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Diagonalization & Eigen vectors proofs

  1. Jan 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Question 1:
    A) Show that if A is diagonalizable then [tex]A^{T}[/tex] is also diagonalizable.

    3. The attempt at a solution
    We know that [tex]A[/tex] is diagonalizable if it's similar to a diagonal matrix.
    So
    [tex]A[/tex]=[tex]PDP^{-1}[/tex]
    [tex]A^{T}[/tex]=[tex](PDP^{-1})^{T}[/tex]
    which gives
    [tex]A^{T}[/tex]=[tex](P^{-1})^{T}DP^{T}[/tex] as [tex]D=D^{T}[/tex]
    Hence [tex]A^{T}[/tex] is diagonalizable

    1. The problem statement, all variables and given/known data
    Question 2
    If A and B are Similar matrices, then show that [tex]A^{2}[/tex] and [tex]B^{2}[/tex]
    are similar

    3. The attempt at a solution
    If A and B are similar then [tex]P^{-1}AP[/tex] = [tex]B[/tex]

    We know that [tex]P^{-1}A^{k}P[/tex] =[tex]D^{k}[/tex]
    let k=2 therefore
    [tex]P^{-1}A^{2}P[/tex] =[tex]B^{2}[/tex]
    hence [tex]A^{2}[/tex] and [tex]B^{2}[/tex] are similar



    1. The problem statement, all variables and given/known data
    Question 3
    Every matrix A is Similar itself

    3. The attempt at a solution
    If A and A are similar then [tex]P^{-1}AP[/tex] =[tex]A[/tex] ? this does not make sense to me.
    Alternatively, do we have to show that A has the same eigenvalues as A? This is obvious, is this then the proof?
     
  2. jcsd
  3. Jan 10, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Looks good.

    You only know that if P is diagonallizable. That is not assumed in this problem.
    Looks like "overkill" to me. If you are asked only about A2 and B2, why use a fact about the kth power? If P-1AP= B, then
    B2= (P-1AP)(P-1AP)= (P-1A)(PP-1)(AP).



    How about just taking P= I?
     
    Last edited: Jan 10, 2008
  4. Jan 14, 2008 #3
    Thanks for the reply, I see where I went wrong.
    I tried to use the method in question 2 and extend it to prove that :
    IF A and B are similar matrices then [tex]A^{k}[/tex] and [tex]B^{k}[/tex] are similar for any non negative integer k.

    This is what I got:
    [tex]B^{k}[/tex]=[tex](P^{-1}AP)[/tex] [tex](P^{-1}AP)[/tex] ......[tex](P^{-1}AP)[/tex] (k times)
    then Multiply the right hand side 2 elements at a time as u did we will end up with [tex]P^{-1}A^{k}P[/tex].
    Is This the correct way to proove it?
     
  5. Jan 14, 2008 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, that works nicely.
     
  6. Jan 14, 2008 #5


    Not quite complete yet. Just write down that transpose of inverse is inverse of transpose.
     
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