Similar Matrices and their Jordan Canonical Forms [duplicate]

Let $A$ and $B$ be two matrices in $M_n$. Is the following ture:

$A$ and $B$ are similar $\iff$ $A$ and $B$ have the same jordan canonical form.

Could someone explain?

It's true. Let $A$ have the same JCF $J$ as $B$ does; say $A = PJP^{-1}$ and $B = Q J Q^{-1}$. Then $J = P^{-1} A P$, so $B = Q P^{-1} A P Q^{-1}$, and so $B$ is similar to $A$.

Suppose $A$ and $B$ are similar, but for contradiction they have different JCFs. Then $A$ reduces to JCF $J$, and $B$ to JCF $K$. But $J, K$ are not similar - this is a contradiction.

If $A=PJP^{-1}$ is the Jordan decomposition of $A$ and $M$ is an invertible matrix such that $A=M^{-1}BM$ then we have $$ B = (MM^{-1})B(MM^{-1})=M(M^{-1}BM)M^{-1} =MAM^{-1}=M(PJP^{-1})M^{-1}=(MP)J(P^{-1}M^{-1})$$

and thus the Jordan normal form of $A$ (namely $J$) is the same as that of $B$.