In symmetric matrix if $A^6=I \implies A^2=I$

Question

Let $A$ be a 3×3 real symmetric matrix such that $A^6 =I.$Then, $A^2 = I.$

Please give me a hint !

Answer 1

There exists inverse matrix $P$ such that $$PAP^{-1}=diag(\lambda_1,\lambda_2,\lambda_3).$$
So $$PA^6P^{-1}=diag(\lambda_1^6,\lambda_2^6,\lambda_3^6)=I.$$ This implies $$\lambda_1^6=\lambda_2^6=\lambda_3^6=1.$$ So $$\lambda_1^2=\lambda_2^2=\lambda_3^2=1;$$ $$PA^2P^{-1}=diag(\lambda_1^2,\lambda_2^2,\lambda_3^2)=I;$$ and $$A^2=I.$$

Answer 2

Hint: $$x^6-1=(x-1)(x+1)(x^2+x+1)(x^2-x+1)$$