Find an integer quartic root of a $3\times 3$ matrix

Question

Find a $3 \times 3 $ matrix $X$ with integer coefficients such that \begin{align*} X^{4} &= 3 \begin{bmatrix} 2 &-1 &-1 \\ -1 &2 &-1 \\ -1 &-1 &2 \end{bmatrix}. \end{align*}

My attempt. Let us consider the matrix \begin{align*} A &= 3 \begin{bmatrix} 2 &-1 &-1 \\ -1 &2 &-1 \\ -1 &-1 &2 \end{bmatrix} \\ &= \begin{bmatrix} 6 &-3 &-3 \\ -3 &6 &-3 \\ -3 &-3 &6 \end{bmatrix} \end{align*} Calculate the roots of characteristic polynomial, i.e calculate the eigenspace $AZ=\lambda Z$, this is given for the equation system $A-\lambda I=0$, where $I$ is $3 \times 3$ identity matrix. \begin{align*} \begin{vmatrix} 6-\lambda & -3 & -3 \\ -3 & 6-\lambda & -3 \\ -3 & -3 & 6-\lambda \end{vmatrix} &= -\lambda \left( \lambda-9\right)^{2} \end{align*} Therefore, the polynomial function, the zero $\lambda=9$ has multiplicity $2$, and $\lambda=0$ has multiplicity $1$ and these special values are called the eigenvalues of the matrix $A$.

We need to know the dimension of the eigenspace generated by this eigenvalue. Thus, solve the system $\left(A-3I\right)Z=0$ where $Z^{T}=\left(x,y,z \right)$ in order to find the eigenvectors. (1) For $\lambda =0$, then $\left(A-3I\right)Z=0Z$. Thus, $x=y=z=0$. Hence, $v_{1}= \left(1,1,1\right)^{T}$ is an eigenvector corresponding to $\lambda=0$.

(2) For $\lambda=9$. Then, we choose $x=0$, $y=1$, then $z=-1$. Hence, $v_{2}= \left(0,1,-1\right)^{T}$. Also, choose $x=1$, $y=-1$, then $z=0$, hence, $v_{3}= \left(1,-1,0\right)^{T}$. Furthermore, $v_{2}$ and $v_{3}$ are eigenvector corresponding to $\lambda=9$.

Thus, we have the matrix $S=\left[v_{1} \ v_{2} \ v_{3} \right]$. Then, \begin{align*} S &= \begin{bmatrix} 1 &0 &1 \\ 1 &1 &-1 \\ 1 &-1 &0 \end{bmatrix} \end{align*} and its inverse \begin{align*} S^{-1} &= \begin{bmatrix} 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & -2/3 \\ 2/3 & -1/3 & -1/3 \end{bmatrix} \end{align*} Thus, $A=SJS^{-1}$, where \begin{align*} J &= \begin{bmatrix} 0 &0 &0 \\ 0 &9 &0 \\ 0 &0 &9 \end{bmatrix} \end{align*} where $J$ is the Jordan canonical form of $A$. Hence, $\displaystyle X=SJ^{1/4} S^{-1}$ \begin{align*} X&=SJ^{1/4}S^{-1} \\ A &= \begin{bmatrix} 1 &0 &1 \\ 1 &1 &-1 \\ 1 &-1 &0 \end{bmatrix} \begin{bmatrix} 0 &0 &0 \\ 0 &9^{1/4} &0 \\ 0 &0 &9^{1/4} \end{bmatrix} \begin{bmatrix} 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & -2/3 \\ 2/3 & -1/3 & -1/3 \end{bmatrix}\\ \end{align*} Now, $9^{1/4}= \sqrt[]{3}, \ - \ \sqrt[]{3}, \ \sqrt[]{3} \ i$, and $\ - \ \sqrt[]{3} \ i$. All these four values can be utilized, for $9^{1/4}$ and accordingly values of $X$ can be changed. All combination can be calculated to find the values of $X$.

