Welcome to our community

Be a part of something great, join today!

Problem of the Week #32 - January 7th, 2013

Status
Not open for further replies.
  • Thread starter
  • Moderator
  • #1

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Here's this week's problem (and the first Graduate POTW of 2013!).

-----

Problem: Prove that if $A$ is a $3\times 3$ matrix over $\mathbb{Q}$ such that $A^8=1$, then in fact $A^4=1$.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Moderator
  • #2

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
This week's problem was correctly answered by Deveno and Opalg. You can find Opalg's solution below:

First, notice that if $A$ is considered as an element of $M_3(\mathbb{C})$ it is diagonalisable because it has finite order. (If its Jordan form had any off-diagonal elements, it could not have finite order.) Its eigenvalues are 8th roots of unity, and we have to show that in fact they are 4th roots of unity.

The irreducible factorisation of $A^8-I=0$ over $\mathbb{Q}$ is $(A-I)(A+I)(A^2+I)(A^4+I) = 0$. The polynomial $(x-1)(x+1)(x^2+1)(x^4+1)$ must therefore be a multiple of the minimal polynomial of $A$ The characteristic polynomial $p(x)$ of $A$ must be a product formed using those same irreducible factors, namely $x\pm1$, $x^2+1$ and $x^4+1.$ It has degree 3, so it cannot involve the factor $x^4+1.$ So it must be formed from copies of the other factors, whose roots are all 4th roots of unity.
 
Status
Not open for further replies.