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Is the matrix Diagonalizable

  1. Jun 21, 2016 #1
    1. The problem statement, all variables and given/known data
    Let A be a 3x3 singular Matrix that satisfy:
    p(A+5I) < p(A)
    p - is the rank of the matrix
    I - is the identity matrix,
    Is A Diagonalizable?
    2. Relevant equations


    3. The attempt at a solution
    I know that A diagonalizable matrix can be Singular from every rank, even at 0 rank, so i can't see how i can conclude anything.
     
    Last edited: Jun 21, 2016
  2. jcsd
  3. Jun 21, 2016 #2

    fresh_42

    Staff: Mentor

    You can exclude many cases by the given assumptions. What does it mean for A to be singular?
    Can A or A+5I equal the zero matrix?
     
    Last edited: Jun 21, 2016
  4. Jun 21, 2016 #3
    singular means it isn't invertible or the columns are linearly independent or the rank is lower than 3.
    i think A + 5I cannot be the zero matrix, but how's that even related to the Diagnolization of A, since any rank of matrix can be diagonliazable.
     
  5. Jun 21, 2016 #4

    fresh_42

    Staff: Mentor

    There is only one possibility for the ranks of A and A + 5I. Assume that A is diagonalizable. What does it mean for the eigenvalues? This gives you a criterium.
     
  6. Jun 21, 2016 #5
    You mean the eigenvectors? they are independent. if not i don't know, i know that if A is diagonalizable it means that the amount of the same eigenvalues should be equal to the dimension of the eigenvectors space that is related to the current eigenvalue.

    also the rank of A<3
    and i can't see how to prove that A+5I > 0 it does look reasonable to me.
    So A+5I > 0 ==> A+5I = > 1
    and then rank of A = 2.
    How is that helps me determine if A Diagonalizable
     
  7. Jun 22, 2016 #6
    Still need help
     
  8. Jun 22, 2016 #7

    fresh_42

    Staff: Mentor

    I meant the eigenvalues.
    ##p(A)=2## and ##p(A+5I)=1## are the only possibilities for the ranks, since ##p(A) < 3## and neither matrix can be zero which is equivalent to rank zero. (If ##A+5I = 0## then ##A = -5I## of rank ##3##. If ##A=0## then there is no way for another rank below that.)

    Now assume ##A## is diagonalizable, i.e. there is a matrix ##S## with ##SAS^{-1} = diag(α,β,γ)##.
    Then ##S(A+5I)S^{-1} = SAS^{-1} + 5I = diag(α+5,β+5,γ+5)## and without loss of generality ##γ=0## because ##p(A)=2.##
    Now we still have ##p(A+5I)=1## which is only possible for ##α = β = -5##.
     
  9. Jun 22, 2016 #8
    So if the rank of the matix nxn is k , then the amount of 0 eigenvalues is n-k ? .
    and also if i get two eigenvalues that are 0, how can i know if the matrix is diagonal or not ?
     
  10. Jun 22, 2016 #9

    Ray Vickson

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    Science Advisor
    Homework Helper

    Why do you say p(A) < 3? All you were told is that p(A+5I) < p(A), and so you can have p(A) = 3, p(A+5I) = 1 or 2.
     
  11. Jun 22, 2016 #10
    Because A is singular, that means it has linearly independent columns, and that means the rank must be less than 3.
     
  12. Jun 22, 2016 #11
    I think i've got it, the rank of P'AP = Rank of A .
    Then Rank of A = Rank of it's Diagonal matrix . therefore the rank - n is equal to the number of 0 on the diagonal.
    But, still haven't figured out if it's Diagonalizable or not.
     
  13. Jun 22, 2016 #12

    fresh_42

    Staff: Mentor

    Amount isn't a good word here. In case of an eigenvalue ##\lambda## we speak of geometric multiplicity (dimension of the eigenspace of ##\lambda##) and algebraic multiplicity (power of ##t-\lambda## in the characteristic polynomial ##det (A-tI) = 0##) . If ##A## is diagonalizable then they are equal.
    In general (in the situation asked) ##n = p(A) + \dim \ker(A) = k + (n-k)## and ## \ker(A) = \{x \in V | Ax=0=0 \cdot x\}## the eigenspace of eigenvalue ##0##.

    You can't. It heavily depends on the scalar domain that make up your coordinates.
     
  14. Jun 22, 2016 #13
    Or maybe A is diagonalizable since the rank of A is 2, and ker of A=1, and it does have only one eigenvalue = 0 , and so the dim of the eigenspace of 0 is 1 .

    How can i see the the dim of the Eigenspace of -5 is 2 ?
     
    Last edited: Jun 22, 2016
  15. Jun 22, 2016 #14

    fresh_42

    Staff: Mentor

    If ##A = diag(-5,-5,0)## then ##A## is diagonal and all conditions are met. We only derived that there can't be other eigenvalues.
     
  16. Jun 22, 2016 #15
    Thanks, think i got it.
     
    Last edited: Jun 22, 2016
  17. Jun 23, 2016 #16
    I wondered one more thing, if i could show that the dim of the eigenspace of -5 is 2 ? or do i must have values of the matrix for that ?
     
  18. Jun 23, 2016 #17

    Ray Vickson

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    If
    [tex] A = \pmatrix{-5 & 1 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 0} [/tex]
    then all the conditions are met and ##A## cannot be diagonalized.
     
  19. Jun 23, 2016 #18
    You mean can be ? if not :
    I = Identity matrix
    I' = Inverse
    I'AI = A
    which means A is diagonlaziable. We just found matrix I that Diagonalize A.
     
    Last edited: Jun 23, 2016
  20. Jun 23, 2016 #19

    Ray Vickson

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    No, it was deliberately chosen to be NON-diagonalizable. Look again: the element ##a_{12} \neq 0##. The matrix ##A## is a 3x3 Jordan canonical form for eigenvalues -5,-5,0, with a non-diagonal Jordan block in the top left 2x2 location.

    The eigenvalue -5 has algebraic multiplicity 2 but geometric multiplicity 1; that means that you cannot find three linearly-independent eigenvectors, only two. You need to include a so-called "generalized" eigenvector in order to have a full 3-dimensional basis. That is why the matrix cannot be diagonalized.
     
  21. Jun 23, 2016 #20
    Where did that a12 =1 came from ?
    and how do you know that the Geometric multiplicity of -5 is 1 ?
     
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