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Let $$ S = \left\{ \begin{pmatrix} x & y \\ z & w \end{pmatrix} \in \mathbb{R}^{2\times2}: \; x,y,z,w \; \text{ is an arithmetic sequence}\right\}. $$ Find all matrices $C \in S$ that satisfying $\;\exists k \in \mathbb{Z}, k\geq 2, C^k \in S$.

It is clear that $$ x\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} $$ satisfies the problem, and it seems that it is the only solution. We then assume that the common difference is not zero. We can easily find that we only need to consider $$ \begin{pmatrix} -3 & -1 \\ 1 & 3 \end{pmatrix} + r\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}. $$ But I don't know what should I do next. By the way I find that $k$ is odd but I can't make use of it.

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    $\begingroup$ "We can easily find that we only need to consider [ ... ]" How? Why? $\endgroup$ Mar 3 at 2:10
  • $\begingroup$ @Zubin write the elements in $S$ in another way $\begin{pmatrix} x-3a & x-a \\ x+a & x+3a \end{pmatrix}$, and the common difference is $2a$. If $a \neq 0$, write the elements in $S$ as $a\begin{pmatrix} -3 & -1 \\ 1 & 3 \end{pmatrix}+x\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$. Notice that $A \in S \implies cA \in S$, where c is a constant. As $a \neq 0$, divide $a\begin{pmatrix} -3 & -1 \\ 1 & 3 \end{pmatrix}+x\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ by $a$, and that is the conclusion. $\endgroup$
    – tys
    Mar 3 at 4:24

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In fact, $x \begin{pmatrix} 1 & 1\\ 1 & 1\\ \end{pmatrix} $ is not the only solution. Note that $ \begin{pmatrix} -3 & -1\\ 1 & 3\\ \end{pmatrix} ^3=8\begin{pmatrix} -3 & -1\\ 1 & 3\\ \end{pmatrix}$, so $x \begin{pmatrix} -3 & -1\\ 1 & 3\\ \end{pmatrix}$ satisfies your problem too. We are going to show that these are the only matrices to meet your condition.

Let $A=\begin{pmatrix} a & a+d\\ a+2d & a+3d\\ \end{pmatrix}\in S$, and $A^n\in S$ . By the Cayley-Hamilton theorem, there is a real quadratic $f(X)$ such that $f(A)=0$, and hence there exists $s,t\in \mathbb R$ such that $A^n=sA+tI$. If $t\neq 0$, then it is clear that $sA+tI\notin S$, a contradiction. Therefore $A^n=sA$.
Case 1: If $A$ is not invertible, then $$0=\det (A)=a(a+3d)-(a+d)(a+2d)=-2d^2,$$ so $d=0$, and we get the first solution $x \begin{pmatrix} 1 & 1\\ 1 & 1\\ \end{pmatrix} $.
Case 2: If $A$ is invertible, then $A^{n-1}=sI$. So all eigenvalues of $A$ must satisfy the equation $\lambda^{n-1}=s$. Since the determinant of $A$ is negative, the two eigenvalues of $A$ must be real and cannot be equal. But the equation $\lambda^{n-1}=s$ has at most 2 solutions in $\mathbb R$, and they are opposite numbers. Hence, the eigenvalues of $A$ are opposite numbers, so $trace(A)=0$, which is to say $2a+3d=0$. We get the second solution $x \begin{pmatrix} -3 & -1\\ 1 & 3\\ \end{pmatrix}$.

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