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Let us consider the matrix $$ A = \left( \begin{array}{crc} 0 & 0 & 3 \\ 81 & 0 & 0 \\ 0 & 3 & 0 \end{array} \right).$$

Does the matrix $A$ admit a square root each of whose entry is an integer? Please help me in this regard.

Thank you very much.

EDIT $:$

I have tried by taking determinant of $A$ but that is not working since $\det (A) =729,$ which is a perfect square. I don't know what other tools should I require to solve this. I think I can also use the trace argument here. If $B$ was the square root of $A$ then I also found that trace of $B$ is $2\sqrt 2 \cos \left (\frac {3 \pi} {8} \right) + 3,$ which is not an integer. Because $\lambda$ is an eigen value of $A$ iff $\sqrt {\lambda}$ is an eigen value of $B.$ So such a $B$ cannot be found.

Please check my argument above whether it holds good or not.

little o
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    What are your thoughts? What have you tried? Where are you stuck? You need to provide context for your question. If you add some appropriate context, we will be happy to help. – Greg Martin Feb 18 '19 at 19:47
  • Actually I have tried to find out such an example but I failed. I first thought by taking determinant. If the determinant was negative or not a perfect square then we were through. But this is not the case here because in this case $\det (A) = 729 = (27)^2.$ – little o Feb 18 '19 at 19:51
  • If one thinks that it is impossible then usually it is smarter to prove that no such example can be found (otherwise one might spend the rest of one's life searching for one). – Fabian Feb 18 '19 at 19:51
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    "What are your thoughts? What have you tried? Where are you stuck?" - Greg Martin – Servaes Feb 18 '19 at 19:52
  • "Actually I have tried to find out such an example but I failed." How did you go about trying to construct these examples? What did that result in? Why wasn't that a working example? Please, edit your question post and tell us these things. – Arthur Feb 18 '19 at 19:54
  • I have mentioned it in my comment above @Arthur. – little o Feb 18 '19 at 19:56
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    That comment was smaller when I made mine. Also, I asked you to edit your question post and put it there. We want new readers of this post to be able to get all relevant information without having to sift through the comments. – Arthur Feb 18 '19 at 20:02
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    Well I have edited my post. See if it's now ok or not. – little o Feb 18 '19 at 20:03
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    Yes, that's much better. Thank you. – Arthur Feb 18 '19 at 20:10
  • It is not quite true that "if $\lambda$ is an eigenvalue of $A$, then $\sqrt\lambda$ is an eigenvalue of $B$". (What is true is that if $\mu$ is an eigenvalue of $B$, then $\mu^2$ is an eigenvalue of $A$.) Exactly how did you compute the trace of $B$? – W-t-P Feb 18 '19 at 20:38
  • Can you provide me an example where it fails @W-t-P? – little o Feb 18 '19 at 20:41
  • Sure. Say, $A=1$ and $B=-1$, as matrices of order $1$. Then $A$ has eigenvalue $\lambda=1$, while $B$ does not have eigenvalue $\sqrt\lambda=1$. – W-t-P Feb 18 '19 at 20:45
  • Then how do I solve my problem? Do you have any other argument? – little o Feb 18 '19 at 20:53
  • There are alltogether $2^3=8$ square roots of the matrix. One of them has trace 6. – user Feb 18 '19 at 20:55
  • Which one do you mean @user? – little o Feb 18 '19 at 20:57
  • $$\begin{pmatrix}2&-1/9&2/3\ 18&2&-3\-3&2/3&2\end{pmatrix}.$$ – user Feb 18 '19 at 21:01
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    @Dbchatto67 I would advise you to delete the comment about a "stupid" example. It actually reveals a stupidity of something (or somebody) else. – user Feb 18 '19 at 21:16

2 Answers2

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$A$ is a pseudo-permutation. Thus $A$ has $3$ distinct eigenvalues and its commutant is $C(A)=span(I,A,A^2)$. If $B^2=A$, then $B\in C(A)$ and

$B=aI+bA+cA^2=\begin{pmatrix}a&9c&3b\\81b&a&243c\\243c&3b&a\end{pmatrix}$. We want that $a,3b,9c \in\mathbb{Z}$.

$B^2-A=0$ is equivalent to $a^2+1458bc=0,2ac+b^2=0,2ab+9\times(81c^2)-1=0$.

If $a\not= 0$, then $bc<0,ac<0,ab<0$, that is contradictory.

If $a=0$, then $b=0$ and $9\times(81c^2)=1$, that is contradictory (because $81c^2\in\mathbb{Z}$).

Conclusion. No solutions.

2

Using the similarity matrix $X=\begin{pmatrix}1&0&0\\0&9&0\\0&0&3\end{pmatrix}$, $A$ is similar to the circulant matrix $A'=\begin{pmatrix}0&0&9\\9&0&0\\0&9&0\end{pmatrix}$. This $A'$ has an easily found square root $B'=\begin{pmatrix}0&3&0\\0&0&3\\3&0&0\end{pmatrix}$. Now, reverse the similarity, and $B=\begin{pmatrix}0&\frac13&0\\0&0&9\\9&0&0\end{pmatrix}$ is a square root of $A$.

So there's a rational square root - but that's not an integer matrix. We need to look deeper. $A'$ can be diagonalized by the Fourier matrix; its eigenvalues are $9w^2$ for each cube root $\omega$ of unity, with eigenvectors $\begin{pmatrix}1&\omega&\omega^2\end{pmatrix}^T$. With this complete set of eigenvectors with different eigenvalues, any square root $B'$ of $A'$ must share those same eigenvectors, and its eigenvalues must be square roots $\pm 3\omega$ of the corresponding eigenvalues of $A'$. That gives us eight possibilities.

First, if we take $+$ signs everywhere or $-$ signs everywhere, we get $\begin{pmatrix}0&3&0\\0&0&3\\3&0&0\end{pmatrix}$ or $\begin{pmatrix}0&-3&0\\0&0&-3\\-3&0&0\end{pmatrix}$. Undoing the similarity, these each lead to square roots of $A$ with one non-integer entry as found before.

Taking a $-$ sign for the eigenvalue $1$ and $+$ signs for the other two eigenvalues leads to $\begin{pmatrix}-2&1&-2\\-2&-2&1\\1&-2&-2\end{pmatrix}$. Reverse the similarity and we get a square root $B=\begin{pmatrix}-2&\frac19&-\frac23\\-18&-2&3\\3&-\frac23&-2\end{pmatrix}$. Of course, taking a $+$ sign at $1$ and a $-$ sign at the other two would lead to the negative of this matrix. (See also @user's comments on the question)

The other four choices come from taking different signs for the two non-real roots. The trace of the square roots we get out of these have nonzero imaginary part $\pm 3\sqrt{3}$, so they can't be rational.

So then, $A$ has four rational square roots. None of them have integer entries, and thus the answer is no.

Your argument, on the other hand, fails; it's an attempt to prove that there aren't any rational square roots, when in fact there are.

jmerry
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