Going through a past paper and I've come across this True or False question:
If $A$ and $B$ have the same Jordan Canonical Form (JCF), $A^2$ and $B^2$ have the same JCF.
I thought it was true, and gave the justification of:
Let $A = P^{-1} J P$ and $B = Q^{-1} JQ$, where $P$ and $Q$ and the respective change of basis matrices, and $J$ the JCF. Then computing $A^2$ and $B^2$ gives them to have the same JCF, namely $J^2$.
I've now come to realise that's incorrect, but now can't see a way of proving this. I've tried to come up with a counter example, but can't seem to think of one, so I'm still assuming the statement is true.
One other thought I've had is letting $Q^{-1} B Q = J = P^{-1} A P$, then multiplying both sides by $P^{-1} A P$ on the left, and doing some substitution to arrive at $P^{-1} A^2 P = Q^{-1} B^2 P$, meaning they will both equal the same JCF, say a matrix $K$. But I'm not sure that that works, and I think it probably misses a subtle point which says you can't do something like that.
Any nudges in the right direction?
I think the other thought you had is good, but what you should do is square both sides of the equation:
$(Q^{-1}BQ)^2=Q^{-1}B^2Q=P^{-1}A^2P=(P^{-1}AP)^2,$
so then $B^2=QP^{-1}A^2PQ^{-1}=(PQ^{-1})^{-1}A^2(PQ^{-1}),$ and we have $B^2$ is similar to $A^2$. Similarity is an equivalence relation, and so they will have the same JCF.