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If (Q^-1)AQ=D, then columns of Q are Eigenvectors of A

  1. Oct 29, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove: if (Q^-1)AQ=D, then each column of Q is an eigenvector of A.

    2. Relevant equations

    A vector v is an eigenvector of A iff there exists a scalar λ such that:

    Av=λv

    3. The attempt at a solution

    Suppose (Q^-1)AQ=D. We need to show each column of Q is an eigenvector of A.

    At this point, should I actually write out (Q^-1)AQ=D in general element form? That is, should I write out (Q^-1),A,Q,D in the form of nxn matrices?

    I'm honestly not even sure how to begin the proof, but any help would be appreciated.

    Thanks!
     
  2. jcsd
  3. Oct 29, 2011 #2
    Well, since Q is invertible, we have that the columns of Q are linearly independent.

    Since Q is nxn, we have a linearly independent list of n vectors in an n-dimension subspace, therefore we have a basis for the subspace.

    Does that help to get you started?
     
  4. Oct 29, 2011 #3
    AQ = QD. The other (maybe better) way to see it is to think about change of basis.
     
  5. Oct 29, 2011 #4

    I like Serena

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    Welcome to PF, blashmet! :smile:

    As deluks917 said: AQ=QD

    Now write Q as (q1 q2 ...) where q1, q2 are the column vectors.
    And write D out as a diagonal matrix with lambda1, lambda2, etcetera.
    Leave A as it is.

    Now write out the matrix multiplication...
     
  6. Oct 29, 2011 #5
    Thanks for the help guys! This is what I have so far:


    Q(Q^-1)AQ=QD iff AQ=QD=diag[d1,...,dn]

    AQ=A[q1,...,qn]=[Aq1,...,Aqn]

    QD=[d1q1,d2q2,...,dnqn]

    so AQ=QD iff [Aq1,...,Aqn]=[d1q1,...,dnqn]


    I'm stuck at this point. How do I proceed from here?

    Thanks! :)
     
  7. Oct 29, 2011 #6

    I like Serena

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    Good! :wink:

    You have to prove that each column of Q is an eigenvector of A.

    q1 is the first column of Q.
    Does it satisfy the criteria for an eigenvector of A?
     
  8. Oct 29, 2011 #7
    Hi Serena!

    In a formal proof of this, do I need to have all of this first?:

    Q(Q^-1)AQ=QD iff AQ=QD=diag[d1,...,dn]

    AQ=A[q1,...,qn]=[Aq1,...,Aqn]

    QD=[d1q1,d2q2,...,dnqn]

    so AQ=QD iff [Aq1,...,Aqn]=[d1q1,...,dnqn]


    Now, is there where I show q1 is an eigenvector for A? If so, I'm confused, because the last line of the proof has [d1q1,...,dnqn]. Don't I need to show d1q1 is an eigenvector?

    Thanks for your help! :smile:
     
  9. Oct 29, 2011 #8
    What is the first column of AQ? What is the first coulumn of QD?
     
  10. Oct 29, 2011 #9
    Hi deluks917,

    i'm not sure how you'd write it, but i know that AQ and QD are nxn square, so it's probably just a shorthand notation for the general element form of the multiplication of 2 nxn square matrices.



    can you answer my other questions? (it's ok if you don't know, i just wasn't sure if you saw them :))
     
  11. Oct 29, 2011 #10
    I'm not sure why you are confused. You wrote AQ = QD implies:

    [Aq1,...,Aqn]=[d1q1,...,dnqn]

    or
    Aq1 = d1q1
    ...
    Aqn = dqn

    Do you see the eigenvectors?
     
  12. Oct 30, 2011 #11
    It looks like the eigenvalues are dn (the diagonal components of the matrix D).

    So I suppose the eigenvectors would be qn (column vectors).

    I guess my problem is this:

    I need to write out a formal proof from beginning to end, and I'm not sure what to add to what I've already written (or if I need to do more than this).

    Can you help me with that? :)
     
  13. Oct 30, 2011 #12

    I like Serena

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    You are done.

    d1...dn are the eigenvalues
    and q1...qn are the eigenvectors,
    since they satisfy Av=λv.
    So each column of Q is an eigenvector of A.
     
  14. Oct 30, 2011 #13
    Does the following suffice as a complete, formal proof?

    "d1...dn are the eigenvalues
    and q1...qn are the eigenvectors,
    since they satisfy Av=λv.
    So each column of Q is an eigenvector of A."


    Thanks Serena! :)
     
  15. Oct 30, 2011 #14

    I like Serena

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  16. Oct 30, 2011 #15
    Ok great! Thanks Serena!
     
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