Question

Let a be a 3Ă—3 real symmetric matrix such that a6 = i. Then, a2 = i

Answer 1

Since AA is real and symmetric, it is diagonalizable. So we may assume that AA is diagonal. In that case, we have a6jj=1ajj6=1 for all jj. So a2jjajj2 is nonnegative and its cube is 11: thus a2jj=1ajj2=1, and A2=IA2=I.

Note that this works for any diagonalizable AA (within Mn(R))Mn(R)); it doesn't have to be symmetric.