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?

  • $\begingroup$ Eigenvalues need not only be integers. $\endgroup$ – Theo Bendit Jan 11 at 15:19
  • $\begingroup$ Why not? The matrix is over $N$. I know that if it were over $C$ it would have eigenvalues. $\endgroup$ – JustAnAmateur Jan 11 at 15:20
  • $\begingroup$ 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 $\mathbb N$. $\endgroup$ – Dave Jan 11 at 15:21
  • $\begingroup$ I have edited my question, my approach clearly doesn't work. $\endgroup$ – JustAnAmateur Jan 11 at 15:43
  • $\begingroup$ I think, $M_n(\Bbb N)$ is a typo. $\endgroup$ – 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. $n=3k$.

  • $\begingroup$ Thank you for your solution ! I only have one question : you said this is a real matrix, but the problem states it is a positive integer matrix. Are we still allowed to use the fact that the trace is the sum of the eigenvalues? I know that this works only over algebraically closed fields. $\endgroup$ – JustAnAmateur Jan 12 at 12:01
  • $\begingroup$ The matrix might have entries in $\mathbb N$, but the eigenvalues are complex numbers. Yes, this is allowed. $\endgroup$ – 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:

Constructing regular integer matrices with distinct integer eigenvalues


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