I was working on the following problem when I stumbled upon an oddity.
If $X=P^{-1}AP$ and $A^3=I$, prove that $X^3=I$
My first approach was to cube both side which led to the following:
$X^3=(P^{-1}AP)^3$ $X^3=(P^{-1})^3A^3P^3$ Since $A^3=I$ $X^3=(P^{-1})^3I^3P^3$ $X^3=(P^{-1})^3P^3$ $X^3=(P^{-1}P)^3$ $X=I^3=I$ However this method seemed to contradict what I had previously learned. Instead I tried this approach: $X=P^{-1}AP$ Now premultiply each side by X and we get: $X^2=P^{-1}APP^{-1}AP=P^{-1}AIAP=P^{-1}A^2P$
We do this once more: $X^3=P^{-1}A^2PP^{-1}AP=P^{-1}A^2IAP=P^{-1}A^3P$
Since $A^3=I$ $X^3=P^{-1}IP=P^{-1}P=I$
Does anyone know which approach is correct?