# Let $A^{27}=A^{64}=I$, show that $A=I$

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

• 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. – Julien Apr 6 '13 at 17:30
• 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 $k=tn$? – Ben Apr 6 '13 at 17:49
• @Ben Wrong direction, I think. $a^n = 1$ does not imply $a = 1$, though the converse is true. Unless I misunderstood... – Thomas Apr 7 '13 at 11:21
• @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 – Ben Apr 8 '13 at 10:32

## 6 Answers

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.

• This is a great way to understand the Euclidean algorithm! – asmeurer Apr 6 '13 at 21:17

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, $A = I$.

• Doesn't this assume that m and n are nonnegative? – templatetypedef Apr 7 '13 at 1:39
• @templatetypedef : I don't know, but here it's OK for $m$ or $n$ to be negative because $A$ is invertible. – Stefan Smith Apr 7 '13 at 21:06

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 $A=I$.

• Easier: Bezout on $27$ and $64$ directly yields $A=A^{u27+v64}=I$. – Julien Apr 6 '13 at 17:24
• Why would you say that $A^{64}=0$? Does $I=0$? – Georgey Apr 6 '13 at 17:29
• @Eran That's called a typo. – Julien Apr 6 '13 at 17:30
• @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 $A=I$. – clark Apr 6 '13 at 18:03
• @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. – Julien Apr 6 '13 at 18:34

Hint: 27 and 64 are coprime and consider Bezout's identity.

• 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? – dato datuashvili Apr 6 '13 at 17:23
• @dato, see julien's comment. – Easy Apr 6 '13 at 17:26

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 $D=I$.

• You really like Jordan normal form, Dominic. I suggest you create a "Jordan normal form" tag. I would use it. – Julien Apr 6 '13 at 17:35
• @julien yeah I really like the jordan normalform :) – Dominic Michaelis Apr 6 '13 at 17:36
• How do you conclude that $N = 0$? – k.stm Apr 30 '13 at 8:40

First of all, since $A^{27}=I$, $A$ is invertible.

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