Similar matrices with same invertible matrix. If $A$ and $B$ are similar matrices, then
$B = P^{-1}AP$ and $A = P^{-1}BP$ for some invertible matrix $P$. True or false.
All I know is if $A, B$ and $C$ are similar matrices then there is equivalence relation within them. Can someone please look into this and let me know whether is it true or false?. If is it true, then why and if false, provide me the justification.
 A: Note that the problem statement says that $B=P^{-1}AP$ and $A=P^{-1}BP$ --- rather than $A=PBP^{-1}$. As it stands, the statement is false.
Consider the case where $A=\operatorname{diag}(1,2,3)$ and $B=S^{-1}AS$, where $S=\pmatrix{0&1&0\\ 0&0&1\\ 1&0&0}$. By construction, the two matrices are similar to each other. Recall that diagonal matrices with distinct diagonal entries only commute with diagonal matrices. If $B=P^{-1}AP$ for some matrix invertible $P$, then $(PS^{-1})A=A(PS^{-1})$. Therefore $PS^{-1}$ is a diagonal matrix, meaning that
$$
P=\pmatrix{0&a&0\\ 0&0&b\\ c&0&0} \text{ for some } a,b,c\ne0.\tag{1}
$$
Now, if we also have $A=P^{-1}BP$, then $A=P^{-1}BP=P^{-1}(P^{-1}AP)P=P^{-2}AP^2$. Therefore $P^2A=AP^2$, i.e. $P^2$ is a diagonal matrix. Yet this is not possible for any $P$ in the form of $(1)$.
A: The answer is no.
If $A = P^{-1}BP$ and $B = P^{-1}BP$ then also $B = PBP^{-1}$ and hence $P^2B = BP^2$.
Take $P = \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 \\ 0 & 3\end{bmatrix}$.
We have
$$P^2B = \begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 2 \\ 0 & 3\end{bmatrix} = \begin{bmatrix} 1 & 8 \\ 0 & 1\end{bmatrix}$$
$$BP^2 = \begin{bmatrix} 1 & 2 \\ 0 & 3\end{bmatrix}\begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 0 & 3\end{bmatrix}$$
so they are not equal.
