Square of a rational matrix of dimension 2

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Square of a rational matrix of dimension 2

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#1 h Feb 17, 2025, 4:53 PM

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The following exercise was posted -two months ago- on the Website StackExchange; cf.
https://math.stackexchange.com/questions/5006488/image-of-the-squaring-function-on-mathcalm-2-mathbbq
There was no solution on Stack.

-Statement of the exercise-
We consider the matrix function $f:X\in M_2(\mathbb{Q})\mapsto X^2\in M_2(\mathbb{Q})$ .
Find the image of $f$ .
In other words, give a method to decide whether a given matrix has or does not have at least a square root
in $M_2(\mathbb{Q})$ ; if the answer is yes, then give a method to calculate at least one of its roots.

This post has been edited 1 time. Last edited by loup blanc, Feb 18, 2025, 1:37 PM

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#2 h Feb 17, 2025, 6:50 PM

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loup blanc wrote:

The following exercise was posted on the Website StackExchange; cf.
look here
There was no solution on Stack.

-Statement of the exercise-
We consider the matrix function $f:X\in M_2(\mathbb{Q})\mapsto X^2\in M_2(\mathbb{Q})$ .
Find the image of $f$ .
In other words, give a method to decide whether a given matrix has or does not have at least a square root
in $M_2(\mathbb{Q})$ ; if the answer is yes, then give a method to calculate at least one of its roots.

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#3 h Feb 21, 2025, 2:51 PM

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This problem also appeared in a regional romanian contest in november, signed by mr. Vlad Mihaly.

This post has been edited 1 time. Last edited by RobertRogo, Feb 21, 2025, 2:54 PM

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Ultimatewarrio7

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Ultimatewarrio7

#4 h Feb 21, 2025, 5:19 PM

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RobertRogo wrote:

This problem also appeared in a regional romanian contest in november, signed by mr. Vlad Mihaly.

Could you please name this contest?

This post has been edited 2 times. Last edited by Ultimatewarrio7, Feb 21, 2025, 5:20 PM

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#5 h Feb 21, 2025, 5:42 PM

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So this is actually a number theory problem. We can look at the characteristic polynomial, and if (and only if) the root is a square in the number field then the matrix have a square root. This will require square determinate but not neccerly sufficient

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#6 h Feb 21, 2025, 10:26 PM

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Ultimatewarrio7 wrote:

RobertRogo wrote:

This problem also appeared in a regional romanian contest in november, signed by mr. Vlad Mihaly.

Could you please name this contest?

Sure, it's called "Marian Țarină"

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#7 h Mar 11, 2025, 3:08 PM

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I see that this question is not more successful here than on StackExchange.
OK, I will give a solution proposal in a week.

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#8 h Mar 13, 2025, 11:53 AM • 2 Y

Y by teomihai, loup blanc

Well, the guy who posted the question on MSE almost answered the question. Anyway, here's the answer.

The image of $f$ is the set $\{qI_2: \ q \in \mathbb{Q}\} \cup S,$ where $S$ is the set of all $Y \in M_2(\mathbb{Q})$ such that $\text{tr} (Y) +2\sqrt{\det Y}=q^2$ or $\text{tr}(Y) -2\sqrt{\det Y}=q^2,$ for some $0 \ne q \in \mathbb{Q}.$

Proof. Let
$X=\begin{pmatrix}a & b \\ c & d\end{pmatrix}, \ \ \ Y=\begin{pmatrix}q_1 & q_2 \\ q_3 & q_4\end{pmatrix}.$ So if $Y=X^2,$ then we'll have the equations
$a^2+bc=q_1, \ \ b(a+d)=q_2, \ \ c(a+d)=q_3, \ \ d^2+bc=q_4. \ \ \ \ \ \ \ \ \ \ (*)$ We now consider two cases.