\begin{align*} X&=SJ^{1/4}S^{-1} \\ A &= \begin{bmatrix} 1 &0 &1 \\ 1 &1 &-1 \\ 1 &-1 &0 \end{bmatrix} \begin{bmatrix} 0 &0 &0 \\ 0 &\sqrt[]{3} &0 \\ 0 &0 &\sqrt[]{3} \end{bmatrix} \begin{bmatrix} 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & -2/3 \\ 2/3 & -1/3 & -1/3 \end{bmatrix}\\ &=\begin{bmatrix} 2/\sqrt[]{3} & -1/\sqrt[]{3} & -1/\sqrt[]{3} \\ -1/\sqrt[]{3} & 2/\sqrt[]{3} & -1/\sqrt[]{3} \\ -1/\sqrt[]{3} & -1/\sqrt[]{3} & 2/\sqrt[]{3} \end{bmatrix} \end{align*} \begin{align*} X&=SJ^{1/4}S^{-1} \\ A &= \begin{bmatrix} 1 &0 &1 \\ 1 &1 &-1 \\ 1 &-1 &0 \end{bmatrix} \begin{bmatrix} 0 &0 &0 \\ 0 &- \ \sqrt[]{3} &0 \\ 0 &0 &- \ \sqrt[]{3} \end{bmatrix} \begin{bmatrix} 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & -2/3 \\ 2/3 & -1/3 & -1/3 \end{bmatrix}\\ &=\begin{bmatrix} -2/\sqrt[]{3} & 1/\sqrt[]{3} & 1/\sqrt[]{3} \\ 1/\sqrt[]{3} & -2/\sqrt[]{3} & 1/\sqrt[]{3} \\ 1/\sqrt[]{3} & 1/\sqrt[]{3} & -2/\sqrt[]{3} \end{bmatrix} \end{align*} \begin{align*} X&=SJ^{1/4}S^{-1} \\ A &= \begin{bmatrix} 1 &0 &1 \\ 1 &1 &-1 \\ 1 &-1 &0 \end{bmatrix} \begin{bmatrix} 0 &0 &0 \\ 0 &\sqrt[]{3} \ i &0 \\ 0 &0 & \sqrt[]{3} \ i \end{bmatrix} \begin{bmatrix} 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & -2/3 \\ 2/3 & -1/3 & -1/3 \end{bmatrix}\\ &=\begin{bmatrix} 2i/\sqrt[]{3} & -i/\sqrt[]{3} & -i/\sqrt[]{3} \\ -i/\sqrt[]{3} & 2i/\sqrt[]{3} &-i/\sqrt[]{3} \\ -i/\sqrt[]{3} & -i/\sqrt[]{3} & 2i/\sqrt[]{3} \end{bmatrix} \end{align*} \begin{align*} X&=SJ^{1/4}S^{-1} \\ A &= \begin{bmatrix} 1 &0 &1 \\ 1 &1 &-1 \\ 1 &-1 &0 \end{bmatrix} \begin{bmatrix} 0 &0 &0 \\ 0 & - \ \sqrt[]{3} \ i &0 \\ 0 &0 & - \ \sqrt[]{3} \ i \end{bmatrix} \begin{bmatrix} 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & -2/3 \\ 2/3 & -1/3 & -1/3 \end{bmatrix}\\ &=\begin{bmatrix} -2i/\sqrt[]{3} & i/\sqrt[]{3} & i/\sqrt[]{3} \\ i/\sqrt[]{3} & -2i/\sqrt[]{3} &i/\sqrt[]{3} \\ i/\sqrt[]{3} & i/\sqrt[]{3} & -2i/\sqrt[]{3} \end{bmatrix} \end{align*}

However, you can see that non of $X$'s have integer coefficients. Any idea where I have messed up something! Any help would be appreciated!

Answer 1

An idea is to find a matrix $A=\begin{pmatrix}0&0&0\\0&a&b\\0&c&d\end{pmatrix}$ which would

1. give $\begin{pmatrix}0&0&0\\0&3&0\\0&0&3\end{pmatrix}$ when being squared;
2. have integer elements after being multiplied by $\begin{pmatrix}1/3&1/3&1/3\\1/3&1/3&-2/3\\2/3&-1/3&-1/3\end{pmatrix}$.

These statements lead to the following conditions for $a$, $b$, $c$ and $d$: \begin{align} a^2+bc&=3,\\ b(a+d)&=0,\\ c(a+d)&=0,\\ d^2+bc&=3,\\ a+2b&=0 \mod 3,\\ a-b&=0 \mod 3,\\ c+2d&=0 \mod 3,\\ c-d&=0 \mod 3. \end{align} If $a+d\ne0$ then $b=c=0$, and the matrix is diagonal, so it has to contain square roots of $3$. So, $a+d=0$. Trying some small values for the unknowns, I came up with $a=2$, $d=-2$, $b=-1$, $c=1$. Finally, the matrix $X$ is $$ X=SAS^{-1}=\begin{pmatrix}-1&1&0\\1&0&-1\\0&-1&1\end{pmatrix}. $$

