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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 $\displaystyle D = P^{-1}AP$

    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 $\displaystyle P$?
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  6. #6
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    Quote Originally Posted by kalagota View Post
    how do you get the inverse of $\displaystyle P$?
    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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    Do you know how to find a matrix P such that $\displaystyle 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 , $\displaystyle PP^{-1}AP = PD \ \ \Rightarrow IAP = PD $

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

    so $\displaystyle IAI = PDP^{-1} $

    given $\displaystyle A = PDP^{-1} $

    Bobak
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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 $\displaystyle D = P^{-1}AP$, there is a much easier way:

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

    $\displaystyle D = \left[\begin{matrix}
    \lambda_1 & 0 & 0 & \cdots & 0\\
    0 & \lambda_2 & 0 & \cdots & 0\\
    0 & 0 & \lambda_3 & \cdots & 0\\
    \vdots & \vdots & \vdots & \ddots & \vdots\\
    0 & 0 & 0 & \cdots & \lambda_n
    \end{matrix}\right]$

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

    Let me give a brief example:

    Let $\displaystyle A = \left[\begin{matrix}
    1 & -1 & -1 \\
    1 & 3 & 1 \\
    -3 & 1 & -1
    \end{matrix}\right]$

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

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

    Putting these eigenvectors into the columns of $\displaystyle P$, we get

    $\displaystyle P = \left[\begin{matrix}
    -1 & 1 & -1\\
    0 &-1 & 1\\
    1 & 4 & 1
    \end{matrix}\right]$

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

    $\displaystyle D = \left[\begin{matrix}
    2 & 0 & 0\\
    0 &-2 & 0\\
    0 & 0 & 3
    \end{matrix}\right]$

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

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

    For example, if we had set

    $\displaystyle P = \left[\begin{matrix}
    1 & -1 & -1\\
    -1 & 0 & 1\\
    4 & 1 & 1
    \end{matrix}\right]$,

    then we would get

    $\displaystyle D = \left[\begin{matrix}
    -2 & 0 & 0\\
    0 & 2 & 0\\
    0 & 0 & 3
    \end{matrix}\right]$
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