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Prove A is Diagonalizable (Actual Question)

  1. Mar 21, 2008 #1
    [SOLVED] Prove A is Diagonalizable (Actual Question)

    1. The problem statement, all variables and given/known data

    Suppose that A [tex]\in[/tex] M[tex]^{nxn}[/tex](F) and has two distinct eigenvalues, [tex]\lambda[/tex][tex]_{1}[/tex] and [tex]\lambda[/tex][tex]_{2}[/tex], and that dim(E(subscript [tex]\lambda[/tex][tex]_{1}[/tex] ))= n-1. Prove that A is diagonalizable.

    3. The attempt at a solution

    So far, I know that dim(E subscript [tex]\lambda[/tex]2) [tex] \geq1[/tex]
    and that
    dim(E subscript [tex]\lambda[/tex]1) + dim(E subscript [tex]\lambda[/tex]2) [tex] \leq[/tex] n.
    So dim(E subscript [tex]\lambda[/tex]2) = 1.

    I am not exactly sure how this helps me to show A is digonalizable. Maybe I am thinking of something else and don't need this to prove A is diagonalizable. Please help.

    (Also, sorry about my prevous blank post; I am new)
  2. jcsd
  3. Mar 21, 2008 #2


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    Welcome to PF!

    Hi pezola! Welcome to PF! :smile:

    Your reasoning seems fine. :smile:

    Can't you now use proof by induction - that is, assume the theorem is true for all numbers up to n - 1, and then prove it for n?
    :smile: … no problemo! … :smile:
  4. Mar 21, 2008 #3


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    A matrix is diagonalizable iff its eigenvectors form a basis for the space (do you understand why?). You've pretty much shown this, maybe just another word or too on why the eigenvectors are independent.
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