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Matrix similarity transformation

  1. May 20, 2012 #1

    sharks

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    1. The problem statement, all variables and given/known data
    For a 3x3 matrix A, i know the eigenvalues and their corresponding 3 eigenvectors.

    Define a matrix P such that ##PAP^{-1}## is a diagonal matrix.

    2. Relevant equations
    Similarity transformation formula: ##D=P^{-1}AP## where D is the diagonal matrix containing the eigenvalues as its trace elements.

    3. The attempt at a solution
    The 3 eigenvectors form a 3x3 matrix P.
    The problem is that i've learned the formula to be: ##D=P^{-1}AP##
    But what is being asked here is: ##PAP^{-1}##

    I've re-arranged the matrix, ##D=P^{-1}AP##, into: ##PDP^{-1}=IAI## which gives ##PDP^{-1}=A## but i have no idea what to do next.
     
  2. jcsd
  3. May 20, 2012 #2
    Well, you can transponse left and right sides of [itex]D=P^{-1}AP[/itex], notice that [itex]D = D^T[/itex]
     
  4. May 20, 2012 #3

    sharks

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    Hi Karamata
    I don't understand what you mean.
    [tex]D^T=PAP^{-1}?[/tex]What would P be then?
     
  5. May 20, 2012 #4
    If you have ##D = P^{-1} A P##, then let ##Q = P^{-1}##. Do you see why this gives you the result you want?
     
  6. May 20, 2012 #5

    sharks

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    If ##Q = P^{-1}## then does ##Q^{-1} = P##? If yes, then ##D=QAQ^{-1}##? Is there any proof for it? I'm not convinced. In that case, what would be the matrix Q?
     
  7. May 20, 2012 #6
    Proving ##(Q^{-1})^{-1} = Q## is easy. What's the definition of the inverse of a matrix?

    Well Q would be the inverse of the matrix P! :P Seriously though, if you have ##D = P^{-1} A P## or equivalently ##PD = AP## then what must the columns of P be? Think about eigenvectors.
     
  8. May 20, 2012 #7

    sharks

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    OK, i think i understand the idea here.

    Let ##Q = P^{-1}##, then ##Q^{-1} = P## and the problem changes to ##D=Q^{-1}AQ##, which corresponds to the standard form for the diagonal, D. Here, Q is the matrix of all 3 eigenvectors, and D is the diagonal matrix with the eigenvalues as its trace elements. Now, i need to find P, where ##P=Q^{-1}##. Is there an easier way to find ##Q^{-1}## other than finding the adjoint matrix of Q and its determinant?

    Instead of using this formula to find ##Q^{-1}##: ##\frac{1}{|Q|}adj(Q)##

    I'm thinking it might be easier to derive it from: ##D=Q^{-1}AQ## but i have no idea how.
     
    Last edited: May 20, 2012
  9. May 20, 2012 #8
    Don't you mean ##D = Q A Q^{-1}##?
    Right, except I think ##Q^{-1}## would be the matrix whose columns are the eigenvectors of A. Try writing out what PD = AP tells you in terms of the columns of P:
    $$
    PD =
    \begin{pmatrix}
    | & | & | \\
    P_1 & P_2 & P_3 \\
    | & | & |
    \end{pmatrix}
    \begin{pmatrix}
    \lambda_1 & 0 & 0 \\
    0 & \lambda_2 & 0 \\
    0 & 0 & \lambda_3
    \end{pmatrix}= \cdots
    $$
    $$
    AP = A
    \begin{pmatrix}
    | & | & | \\
    P_1 & P_2 & P_3 \\
    | & | & |
    \end{pmatrix} = \cdots
    $$
     
  10. May 20, 2012 #9

    sharks

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    I think you misunderstood the problem:
    So, ##D=Q^{-1}AQ## and Q is the matrix of eigenvectors of A.
     
  11. May 20, 2012 #10
    Ah, sorry I hadn't noticed you had edited your post!

    Since you already know the eigenvalues and eigenvectors, just put them in a matrix and invert it using whatever method you prefer. You can use Cramer's rule, the equation for which you've listed below, or write the augemented matrix P|I and then row-reduce the left-hand side to the identity.

    I don't see any shortcut using the equation ##D = Q A Q^{-1}##. Certainly you know what ##Q^{-1}## must be, but I think you just have to calculate its inverse as usual.

    Edit: I think we're confusing each other about what P is and what Q is. I took P to be as in the similarity formula you listed in the OP, ##D = P^{-1} A P##, which would then make ##Q = P^{-1}## the solution you're looking for.
     
  12. May 20, 2012 #11

    sharks

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    I know. :smile: So, if i find ##Q^{-1}## then i'll know the required matrix P. According to my calculations:
    [tex]Q^{-1}= \begin{bmatrix}-1 & 0 & -2 \\ 1 & 1 & 3 \\ -1 & -1 & -2\end{bmatrix}[/tex]
    I tested the answer and got ##QQ^{-1}=I##

    Thanks for the help.
     
  13. May 20, 2012 #12
    Great! And just to make sure, you might want to calculate ##P A P^{-1}## and verify that you get the diagonal matrix D. And if it doesn't give you D, then I've succeeded in confusing you and you should see if ##P^{-1} A P## works! :tongue:
     
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