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When a matrix isn't diagonalizable

  1. Dec 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Determine if the matrix is diagonalizable or not.

    A=
    [ 3 -1 ]
    [ 1 1 ]
    2. Relevant equations
    Eigenvalues = det(A-Iλ)
    determinant of a 2x2 matrix = ad-bc

    3. The attempt at a solution
    Eigenvalues = det(A-Iλ)

    [ 3 -1 ] - [ λ 0 ] = [ 3 -λ -1 ]
    [ 1 1 ] [ 0 λ ] [ 1 1-λ ]

    det =

    (3 -λ*1-λ) -1*(-1)
    λ2-4λ+4
    (λ-2)(λ-2) = 0
    λ = 2

    Only one eigenvalue -- for this matrix A to be diagonalizable we need two eigenvalues so that

    D=
    [ λ1 0 ]
    [ 0 λ2 ]

    where λ1 and λ2 are 2 eigenvalues.

    Am I correct in my assumptions?
     
  2. jcsd
  3. Dec 7, 2015 #2

    Samy_A

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    Science Advisor
    Homework Helper

    Take the matrix $$B=\begin{pmatrix}
    1 & 0 \\
    0 & 1 \\
    \end{pmatrix}$$

    What are the eigenvalues of B?
    Is B diagonalizable?

    Not saying that your final conclusion is necessarily wrong, but what about the argumentation ?
     
  4. Dec 7, 2015 #3
    The eigenvalue of your question would be = 1.
     
  5. Dec 7, 2015 #4

    Mark44

    Staff: Mentor

    Not necessarily. Sometimes a single eigenvalue can be associated with more than one eigenvector.
     
  6. Dec 7, 2015 #5

    Samy_A

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    Yes, and surely you'll accept that B is diagonalizable.
    This shows that not having 2 eigenvalues doesn't necessarily mean that a 2*2 matrix is not diagonalizable.

    What are the eigenvectors of your matrix A?
     
  7. Dec 7, 2015 #6
    In my question I get an eigenvalue = 2

    If I plug that into (A-λI)*v1 = 0
    [ 3 -1 ] - [ 2 0 ]
    [ 1 1 ] [ 0 2 ] * v1 = 0

    [ 1 -1]
    [ 1 -1] = 0

    [1 -1] (x)
    [1 -1] (y) = 0

    x-y = 0
    x-y = 0

    iff x = 1 y = 1

    eigenvector 1 = (1,1)

    Just reading some more theory now. Because the matrix is 2x2, this means if it is to be diagonalizable it needs to have 2 linearly independent eigenvectors. I only have one.
     
  8. Dec 7, 2015 #7

    Samy_A

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    Homework Helper

    I think this is essentially correct.
    But it is so difficult to parse what you write. :frown:

    What you proved is that the eigenspace of the eigenvalue 2 has dimension 1. And it should have been 2 for the matrix to be diagonalizable.
     
  9. Dec 7, 2015 #8

    Mark44

    Staff: Mentor

    Yes, that's correct. Since you have only one eigenvector, but need two of them, your matrix is not diagonalizable.
     
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