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Linear Algebra - Diagonalizable and Eigenvalue Proof

  1. Apr 7, 2009 #1
    1. The problem statement, all variables and given/known data

    "Let A be a diagonalizable n by n matrix. Show that if the multiplicity of an eigenvalue lambda is n, then A = lambda i"

    2. Relevant equations



    3. The attempt at a solution

    I had no idea where to start.
     
  2. jcsd
  3. Apr 7, 2009 #2
    Since [tex] A [/tex] is diagonalizable, we can choose some invertible matrix [tex] S [/tex] such that [tex] A = S D S^{-1} [/tex], where [tex] D [/tex] is diagonal and the diagonal entries of [tex] D [/tex] are the eigenvalues of [tex] A [/tex]. We can translate the assumption regarding the multiplicity of [tex] \lambda [/tex] into a statement about [tex] D [/tex], after which the result follows by using [tex] A = S D S^{-1} [/tex].
     
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