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How can I solve in $\mathcal{M}_{2}(\mathbb{Z})$ the equation $$A^3-3A=\begin{pmatrix}-7 & -9\\ 3 & 2\end{pmatrix}?$$

I try to use $$A^2-Tr(A)A+detA\cdot I_2=O_2$$ but I don't still obtain anything.

thanks.

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    $\begingroup$ Diagonalize the matrix. And note that $(PXP^{-1})^n=PX^nP^{-1}$. $\endgroup$ – xavierm02 Oct 23 '13 at 16:02
  • $\begingroup$ and what can you do, if you can't use this trick? $\endgroup$ – Iuli Oct 23 '13 at 16:03
  • $\begingroup$ What can't you use? $\endgroup$ – xavierm02 Oct 23 '13 at 16:04
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    $\begingroup$ It implies that you can use the theory you built on $\Bbb C$ even though $\Bbb Z$ isn't a field. $\endgroup$ – xavierm02 Oct 23 '13 at 16:14
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    $\begingroup$ You know the Cayley–Hamilton theorem but not how to diagonalize a matrix? $\endgroup$ – xavierm02 Oct 23 '13 at 16:53
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$$ A=\pmatrix{{2}&{3}\\{-1}&{-1}}.$$

Note that $a^3+b^3=(a+b)^3-3ab(a+b)$.

Let $a,b$ be eigenvalues of $A$. Then $trA=a+b$, $detA=ab$, and they are integers.

Let $X=trA$.

We see from $A(A^2-3I)$ has determinant $13$, we have $detA | 13$.

From the identity in the beginning, we have $X^3-3(ab+1)X+5=0$ , which we can obtain from taking traces of original equation.

Just plug in the four possible values for $ab$, namely $\pm 1, \pm 13$, we find that only possible value is $ab=1$, and consequently

$X^3-6X+5=0$.

Solving for integers, we obtain $X=1$.

Thus, $A$ should satisfy $trA=1$, $detA=1$.

By Cayley-Hamilton, we have $A^2-A+I=0$, this forces $A^3=-I$.

Now, plug in to original equation to get $$ -3A=\pmatrix{{-6}&{-9}\\{3}&{3}}.$$ Then we arrive at the answer.

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Let $t=\operatorname{trace}(A)$ and $d=\det(A)$. By Cayley-Hamilton theorem, $A^2 = tA - dI$ and hence $$ \pmatrix{-7&-9\\ 3&2} = A^3 - 3A = tA^2 - (d+3)A = (t^2-d-3)A - tdI.\tag{1} $$ Taking traces on both sides, we get $-5 = (t^2-d-3)t - 2td$. Hence $t$ divides $5$, i.e. $t=1,-1,5,-5$, and we can express $d$ in terms of $t$ as $d = \frac13\left(t^2 + \frac5t\right) - 1$. Since $d$ must be an integer, $(t,d)=(1,1)$ or $(-5,7)$. However, by inspecting the $(2,1)$-th entries on both sides of $(1)$, we see that $(t^2-d-3)\mid3$. Therefore $(t,d)=(1,1)$. So, from $(1)$ we obtain $$ \pmatrix{-7&-9\\ 3&2} = -3A - I\quad\Longrightarrow\quad A=\pmatrix{2&3\\ -1&-1}. $$

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  • $\begingroup$ (+1) Easier than mine. Thought about this way after posting my solution. $\endgroup$ – Sungjin Kim Oct 24 '13 at 2:12
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Here's an approach that works even if the left hand side is a more complicated expression of the form $AP(A)$, with an arbitrary polynomial $P$ (with integer coefficients).

Let $M$ denote the matrix on the right hand side. The equality implies that $A$ and $M$ commute. If we write this as

$$\pmatrix{a & b\\ c & d}\pmatrix{-7 & -9\\ 3 & 2}=\pmatrix{-7 & -9\\ 3 & 2}\pmatrix{a & b\\ c & d}$$

we find that

$$b=-3c\text{ and }d=a+3c$$

so

$$A=\pmatrix{a & -3c\\c & a+3c}$$

with

$$\det(A)=a^2+3ac+3c^2={(2a-3c)^2+3c^2\over4}$$

Since we're working with integers, $AP(A)=M$ implies $\det(A)$ divides $\det(M)=13$, which limits the possibilites to integer solutions to

$$(2a-3c)^2+3c^2=4\text{ or } 52$$

of which there are clearly a (fairly small) finite number.

More generally, if

$$M=\pmatrix{p & q\\ r & s}$$

this approach shows that $rb=qc$, $rd=ra+(s-p)c$, and

$$r(ps-qr)=k(ra^2+(s-p)ac-qc^2)$$

where the inequality that makes the pertinent quadratic form on the right hand side positive definite (thereby limiting the number of possible solutions) is

$$(p+s)^2-4(ps-qr)\lt0$$

which is to say, $M$ should have complex eigenvalues. In sum, if $M$ has complex eigenvalues, then the commutativity condition $AM=MA$ and the divisibility condition $\det(A)|\det(M)$ all by themselves limit the possibilities for $A$ to a finite set, regardless of the polynomial $P(A)$ on the left hand side.

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