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Diagonalizability of a matrix containing smaller diagonalizable matrices

  1. Jun 26, 2012 #1
    Please don't mind my math english, I'm really not used to it yet..

    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 - 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 [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 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!
     
  2. jcsd
  3. Jun 26, 2012 #2
    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.
     
  4. Jun 27, 2012 #3
    Thanks for your reply.
    I still dont 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?
     
  5. Jun 28, 2012 #4
    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?
     
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