Diagonalizability of a matrix containing smaller diagonalizable matrices

[oferon]

Please don't mind my math english, I'm really not used to it yet..

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 - Prove that R is diagonalizable.

Ok, So what i did was 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 managed to do the same using D. So now I have a set of eigenvalues and vectors of R.
My question is - How can I tell that R has no other eigenvalues other than those of A and D, and finish my proof... Thanks!

 

[ougoah]

So you've shown that $R$ has $n$ linearly independent eigenvectors, since they come from different two matrices with no overlapping eigenvalues, and also where an eigenvalue does repeat within either $A$ or $D$, the geometric multiplicity of the eigenvalue is equal to its algebraic multiplicity (because $A$ and $D$ are both diagonalisable). Therefore, we can already write $R$ in the form $PD P^{-1}$ in the standard way; that is, $R$ is diagonalisable.

On the other hand, suppose there exists another eigenvalue of $R$ not covered by the eigenvalues of $A$ and $D$. Then there must correspond an eigenvector that is linearly independent of all of the other eigenvectors of $R$. But this means the eigenvectors of $R$ spans $n+1$ dimensions, which is not possible. So you cannot have another eigenvalue.

 

[oferon]

Thanks for your reply.
I still don't get it - I never said I found n eigenvectors. I said I found vectors for all eigen values of A and D.
How can you tell A gives total of n1 and D gives total of n2 vectors?

 

[ougoah]

If an $n\times n$ matrix is diagonalisable, then it must have $n$ linearly independent eigenvectors. If you have a repeated eigenvalue with multiplicity $k$, then the dimension of the corresponding eigenspace must also be $k$. Is it possible you haven't found all the eigenvectors corresponding to each eigenvalue?