# Find $\det A$ and $\operatorname{Tr} A$ if $\det(A-\sqrt[n]{3}I_n)=0$

$$A \in M_{n}(\mathbb{Q})$$ 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.

• 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? Dec 22, 2018 at 20:52
• Yes,I made an edit,sorry for the typo Dec 22, 2018 at 20:56

$$\det (A+\lambda I_n)$$ 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).
• Why must the polynomial be a multiple of $\lambda ^n -3$? Dec 22, 2018 at 20:48
• 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 $n$. Dec 22, 2018 at 20:50
• @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 $0$? Dec 22, 2018 at 21:02