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Direct sum of the eigenspaces equals V?

  1. Aug 6, 2013 #1
    I am following Friedberg's text and having some trouble understanding some of the theorems regarding diagonalizability. The proofs seem to skip some steps, so I guess I need to work through them a bit more slowly.

    Given a linear operator ## T:V → V ##, with eigenspaces ## \{ E_{ \lambda_{1}},E_{ \lambda_{2}},...E_{ \lambda_{k}} \} ##, is it true that ## E_{ \lambda_{1}} \bigoplus E_{ \lambda_{2}} \bigoplus E_{ \lambda_{2}} ... \bigoplus E_{ \lambda_{k}} = V ## ?

    What if T is diagonalizable? This is just a thought which may perhaps help me to understand diagonalizability a bit better. I simply need to know whether this is right or not. Please no explanations, as I will prove it myself. Thanks!

    BiP
     
  2. jcsd
  3. Aug 6, 2013 #2

    mathman

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    Assuming all eigenvalues are distinct (V is k dimensional), it is correct.

    Things get slightly more complicated if an eigenvalue has multiplicity. The eigenvectors for such an eigenvalue define a multidimensional space (dimension = multiplicity of eigenvalue).
     
  4. Aug 7, 2013 #3
    Suppose the eigenvalues are distinct, but dim(V) > k so that the eigenvalues all have multiplicity. Then what?

    BiP
     
  5. Aug 7, 2013 #4

    HallsofIvy

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    If the n by n matrix, V, is diagonalizable then we have n independent eigenvectors. If a given eigenvalue has multiplicity i then we can find i independent vectors that span its "eigenspace". The whole point of "diagonal" is that there exist a basis such that all vectors in the basis are eigenvectors.
     
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