Show that a block matrix is similar to another block matrix if and only if their blocks are similar. One of the problems I am trying to work is

Let $A \in M_n$ and $B,C \in M_m$ be given. Show that $\begin{pmatrix} A & 0\\
0 & B \end{pmatrix} \in M_{m+n}$ is similar to $\begin{pmatrix} A & 0\\
0 & C \end{pmatrix} \in M_{m+n}$ if and only if $B$ is similar to $C$.

$\Longleftarrow$: For these,  I was thinking I could do it directly: If $B \sim C$, then $B=S^{-1}CS$ and $C=SBS^{-1}$. Then we have that $\begin{pmatrix}A & 0\\
0 & B \end{pmatrix}=\begin{pmatrix}A & 0\\
0 & S^{-1}CS \end{pmatrix}= \begin{pmatrix}A & 0\\
0 & S^{-1}(SBS^{-1}) \end{pmatrix}$... I don't see where to go from here (if there is even a direction for it).
Now, for the $\Longrightarrow$ direction, I was thinking something similar. Can I use entries of $S$, $A$, $B$, and $C$ to multiply these matrices? Is there a better way?
 A: $\Longrightarrow$ Assuming your matrices are over an algebraically closed field, Jordan normal theory says that two square matrices $A$ and $B$ are similar if and only if their Jordan matrices have the same Jordan blocks. Let $A'$ be the Jordan matrix of $A$, with Jordan blocks $\{A_i\}$, $B'$ the Jordan matrix of $B$ with Jordan blocks $\{B_j\}$ and $C'$ the Jordan matrix of $C$ with Jordan blocks $\{C_k\}$. Then the Jordan matrix $J$ of $\begin{pmatrix}A & 0 \\ 0 & B\end{pmatrix}$ has Jordan blocks $\{A_i\} \sqcup \{B_j\}$, and is similar to $\begin{pmatrix}A & 0 \\ 0 & C\end{pmatrix}$ whose Jordan matrix is also $J$. Hence the Jordan blocks $\{C_k\}$ of $C$ are the Jordan blocks $\{B_j\}$, of $B$. As the Jordan blocks of $B$ and $C$ are the same, they are similar.  
$\Longleftarrow$ If $C=P^{-1}BP$ for some invertible matrix $P$ (i.e. $B$ and $C$ are similar), then $\begin{pmatrix}A & 0 \\ 0 & B\end{pmatrix}$ is similar to $\begin{pmatrix}A & 0 \\ 0 & C\end{pmatrix}$ via the matrix $\begin{pmatrix}I_{m} & 0 \\ 0 & P\end{pmatrix}$.  $\:\square$
A: The result remains true over an arbitrary field: Theorem 7.7 in this book shows that the (multiset) union of elementary divisors of $A$
 and $B$ (respectively $A$ and $C$) is the set of elementary divisors of $$\begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix} \qquad \left ( \text{respectively} \begin{bmatrix} A & 0 \\ 0 & C \end{bmatrix} \right ).$$ So if the two block diagonal matrices are similar, $B$ and $C$ must have the same elementary divisors, and therefore must be similar.
It is easy to see that the converse provided by Edward ffitch remains true over an arbitrary field.
