If $A$ is a square matrix such that $A^{27}=A^{64}=I$ then $A=I$.

What I did is to subtract I from both sides of the equation:



\begin{align*} A^{27}-I &= (A-I)(A+A^2+A^3+\dots+A^{26})=0\\ A^{64}-I &= (A-I)(A+A^2+A^3+\dots+A^{63})=0. \end{align*}

So from what I understand, either $A=I$ (as needed) or $A+A^2+A^3+\dots+A^{26}=0$ or $A+A^2+A^3+\dots+A^{63}=0$.

At this point I got stuck. By the way, I found out that $A$ is an invertible matrix because if $A^{27}=I$ then also $A^{26}A=AA^{26}=I$ then $A^{26}=A^{-1}$.

Also I thought to use the contradiction proving by assuming that $A+A^2+A^3+\dots+A^{63}=0$, but because $A^{27}=I$, then: $$A+A^2+A^3+\dots+A^{26}+I+A^{28}+\dots+A^{53}+I+A^{55}+\dots+A^{63}=0$$ but yet nothing.

Would appreciate your guidance, thanks!

  • 2
    $\begingroup$ Note, if the product of two matrices is zero, it is not necessarily the case that one of the matrices is zero. That is, $AB = 0 \nRightarrow A = 0$ or $B = 0$. $\endgroup$ – Michael Albanese Mar 20 '16 at 18:18
  • $\begingroup$ @MichaelAlbanese good point. wondering why I didn't notice. thanks $\endgroup$ – Ami Gold Mar 20 '16 at 18:20
  • $\begingroup$ Hint: Compute $$\left(A^{64}\right)^8\cdot A=A^{64\times8+1}=A^{513}=A^{27\times19}=\left(A^{27}\right)^{19}.$$ $\endgroup$ – Did Mar 20 '16 at 18:41

\begin{align*} & I = A^{64} = A^{2(27) + 10} = (A^{27})^2A^{10} = A^{10}\\ \implies & I = A^{27} = A^{2(10) + 7} = (A^{10})^2A^7 = A^7\\ \implies & I = A^{10} = A^{1(7)+3} = (A^7)^1A^3 = A^3\\ \implies & I = A^7 = A^{2(3) + 1} = (A^3)^2A = A. \end{align*}

This is nothing more than the Euclidean algorithm applied to the exponents. The same procedure can be used to show that if $A^p = I$ and $A^q = I$ with $p$ and $q$ coprime, then $A = I$.

  • $\begingroup$ wow this is tricky! liked your solution, however I wonder I could think about it all by myself. what was your motivation for such a solution? $\endgroup$ – Ami Gold Mar 20 '16 at 18:30
  • 1
    $\begingroup$ Well, I knew I could write $A^{64}$ as $A^{64-27}A^{27} = A^{37}$, so $A^{37} = I$. But then I could just keep repeating this trick, each time getting a smaller power of $A$ which was the identity, until that power was $1$. $\endgroup$ – Michael Albanese Mar 20 '16 at 18:34

Firstly, since $A A^{26} = A^{26}A=I$, then $A$ is an invertible matrix.

Use the fact that $\gcd(27,64)=1$: hence there exist some $a,b \in \Bbb{Z}$ such that $1=27a+64b$. Now, compute $$A=A^1=A^{27a+64b}=(A^{27})^a(A^{64})^b=I^aI^b=I$$

  • $\begingroup$ Nice. This is much quicker than the method I suggested. $\endgroup$ – Michael Albanese Mar 20 '16 at 18:19

Hint: if matrix $A$ has eigenpair $(\lambda, v )$, then $A^{27}$ and $A^{64}$ have eigenpairs $(\lambda^{27}, v)$ and $(\lambda^{64}, v)$. At the same time $A^{27} v = A^{64} v = 1 \cdot v$. Could you proceed from here?


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.