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How to prove that if A is a diagonalizable matrix, then the rank of A

  1. Mar 15, 2010 #1
    How to prove that if A is a diagonalizable matrix, then the rank of A is the number of nonzero eigenvalues of A.
    Thanks and regard.
     
  2. jcsd
  3. Mar 15, 2010 #2

    Landau

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    Re: Proving

    Let's say D is the diagonalized form of A. Then the diagonal elements of D are precisely the eigenvalues of A. The rank of D is the number of linearly independent columns. Obviously this equals the number of non-zero eigenvalues. Since the rank of A and the rank of D are the same, the conclusion follows.
     
  4. Mar 17, 2010 #3
    Re: Proving

    If [tex]A[/tex] is similar to [tex]B[/tex] then [tex]\textrm{rk}(A)=\textrm{rk}(B)[/tex], then consider the rational canonical form, and it follows as Landau stated above.
     
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