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Theorectical diagonalazation question

  1. Mar 2, 2009 #1
    there is
    [tex]
    A\epsilon M_{2x2}(Q)
    [/tex]
    I am given that A is diagonazable
    prove that
    A)
    [tex]
    A^{10}+12A
    [/tex]
    is diagonizable too

    B)give an example for a matrix[tex] B\epsilon M_{2x2}(Q)[/tex]
    that is not diagonizable,but b^2 is diagonisable
    ??

    i know that the eigenvalues of a matrix are the same as for every matrix
    like A^10 or A^3+2A+3I etc..

    but i dont now how to show what they ask
    ??
     
    Last edited: Mar 3, 2009
  2. jcsd
  3. Mar 2, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What do you mean by [itex]A^{10}+ 12A[/itex]?
     
  4. Mar 3, 2009 #3
    i mean operator of A n the power 10 plus 12 A
    that matrix is diagonizable too
    ?
     
  5. Mar 3, 2009 #4
    A) If A is diagonalizable, then [tex]D=T^{-1}AT[/tex], where D is diagonal. Now, note that sums and powers of diagonalizable matrices are also diagonal.

    B) Hint: a certain rotation in the 2-D plane...
    To show that a matrix is not diagonalizable over Q, it is sufficient to show that the characteristic polynomial has no rational roots.
     
  6. Mar 3, 2009 #5
    regarding A:
    i tried like this:
    A=p^-1*D*P
    P(A^10+12A)P^-1=P*A^10*P^-1 +12 P*A*P^-1
    next..
    why
    P*A^10*P^-1=(P*A*P^-1)^10
    ??

    regarding b:
    i took matrix which polinomial is b=x^2+x+1 =[1,1,1,0]

    b^2=[2,1,1,1]=x^3+x^2+x+2
    which has only one eigen value of -2
    how to deside ith square polinomial that i got is diagonizable??
     
  7. Mar 3, 2009 #6

    Office_Shredder

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    Gold Member

    If you only get one eigenvalue, you need to check if there are two linearly independent eigenvectors
     
  8. Mar 3, 2009 #7
    actually -2 is not an eigen value
    so it crashes the theory proposed by YYAT
     
  9. Mar 3, 2009 #8

    Mark44

    Staff: Mentor

    You can't possibly get a 3rd degree characteristic polynomial from a 2x2 matrix. Furthermore, a matrix is not equal to its characteristic polynomial, as you show above.
     
  10. Mar 3, 2009 #9
    "yes i can"

    we got 4 coordinated each one represents a member of a polinomial
    so there is no much room for desitions
    {1,x,x^2,x^3}
     
  11. Mar 3, 2009 #10

    Mark44

    Staff: Mentor

    Then it is apparent that you don't know how to calculate the characteristic polynomial of a matrix. The characteristic polynomial of an n x n matrix A comes from the equation
    [itex]det(A - \lambda I) = 0[/itex].
     
  12. Mar 3, 2009 #11
    i know this thing 100%
    give me a way to find a polinomial which doesnt have eigen values
    but its square does have
    ??
     
  13. Mar 3, 2009 #12
    regarding A:
    i tried like this:
    A=p^-1*D*P
    P(A^10+12A)P^-1=P*A^10*P^-1 +12 P*A*P^-1
    next..
    why
    P*A^10*P^-1=(P*A*P^-1)^10
    ??
     
  14. Mar 3, 2009 #13

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You mean "matrix" that doesn't have eigenvalues, not "polynomial". Every square matrix has eigenvalues over the complex numbers so I assume here you are talking about real numbers. Try
    [tex]\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}[/tex]
     
  15. Mar 3, 2009 #14
    [tex](PAP^{-1})^2=PAP^{-1}PAP^{-1}=PA^2P^{-1}[/tex]

    since [tex]P^{-1}P=I[/tex]. I think you can see the general pattern.
     
  16. Mar 3, 2009 #15

    the polinomial that comes from it is
    x-x^2
    it has an eigen values
    x(1-x) x=0 x=1
    i need a matrix which polinomial doesnt have eigen values but the square of this matrix has
    ??
     
  17. Mar 3, 2009 #16
    The characteristic polynomial is

    det(A-xI)=x^2+1

    so the matrix has no real eigenvalues.
     
  18. Mar 3, 2009 #17
    how to find the matrix of
    x^2+x+3

    ??
     
  19. Mar 3, 2009 #18

    Mark44

    Staff: Mentor

    I'm just being curious, but why is this important to find? I don't see anything in this thread that seems related to this particular polynomial.
     
  20. Mar 4, 2009 #19
    i dont know how to think of a matrix which polynomial doesnt have eigenvalues.
    on the other hand i can think of such polynomial but how to get its matrix.
    ??
     
  21. Mar 4, 2009 #20

    Mark44

    Staff: Mentor

    Well, you're dealing with 2 x 2 matrices, with entries [a b; c d] (reading row by row). Maybe you can fiddle with these values to make (a - lambda)(d - lambda) - bc come out the way you want. The preceding quadratic in lambda is det(A - lambda*I).
     
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