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#1 h Apr 2, 2025, 7:06 AM

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Does the matrix equation $X^3=mI_2$ is solvable over $M_{2}(\mathbb{Z})$ for every $m\in \mathbb{Z}$ . Here $X\in M_{2}(\mathbb{Z})$ , $I_2=\begin{pmatrix} 1& 0\\0 & 1\end{pmatrix}$ .

This post has been edited 1 time. Last edited by ay19bme, Apr 2, 2025, 7:08 AM
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#2 h Apr 2, 2025, 8:45 AM

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Take $m=2$ as an example. Each eigenvalue $t$ of $X$ is a root of $x^3-2$ which is irreducible by Eisenstein. And also because it has no rational root. Also $[Q(t):Q]\le 2$ since $F_X(t)$ has degree 2. Contradiction. Hence for $m=2$ we have no solution in $M_2(Z)$ . The equation is solvable iff $m$ is the cube of an integer.

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#3 h Apr 2, 2025, 8:49 AM

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I would say the following about this:
Based on the Cayley-Hamilton formula
${{X}^{2}}-tX+d{{I}_{2}}={{O}_{2}}$ where $t=Tr(X),d=\det (X).$
This can also be written as:
${{X}^{3}}-t{{X}^{2}}+dX={{O}_{2}}.$
After backsubstitution we get:
$({{t}^{2}}-d)X=(m+td){{I}_{2}}$
Now try to draw the conclusion from here!

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#4 h Apr 2, 2025, 8:51 AM

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Maybe a more elementary solution: The equation implies
$(\det X)^3 = \det (X^3) = \det (m I_2)=m^2.$ Hence $m^2$ , and consequently also $m$ , is the cube of an integer.
The same proof works in any dimension $n$ not divisible by $3$ .
If the dimension is divisible by $3$ , however, the equation is always solvable.

This post has been edited 1 time. Last edited by Gryphos, Apr 2, 2025, 8:52 AM

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#5 h Apr 2, 2025, 10:58 AM • 1 Y

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Thanks..So $m$ must be the cube of some integer.

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#6 h Apr 2, 2025, 6:13 PM

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The sequel of the Gryphos' post #4.

i) If $3$ doesn't divide $n$ , then $m=s^3$ and $X^3=s^3I_n$ .

$\bullet$ If $s=0$ then $X^3=0$ and $X=Q^{-1}diag(0,\cdots,0,J_2,\cdots,J_2,J_3,\cdots,J_3)Q$ , where $Q\in GL_n(\mathbb{Q})$ .

But what are all the matrices $Q$ so that $X$ is an integer matrix ? (In general, the integer matrices $Q$ are not sufficient).

$\bullet$ If $s\not= 0$ then $X$ is diagonalizable over $\mathbb{C}$ and $spectrum(X)\subset\{s,sj,sj^2\}$ , where $j=\exp(2i\pi/3)$ .

Then $X=Q^{-1}diag(sI_p,s\begin{pmatrix}0&-1\\1&-1\end{pmatrix},\cdots,s\begin{pmatrix}0&-1\\1&-1\end{pmatrix})Q$ .

But what are all the matrices $Q$ so that $X$ is an integer matrix ?

ii) If $3$ divides $n$ and $m\not= 0$ , then $X^3=mI_{3p}$ , $X$ is diagonalizable over $\mathbb{C}$ and $spectrum(X)\subset\{m^{1/3},m^{1/3}j,m^{1/3}j^2\}$ .

If $m$ is not a cube, then $x^3-m$ is irreducible over $\mathbb{Q}$ and we can group the eigenvalues $3$ by $3$ , that is, $p\times$ $(m^{1/3},m^{1/3}j,m^{1/3}j^2)$ .

Then $X=Q^{-1}diag(U_1,\cdots,U_p)Q$ , where $U_i=\begin{pmatrix}0&0&m\\1&0&0\\0&1&0\end{pmatrix}$ .

But what are all the matrices $Q$ so that $X$ is an integer matrix ?

This post has been edited 3 times. Last edited by loup blanc, Apr 2, 2025, 6:20 PM

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