Diagonalizability of a matrix containing smaller diagonalizable matrices

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Given [itex]R\in M_n(F)[/itex] and two matrices [itex]A\in M_{n1}(F)[/itex] and [itex]D\in M_{n2}(F)[/itex] where [itex]n1+n2=n[/itex]
[itex]R = \begin{pmatrix} A & B \\ 0 & D \end{pmatrix}[/itex]
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 [itex]λ_1 [/itex] is eigenvalue of A with eigenvector [itex]V = \begin{pmatrix} V1 \\ V2 \\. \\. \\. \end{pmatrix}[/itex], then taking [itex]U = \begin{pmatrix} V1 \\ V2 \\. \\. \\0 \\0\\0\end{pmatrix}[/itex] would give [itex]R*U = λ_1*U[/itex] 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?
 
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StoneTemplePython
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Given [itex]R\in M_n(F)[/itex] and two matrices [itex]A\in M_{n1}(F)[/itex] and [itex]D\in M_{n2}(F)[/itex] where [itex]n1+n2=n[/itex]
...
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##
 

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