The matrix $X$ isn't unique. There are quite a few steps which can be done differently. First, instead of $\begin{pmatrix}0&0&0\\0&3&0\\0&0&3\end{pmatrix}$ one can use any matrix which square equals $\begin{pmatrix}0&0&0\\0&9&0\\0&0&9\end{pmatrix}$, for example $\begin{pmatrix}0&0&0\\0&-3&0\\0&0&-3\end{pmatrix}$ (just make sure it has the two eigenvalues of the same sign, otherwise you wouln'd be able to find the square root of it). Second, the condition 2 is a bit too strict. You'd want matrix $SAS^{-1}$ to be integer, and integrality of $AS^{-1}$ is a sufficient condition for that (in general it isn't necessary, though for our case it is). Third, there are multiple solutions for $a$, $b$, $c$, $d$ even for the listed case. For example, $a=4$, $d=-4$, $b=13$, $c=-1$.

Answer 2

Here is another approach. Let $E_k$ be the $k\times k$ matrix of ones. We are going to find all integer matrix solutions of $X^4=A:=3(3I_3-E_3)$.

The eigenvalues of $A$ are $0,9,9$. Therefore $X$ has a simple eigenvalue $0$ and $\pm\sqrt{3},\pm i\sqrt{3}$ are the only possible nonzero eigenvalues of $X$. However, as the characteristic polynomial of $X$ has integer coefficients, the spectrum of $X$ must be $\{0,\sqrt{3},-\sqrt{3}\}$ or $\{0,i\sqrt{3},-i\sqrt{3}\}$.

The characteristic polynomial of $X$ is therefore $x^3\pm3x$. Hence $X^3=\pm3X$ and $3(3I_3-E_3)=A=X^4=\pm3X^2$. Consequently, we must have $$ X^2=\pm(3I_3-E_3).\tag{1} $$ As $X$ must share the same null space as $X^4=3I_3-E_3$, all row sums and column sums of $X$ are zero. Therefore, if we define $e=(1,1)^T$, $X$ must be in the form of $$ X=\pmatrix{P&-Pe\\ -e^TP&e^TPe}.\tag{2} $$ for some $2\times2$ matrix $P$. So, we need to solve $$ X^2=\pmatrix{P(I_2+E_2)P&-P(I_2+E_2)Pe\\ -e^TP(I_2+E_2)P&e^TP(I_2+E_2)Pe} =\pm(3I_3-E_3)=\pm\pmatrix{3I_2-E_2&-e\\ -e^T&2}. $$ This boils down to solving $P(I+E)P = \pm(3I_2-E)$, i.e. $$ P\pmatrix{2&1\\ 1&2}P=\pm\pmatrix{2&-1\\ -1&2}.\tag{3} $$ At this point, we can already obtain an obvious solution $P=\operatorname{diag}(1,-1)$, which translates back to $$ X=\pmatrix{1&0&-1\\ 0&-1&1\\ -1&1&0},\tag{4} $$ but we shall continue our search for other solutions. Note that the two constant matrices in $(3)$ have identical eigenspaces, so it's a good idea to perform a change of basis. Let $L=\pmatrix{1&1\\ 1&-1}$. Then $L^{-1}=L/2$. If we define $Q=\pmatrix{a&b\\ c&d}:=LPL^{-1}$, then $$ P=L^{-1}QL =\frac12\pmatrix{a+b+c+d&a-b+c-d\\ a+b-(c+d)&a-b-(c-d)}.\tag{5} $$ By considering the values of $(1,\pm1)P(1,\pm1)^T$, we see that if $P$ is an integer matrix, $a,b,c,d$ must be integers, and if $a,b,c,d$ are integers, $P$ is an integer matrix if and only if $a+b+c+d$ is even. Also, by conjugation by $L$ on both sides of $(3)$, we get $$ Q\pmatrix{3&0\\ 0&1}Q=\pmatrix{3a^2+bc&(3a+d)b\\ (3a+d)c&3bc+d^2}=\pm\pmatrix{1&0\\ 0&3}.\tag{6} $$ Therefore, all solutions to $(3)$ are given by all integers $a,b,c,d$ such that \begin{cases} a+b+c+d\text{ is even},\\ 3a^2+bc=\pm1,\\ 3a+d=0. \end{cases} As $3a+d=0$, the condition that $a+b+c+d$ is even can be further reduced to that $b+c$ is even. Hence $b$ and $c$ have the same parity. In summary, all integer matrix solutions to $(3)$ and hence to $(1)$ are given by integers $a,b,c,d$ such that \begin{cases} bc=\pm1-3a^2,\\ b,c \text{ have the same parity},\\ d=-3a. \end{cases}