If A is involutory matrix then every natural power of A is involutory. Its written on Wikipedia I didn't get it. Suppose $A^7=A^6.A \ne I$.
So how come A is involutory ?
https://en.wikipedia.org/wiki/Involutory_matrix Fourth property.
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1$\begingroup$ You are thinking of the converse. The property is: If $A^2=I$ then $(A^n)^2=(A^2)^n=I$. The converse it not true. If $A=(e^{2\pi i/4})$, then $A^4=(1)=I$, but $A^2=(e^{\pi i})=(-1)\neq (1)=I$. $\endgroup$– user580373Aug 7, 2018 at 13:19
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$\begingroup$ @b00nheT That I read. The first line didn't make sense to me. $\endgroup$– DamanAug 7, 2018 at 13:20
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$\begingroup$ @spiralstotheleft Only even power makes it involuntary matrix right? $\endgroup$– DamanAug 7, 2018 at 13:23
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$\begingroup$ No, the converse is just not true. $\endgroup$– user580373Aug 7, 2018 at 13:25
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2 Answers
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In the article, there is only one direction specified:
If $\mathbf{A}$ is involutory, then $\forall n\in\mathbb{N}:\mathbf{A}^n$ is involutory, and
if $n$ is odd, then $\mathbf{A}^n=\mathbf{A}$
if $n$ is even, then $\mathbf{A}^n=\mathbf{I}$
In other words, every power is again involutory, as they relate to either $\mathbf{A}$ or $\mathbf{I}$ and $\mathbf{A}$ was supposed to and $\mathbf{I}$, the identity, trivially is.
Where your confusion comes from, is that you've read the statement as the converse implication, i.e. that if $\mathbf{A}^n$ is involutory, then $\mathbf{A}$ is which is generally not true.
EDIT: I give a proof the above mentioned statement: Let $\mathbf{A}$ be an involution, i.e. $\mathbf{A}^2=\mathbf{I}$. Let's look a the power $\mathbf{A}^n$ for $n\in\mathbb{N}$.
Let $n$ be even, i.e. $n=2k$. Now, we can write $\mathbf{A}^n=\mathbf{A}^2\dots\mathbf{A}^2$($k$ times). But now, $$\mathbf{A}^n=\mathbf{A}^2\dots\mathbf{A}^2=\mathbf{I}\dots\mathbf{I}=\mathbf{I}$$ For $n$ odd, i.e. $n=2k+1$, we may write $\mathbf{A}^n=\mathbf{A}^2\dots\mathbf{A}^2\mathbf{A}$ with k many $\mathbf{A}^2$. Now, $$\mathbf{A}^n=\mathbf{A}^2\dots\mathbf{A}^2\mathbf{A}=\mathbf{I}\dots\mathbf{I}\mathbf{A}=\mathbf{A}$$.
As $\mathbf{A}$ and $\mathbf{I}$ are involutory and either$\mathbf{A}^n=\mathbf{A}$ or $\mathbf{A}^n=\mathbf{I}$, we have that $\mathbf{A}^n$ with the desired equalities shown before.
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$\begingroup$ Meaning that if A is involutory and any power raised to it will be involutory as well. Suppose I say A is involutory how will you prove $A^7$ is involuntary. I am not just getting it. $\endgroup$– DamanAug 7, 2018 at 13:40
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$\begingroup$ @Damn1o1 I proved it in my comment above. When $A$ is involutory, this is when $A^2=I$, then $A^7$ is involutory because $(A^7)^2=(A^2)^7=I^7=I$. $\endgroup$– user580373Aug 7, 2018 at 13:43
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$\begingroup$ @spiralstotheleft Ohh I got it now! Thanks for the effort. $\endgroup$– DamanAug 7, 2018 at 13:45
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$\begingroup$ Oh, I just saw your comment, but I've included a proof of the statement in the edit anyway. $\endgroup$– blubAug 7, 2018 at 13:48
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A matrix $A$ is involutory if $A^2=I$. In particular $A$ is invertible.
An elementary property of powers is that $(A^m)^n=A^{mn}=(A^n)^m$. If $A$ is involutory, then $$ (A^n)^2=(A^2)^n=I^n=I $$ so $A^n$ is involutory as well.
Actually, if $n=2k+1$ is odd, then $A^n=A^{2k+1}=A^{2k}A=IA=A$; if $n=2k$ is even, then $A^n=A^{2k}=I$.
If $A^7=A^6$ and $A$ is invertible, then $A=I$.
