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  1. #1
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    Matrix

    Find an orthogonal matrix P and a diagonal matrix D such that

    A = PDP^(-1).

    I found P but how do I find D? Is it correct to do D = PAP^(-1)? Thanks in advance.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Air View Post
    Find an orthogonal matrix P and a diagonal matrix D such that

    A = PDP^(-1).

    I found P but how do I find D? Is it correct to do D = PAP^(-1)? Thanks in advance.
    yes, you can do that.. but i think, it is D = P^{-1}AP

    oh well, as long as they are inverses..
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  3. #3
    Junior Member rednest's Avatar
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    Quote Originally Posted by Air View Post
    Find an orthogonal matrix P and a diagonal matrix D such that

    A = PDP^(-1).

    I found P but how do I find D? Is it correct to do D = PAP^(-1)? Thanks in advance.
    No you cannot do that I think because matrices are not commutative.
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  4. #4
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    Quote Originally Posted by rednest View Post
    No you cannot do that I think because matrices are not commutative.

    So, how do I work out D?
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  5. #5
    MHF Contributor kalagota's Avatar
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    how do you get the inverse of P?
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  6. #6
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    Quote Originally Posted by kalagota View Post
    how do you get the inverse of P?
    It's just Transpose.
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  7. #7
    MHF Contributor kalagota's Avatar
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    yeah.. so apply the corrected equation i gave..
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  8. #8
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    Do you know how to find a matrix P such that P^{-1}AP = D ?

    If your familiar with that all you need to do is some basic matrix algebra

    first pre multiply both sides by P , PP^{-1}AP = PD \ \ \Rightarrow IAP = PD

    then post multiply by P^{-1} so  IAPP^{-1} = PDP^{-1}

    so IAI = PDP^{-1}

    given A =  PDP^{-1}

    Bobak
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  9. #9
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    Quote Originally Posted by Air View Post
    Find an orthogonal matrix P and a diagonal matrix D such that

    A = PDP^(-1).

    I found P but how do I find D? Is it correct to do D = PAP^(-1)? Thanks in advance.
    Although you could do D = P^{-1}AP, there is a much easier way:

    In the process of finding P, you probably found the eigenvalues and eigenvectors of A. We know that if the eigenvalues of A are \lambda_1,\;\lambda_2,\;\dots,\;\lambda_n, and each eigenvalue \lambda_i corresponds to an eigenvector \textbf{p}_i such that \textbf{p}_1,\;\textbf{p}_2,\;\dots,\;\textbf{p}_n are linearly independent, then the matrix P = \left[\begin{array}{c|c|c|c}\textbf{p}_1 & \textbf{p}_2 & \cdots & \textbf{p}_n\end{array}\right] will diagonalize A, and A = PDP^{-1}. Now, it turns out that

    D = \left[\begin{matrix}<br />
\lambda_1 &       0       &        0      & \cdots & 0\\<br />
      0       & \lambda_2 &        0      & \cdots & 0\\<br />
      0       &       0       & \lambda_3 & \cdots & 0\\<br />
   \vdots   &   \vdots    &     \vdots  & \ddots & \vdots\\<br />
      0       &       0       &        0      & \cdots & \lambda_n<br />
\end{matrix}\right]

    Therefore, if you diagonalized A with P to get A = PDP^{-1}, then D will be a diagonal matrix whose diagonal entries are the eigenvalues of A, listed in the same order as you placed the eigenvectors of A into P.

    Let me give a brief example:

    Let A = \left[\begin{matrix}<br />
1 & -1 & -1 \\<br />
1 &  3 &  1 \\<br />
-3 & 1 & -1<br />
\end{matrix}\right]

    We may easily determine the eigenvalues of A to be \lambda_1 = 2,\;\lambda_2 = -2,\;\lambda_3 = 3, and their corresponding eigenvectors:

    \begin{array}{lcrl}<br />
\lambda_1 & = & 2: & \left[\begin{matrix}-1\\0\\1\end{matrix}\right]\medskip\\<br />
\lambda_2 & = & -2: & \left[\begin{matrix}1\\-1\\4\end{matrix}\right]\medskip\\<br />
\lambda_3 & = & 3: & \left[\begin{matrix}-1\\1\\1\end{matrix}\right]<br />
\end{array}

    Putting these eigenvectors into the columns of P, we get

    P = \left[\begin{matrix}<br />
-1 & 1 & -1\\<br />
 0 &-1 & 1\\<br />
 1 & 4 & 1<br />
\end{matrix}\right]

    so A = PDP^{-1} for some diagonal D. And if you find D, you will find that it is

    D = \left[\begin{matrix}<br />
 2 & 0 & 0\\<br />
 0 &-2 & 0\\<br />
 0 & 0 & 3<br />
\end{matrix}\right]

    (try verifying this) and the entries along the main diagonal correspond exactly to the eigenvalues we found for A.

    But, if you were to change the order of the eigenvectors in P, the entries in D will also be in different order.

    For example, if we had set

    P = \left[\begin{matrix}<br />
1 & -1 & -1\\<br />
-1 & 0 & 1\\<br />
4 &  1 & 1<br />
\end{matrix}\right],

    then we would get

    D = \left[\begin{matrix}<br />
 -2 & 0 & 0\\<br />
 0 & 2 & 0\\<br />
 0 & 0 & 3<br />
\end{matrix}\right]
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