TITLE: Minimal polynomial and diagonalizable matrix
QUESTION [7 upvotes]: Let $B,C$ square matrices above a field,and $D$ Rectangle matrix in the correct size above the same field. $A=\begin{pmatrix}
B &D \\ 
 0& C
\end{pmatrix}$ 
I need to prove that if $A$ is Diagonalizable so Do $C$ and $B$,
And if $B$ and $c$ are Diagonalizable matrices with no common eigenvalue, so $A$ is Diagonalizable.
So I want to use the claim that says that if a matrix is Diagonalizable so it's Minimal polynomial splits to different factors with degree 1.
The second claim is easy because We know that $M_{A}(x)=M_{B}(x)M_{C}(x)$ and it is easily can be shown, but is this claim works the other way? Can I use it also the other way? And if not, what should I do to prove it?
(Please write simple and clear as you can cause I lose it with explanations of Linear algebra in English, differently from calculus, that for some reason it's more easy and fluent to me)
Thank you very much!

REPLY [3 votes]: I would like to clarify a small part of Arturo's excellent answer.
Let $p$ be any polynomial.  Because $A^n=\pmatrix{ B^n & * \\ 0 & C^n}$ (where $*$ stands for some unknown entry), we have that $p(A)=\pmatrix{ p(B) & * \\ 0 & p(C)}$.  Now, let $M_A$ be the minimal polynomial of $A$.  Our computation shows that $M_A(B)=0$ and $M_A(C)=0$.  Therefore, the minimal polynomial of $B$ and the minimal polynomial of $C$ ($M_B$ and $M_C$) both divide $M_A$.  However, we do NOT have that $M_A=M_B M_C$ unless the eigenvalues of $B$ are distinct from the eigenvalues of $C$.  For example, take $A=I_{2n}$ and $B=C=I_n$.  We have that $M_A=M_B=M_C=(x-1)$.