8
$\begingroup$

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!!

$\endgroup$
  • 1
    $\begingroup$ 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 '20 at 18:57
  • 1
    $\begingroup$ 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 '20 at 19:00
  • 1
    $\begingroup$ 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 '20 at 19:07
  • 1
    $\begingroup$ 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 '20 at 19:16
  • 1
    $\begingroup$ @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. $\endgroup$ – Cool Cloud Nov 30 '20 at 19:16
12
$\begingroup$

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 $A$: $$x^{2} - (x-1)(x+1) = 1.$$

$\endgroup$
  • 1
    $\begingroup$ Hi, thanks for remaining patient and answering, I understand what you mean ! $\endgroup$ – Cool Cloud Nov 30 '20 at 19:58
  • $\begingroup$ Hi again, just wanted to know, is there anything like this related to 3x3 matrix? $\endgroup$ – Cool Cloud Dec 1 '20 at 6:41
  • 1
    $\begingroup$ @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 '20 at 7:03
  • $\begingroup$ @mr_e_man That would be identity matrix right? $\endgroup$ – Cool Cloud Dec 1 '20 at 10:49
  • 2
    $\begingroup$ @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 '20 at 20:06
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @mr_e_man Corrected, thanks. $\endgroup$ – saulspatz Nov 30 '20 at 19:33
  • $\begingroup$ Thanks for the reply, I understand this better now. $\endgroup$ – Cool Cloud Nov 30 '20 at 19:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.