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Let $A$ be a square matrix, $A^{27}=A^{64}=I$, show that $A=I$

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    $\begingroup$ I'll remark, as Martin Brandenburg would probably do, that this has nothing to do with square matrices in particular. This is true in any multiplicative group. $\endgroup$ – Julien Apr 6 '13 at 17:30
  • $\begingroup$ Can we not just use the fact that if $ g^t=e $ where $ e $ is neutral and $ t $ is minimal and if $g^k=e $ then $ k=tn $? $\endgroup$ – Ben Apr 6 '13 at 17:49
  • $\begingroup$ @Ben Wrong direction, I think. $a^n = 1$ does not imply $a = 1$, though the converse is true. Unless I misunderstood... $\endgroup$ – Thomas Apr 7 '13 at 11:21
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    $\begingroup$ @Thomas I mean since 64 is a multiple of the order as is 27 is it not obvious? If we let $64 = k_1 * t$ and $27 = k_2 * t$ so $t$ has to be one $\endgroup$ – Ben Apr 8 '13 at 10:32
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Use the fact that $A^{x} * A^{y}=A^{x+y}$

$A^{27}=I$ implies $A^{54}=I$.

$A^{54}=I$ and $A^{64}=I$ imply $A^{10}=I$.

$A^{10}=1$ implies $A^{30}=I$

$A^{30}=I$ and $A^{27}=I$ imply $A^{3}=I$.

$A^{3}=1$ implies $A^{9}=I$

$A^{9}=I$ and $A^{10}=I$ imply $A^{1}=A=I$.

This is essentially applying the euclidean algorithm to find the gcd of 27 and 64.

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    $\begingroup$ This is a great way to understand the Euclidean algorithm! $\endgroup$ – asmeurer Apr 6 '13 at 21:17
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Because $\gcd(27, 64) = 1$, there are integers $m, n$ such that $27m = 64n + 1$.

We know that $$A^{27m} = \left(A^{27}\right)^m = I^m = I$$ but at the same time, we have $$A^{27m} = A^{64n + 1} = AA^{64n} = A\left(A^{64}\right)^n = AI^n = A$$

Thus, $A = I$.

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  • $\begingroup$ Doesn't this assume that m and n are nonnegative? $\endgroup$ – templatetypedef Apr 7 '13 at 1:39
  • $\begingroup$ @templatetypedef : I don't know, but here it's OK for $m$ or $n$ to be negative because $A$ is invertible. $\endgroup$ – Stefan Smith Apr 7 '13 at 21:06
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Since $A^{27}-I=0$ and $A^{64}-I=0$, the minimal polynomial $m_A$ of $A$ divides both $x^{27}-1$ and $x^{64}-1$. Since 27 and 64 are co-prime, the greatest common divisor of these two polynomials is $x-1$ and thus $A-I=0$ or $A=I$.

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    $\begingroup$ Easier: Bezout on $27$ and $64$ directly yields $A=A^{u27+v64}=I$. $\endgroup$ – Julien Apr 6 '13 at 17:24
  • $\begingroup$ Why would you say that $A^{64}=0$? Does $I=0$? $\endgroup$ – Georgey Apr 6 '13 at 17:29
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    $\begingroup$ @Eran That's called a typo. $\endgroup$ – Julien Apr 6 '13 at 17:30
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    $\begingroup$ @julien Perphaps it is easier but it is a different argument worth mentioning. $P_1(A)=P_2(A)=0$ and $gcd(P_1(x),P_2(x))=x-1$ it implies $A=I$. $\endgroup$ – clark Apr 6 '13 at 18:03
  • $\begingroup$ @clark I did not really say it was not worth mentioning. And I understood the argument. It is just that when I wrote my comment, I had not seen Easy's answer. $\endgroup$ – Julien Apr 6 '13 at 18:34
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Hint: 27 and 64 are coprime and consider Bezout's identity.

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  • $\begingroup$ so it means that if two matrix is equal to each other at given powers and these powers are coprime,then each matrix is identity matrix? $\endgroup$ – dato datuashvili Apr 6 '13 at 17:23
  • $\begingroup$ @dato, see julien's comment. $\endgroup$ – Easy Apr 6 '13 at 17:26
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Use the Jordan normal form and you have $$ (D+N)^{27}= (D+N)^{64}$$ Hence $N=0$ As they are coprimes they can't be root of unities. Hence $D=I$.

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    $\begingroup$ You really like Jordan normal form, Dominic. I suggest you create a "Jordan normal form" tag. I would use it. $\endgroup$ – Julien Apr 6 '13 at 17:35
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    $\begingroup$ @julien yeah I really like the jordan normalform :) $\endgroup$ – Dominic Michaelis Apr 6 '13 at 17:36
  • $\begingroup$ How do you conclude that $N = 0$? $\endgroup$ – k.stm Apr 30 '13 at 8:40
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First of all, since $A^{27}=I$, $A$ is invertible.

Since $19\cdot27-8\cdot64=1$, we have $$ A^1=\left(A^{27}\right)^{19}\left(A^{64}\right)^{-8}=I^{19}I^{-8}=I $$

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