# A positive integer matrix with no integer eigenvalues

Let $$n \in \mathbb{N}$$ and $$A\in M_n(\mathbb{N})$$ with $$Tr(A)=0$$ and $$A^3+A-2I_n=O_n$$.Prove that $$n$$ is a multiple of $$3$$ and $$\det(A^2)=\det(A^2+I_n)$$.
I tried to find $$A$$'s eigenvalues, but the equation $$x^3+x-2=0$$ has only one integer root, $$1$$,and this contradicts $$Tr(A) =0$$.
EDIT:My approach doesn' t work as pointed out in the comments, how should this be solved?

• Eigenvalues need not only be integers. – Theo Bendit Jan 11 at 15:19
• Why not? The matrix is over $N$. I know that if it were over $C$ it would have eigenvalues. – JustAnAmateur Jan 11 at 15:20
• The eigenvalues of a matrix with entries in, for instance $\mathbb N$, are solutions to a polynomial equation whose coefficients are in $\mathbb N$. These solutions need not (and in fact are often not) in $\mathbb N$. – Dave Jan 11 at 15:21
• I have edited my question, my approach clearly doesn't work. – JustAnAmateur Jan 11 at 15:43
• I think, $M_n(\Bbb N)$ is a typo. – Dietrich Burde Jan 11 at 16:23

Since $$A^3 + A - 2 I = 0$$, all eigenvalues of $$A$$ are roots of the polynomial $$x^2 + x - 2$$, thus $$1$$ or $$-1/2 \pm \sqrt{7} i/2$$. Since it's a real matrix, the two non-real eigenvalues have equal algebraic multiplicities. Since the trace (which is the sum of the eigenvalues) must be $$0$$, all three eigenvalues have equal multiplicities. Thus if this multiplicity is $$k$$, there are $$3k$$ eigenvalues counted by algebraic multiplicity, i.e. $$n=3k$$.
• The matrix might have entries in $\mathbb N$, but the eigenvalues are complex numbers. Yes, this is allowed. – Robert Israel Jan 13 at 6:46
Take the matrix in $$GL_2(\Bbb Z)$$ $$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix},$$ it has not integers as eigenvalues, but the golden ratio $$\frac{1\pm \sqrt{5}}{2}$$. If you want to construct matrices with integer eigenvalues, then see for example here: