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I was going through my school papers and found an interesting question, so I experimented a bit more and found out a pattern, so I made a formula for such matrices.

$$A = \begin{bmatrix}x&-(x-1)\\x+1&-x\end{bmatrix}$$ where $x > 0$ is an integer

$$A^n = \begin{cases} I, & \text{if $n$ is even} \\ A, & \text{if $n$ is odd} \end{cases}$$ where $I$ is identity matrix of order 2.

I just wanted to know if this has been found before or whether it has a name too or if there are some cases that does not obey this.

Hope someone can format my question properly, I'm new to this community. Hope this is the correct way of putting things together too. Thanks!!

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    $\begingroup$ Do you mean for $x$ to be an integer? What is $f$, other than the definition you give? I don't understand what pattern you see. $\endgroup$
    – saulspatz
    Nov 30, 2020 at 18:57
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    $\begingroup$ What is the context/motivation for defining $A$ and $f$? There's no reason why you can't define a function $f\colon \mathbb{Z}^{+} \to \mathbb{Z}^{2\times 2}$, but from your question there's also not a clear reason why you would want to. $\endgroup$
    – DMcMor
    Nov 30, 2020 at 19:00
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    $\begingroup$ That's all very well, but it's not clear to me what you have noticed. Can you give a concrete example? $\endgroup$
    – saulspatz
    Nov 30, 2020 at 19:07
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    $\begingroup$ If you are referring to the property of matrix $A$, it simply means that $A$ is its own inverse, that is, $A^{-1}=A$. $\endgroup$
    – NoName
    Nov 30, 2020 at 19:16
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    $\begingroup$ @DMcMor Is it still unclear? What i mean is if you make a matrix by putting in a value for x, and call it matrix A. when this A is raised to even numbers, it will give identity matrix and when it is raised to odd numbers it will give the same matrix A. $\endgroup$ Nov 30, 2020 at 19:16

2 Answers 2

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Thank you for addressing our comments and clarifying your question! Any square matrix which satisfies your equation $$A^n = \begin{cases} I, & \text{if $n$ is even} \\ A, & \text{if $n$ is odd} \end{cases}$$ is its own inverse, i.e. it is an involutory matrix. This is because if $n = 2$ you have $AA = I$, which by definition means that $A$ is its own inverse. From that simple statement you can extrapolate your formula, because if $n$ is even then $$A^{n} = A^{2m} = (AA)^{m} = I^{m} = I$$ for some integer $m$, and if $n$ is odd then $$A^{n} = A^{2k+1}=A^{2k}A=(AA)^{k}A = I^{k}A=IA=A$$ for some integer $k$. In fact, for any $2\times 2$ matrix $$\begin{pmatrix} a & b\\c & -a\end{pmatrix},$$ of which your matrix is an example, such a matrix will be involutory if $a^{2} + bc = 1$. We can verify this for your matrix $A$: $$x^{2} - (x-1)(x+1) = 1.$$

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    $\begingroup$ Hi, thanks for remaining patient and answering, I understand what you mean ! $\endgroup$ Nov 30, 2020 at 19:58
  • $\begingroup$ Hi again, just wanted to know, is there anything like this related to 3x3 matrix? $\endgroup$ Dec 1, 2020 at 6:41
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    $\begingroup$ @CoolCloud - Of course. There are $n\times n$ involution matrices for any $n$. For example, just take your $2\times2$ matrix and expand it with $1$'s on the diagonal and $0$'s everywhere else. $\endgroup$
    – mr_e_man
    Dec 1, 2020 at 7:03
  • $\begingroup$ @mr_e_man That would be identity matrix right? $\endgroup$ Dec 1, 2020 at 10:49
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    $\begingroup$ @CoolCloud - I mean take $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ where $A^2=I_{2\times2}$ and make $$B=\begin{bmatrix}a&b&0&0&\cdots&0\\c&d&0&0&\cdots&0\\0&0&1&0&\cdots&0\\0&0&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&0&\cdots&1\end{bmatrix}$$ so that $B^2=I_{n\times n}$. $\endgroup$
    – mr_e_man
    Dec 3, 2020 at 20:06
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You simply have $A^2=I$, from which it follows that $A^3=AA^2=AI=A$, $A^4= AA^3=AA=A^2=I$ and so on. Of course $A^2=I$ is the same as saying that $A=A^{-1}$ so that $A$ is its own inverse.

This is sometimes stated as $A$ is self-inverse, or that $A$ is an involution, in more highfalutin language.

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  • $\begingroup$ @mr_e_man Corrected, thanks. $\endgroup$
    – saulspatz
    Nov 30, 2020 at 19:33
  • $\begingroup$ Thanks for the reply, I understand this better now. $\endgroup$ Nov 30, 2020 at 19:59

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