Case 1: $\text{tr}(X)=0.$ So $a+d=0$ and hence, by $(*), \ q_2=q_3=0, \ q_1=q_4,$ which gives $Y=q_1I_2.$ Conversely, if $Y=qI_2, \ q \in \mathbb{Q},$ then $Y=X^2,$ where $X=\begin{pmatrix}0 & q \\ 1 & 0\end{pmatrix}.$

Case 2: $\text{tr}(X) \ne 0.$ Let $t:=\text{tr}(X)=a+d.$ Then, by $(*),$
$b=\frac{q_2}{t}, \ \ \ \ c=\frac{q_3}{t}.$ Also, $a^2-d^2=q_1-q_4,$ which gives $a-d=\frac{q_1-q_4}{t},$ which together with $a+d=t$ give
$a=\frac{t^2+q_1-q_4}{2t}, \ \ \ \ d= \frac{t^2-q_1+q_4}{2t}.$ So if we put what we got for $a,b,c$ in the identity $a^2+bc=q_1$ in $(*)$ and simplify, we'll get
$t^4-2(q_1+q_4)t^2+(q_1-q_4)^2+4q_2q_3=0$ and so $t^2=q_1+q_4 \pm 2\sqrt{q_1q_4-q_2q_3}=\text{tr}(Y) \pm 2\sqrt{\det Y}.$ Conversely, if $\text{tr}(Y) + 2\sqrt{\det Y}=q^2$ or $\text{tr}(Y) - 2\sqrt{\det Y}=q^2,$ for some $0 \ne q \in \mathbb{Q},$ then $Y=X^2,$ where
$X=\frac{1}{2q}\begin{pmatrix}q^2+q_1-q_4 & 2q_2 \\ 2q_3 & q^2-q_1+q_4 \end{pmatrix}.$

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#9 h Saturday at 6:05 PM

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$\textbf{Remark 1.}$ When $A$ is a rational scalar matrix, then it has an infinity of square roots over $\mathbb{Q}$ .
Otherwise, it has $0,2$ or $4$ square roots.
In the generic case (choose randomly $A$ ), its eigenvalues are in the form $u+\sqrt{v}\in K=\mathbb{Q}[\sqrt{v}]$ , where $u,v\in\mathbb{Q}$
and $v<0$ or ( $v>0$ and is not a square). If $A$ is a square, then necessarily its eigenvalues are squares in $K$ , that is,
$u+\sqrt{v}=(a+b\sqrt{v})^2$ , where $a,b\in\mathbb{Q}$ . In general, this property is not satisfied.
Let $A=\begin{pmatrix}0&-4\\1&q\end{pmatrix}$ . If $q=7$ then $A$ has no square root because $\dfrac{7+\sqrt{33}}{2}$ is not a square.
If $q=8$ , then $4+2\sqrt{3}=(1+\sqrt{3})^2$ ; there is no obstruction and we can verify that $A$ has only $2$ square roots.
If $q=5,4$ , then $A$ has $4,0$ square roots.
An example with $v<0$ . $A=\begin{pmatrix}0&-9\\1&-2\end{pmatrix}$ ; $-1+2i\sqrt{2}=(1+i\sqrt{2})^2$ and $A$ has $2$ square roots.
$\textbf{Remark 2.}$ If we randomly choose $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ with $a,b,c,d$ integers in $[[-99..99]]$ , with an uniform
distribution of probability, then the probability that $A$ is a square is $\approx 6.2\times 10^{-4}$ .

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#10 h Yesterday at 1:28 AM

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As someone on MSE pointed out, the relation $(\text{tr}(X))^2=\text{tr}(Y) \pm 2\sqrt{\det Y},$ where $Y=X^2,$ can be quickly proven using the Cayley-Hamilton theorem; i.e. by taking the trace of both sides of $X^2-\text{tr}(X)X+(\det X)I_2=0.$ But the problem with this method is that it's not helpful for proving the converse; i.e. if $\text{tr}(Y) \pm 2\sqrt{\det Y}=p^2,$ for some $0 \ne p \in \mathbb{Q},$ then there exists $X \in M_2(\mathbb{Q})$ such that $Y=X^2.$ The method I gave in my previous post gives an $X.$

This post has been edited 1 time. Last edited by ysharifi, Yesterday at 1:29 AM
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