  • $\begingroup$ That is because the minimal polynomial of $n$-th root of $3$ is $x^n-3$, i.e. there is no polynomial that admits $n$-th root of $3$ as a root and has degree less than $n$. $\endgroup$
    – Levent
    Dec 22, 2018 at 20:50
  • $\begingroup$ @Arthur you are most definitely right, I guess I just automatically made some assumptions. However in this case there are matrices that satisfy the condition and have different trace and determinant which makes me believe that OP may have meant rational entries. $\endgroup$
    – Levent
    Dec 22, 2018 at 20:54
  • $\begingroup$ @Levent one more question : if the characteristic polynomial is a multiple of $\lambda^n -3$,why do we still get that $\det$ and $Tr$ must be $(-1)^n\cdot -3$ and $0$?Shouldn't they be a multiple of $(-1)^n\cdot -3$ and $0$? $\endgroup$
    – Math Guy
    Dec 22, 2018 at 21:02
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    $\begingroup$ You are welcome. If you have troubles with any of the facts I pull out of the hat, I can edit the answer to clarify. $\endgroup$
    – Levent
    Dec 22, 2018 at 21:09

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