Diagonalizability of a matrix containing smaller diagonalizable matrices

[oferon]

Given $R\in M_n(F)$ and two matrices $A\in M_{n1}(F)$ and $D\in M_{n2}(F)$ where $n1+n2=n$
$R = \begin{pmatrix} A & B \\ 0 & D \end{pmatrix}$
Given A,D both diagonalizable (over F), and don't share any identical eigenvalues - show R is diagonalizable.

I'm building eigenvectors for R, based on the eigenvalues and eigenvectors of A and D.
for example, suppose $λ_1$ is eigenvalue of A with eigenvector $V = \begin{pmatrix} V1 \\ V2 \\. \\. \\. \end{pmatrix}$, then taking $U = \begin{pmatrix} V1 \\ V2 \\. \\. \\0 \\0\\0\end{pmatrix}$ would give $R*U = λ_1*U$ thus λ1,U are eigenvalue and vector of R

I do the same using D. So now I have a set of eigenvalues and vectors of R.
But, how can I tell that R has no other eigenvalues other than those of A and D and thus is diagonalizable?

 

[StoneTemplePython]

Given $R\in M_n(F)$ and two matrices $A\in M_{n1}(F)$ and $D\in M_{n2}(F)$ where $n1+n2=n$
...
But, how can I tell that R has no other eigenvalues other than those of A and D and thus is diagonalizable?

because the algebraic multiplicity is $n$ and an n degree monic (characteristic) polynomial has at most n roots. There can be no more eigenvalues than this.
- - - - -
It is a standard, though perhaps subtle, exercise to show that your problem is equivalent to solvability of the following Sylvester Equation

$\mathbf {AX} - \mathbf{XD} = \mathbf B$