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Math Help - Orthogonal Matrix

  1. #1
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    Orthogonal Matrix

    Can anyone help me out with these questions:

    Given: Matrix(P)=
    0-4-1
    0 3 7
    1 0 0





    Such That: P-1AP=D is diagonal arranged so that the (1,1)-entry of D is different from the other diagonal entries.

    Find an orthogonal matrix Q so that transpose(Q)AQ=D.
    ================================================== ================================================== ===========


    2)


    Given: matrix(A)=


    -14/25 48/25 0
    48/25 14/25 0
    0 0 0

    Find a diagonal matrix D and an orthogonal matrix Q so that the given matrix A = Q D transpose(Q).

    Plz Help
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  2. #2
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    Quote Originally Posted by yakuut View Post
    Can anyone help me out with these questions:

    Given: Matrix(P)=
    0-4-1
    0 3 7
    1 0 0

    Such That: P-1AP=D is diagonal arranged so that the (1,1)-entry of D is different from the other diagonal entries.

    Find an orthogonal matrix Q so that transpose(Q)AQ=D.
    The question says that "(1,1)-entry of D is different from the other diagonal entries", but I think that in order to answer the question you need to know more than that, namely that the other two diagonal entries of D are the same, so that D is of the form \begin{bmatrix}\lambda&0&0\\ 0&\mu&0\\ 0&0&\mu\end{bmatrix}. In the absence of any information at all about the matrix A, you will surely need to make that assumption about D.

    If P^{-1}AP = D then A = PDP^{-1}. The idea is to write P in the form P=QR, where Q is orthogonal and R is a positive definite matrix that commutes with D. Then A = PDP^{-1} = QRDR^{-1}Q^{\textsc{t}} = QDQ^{\textsc{t}}, from which Q^{\textsc{t}}AQ = D.

    The factorisation P = QR is called the polar decomposition of P. The positive definite factor R is the square root of P^{\textsc{t}}P. So start by calculating P^{\textsc{t}}P = \begin{bmatrix}1&0&0\\ 0&25&25\\ 0&25&50\end{bmatrix}. This has square root R = \begin{bmatrix}1&0&0\\ 0&0&-5\\ 0&-5&-5\end{bmatrix}. (Fortunately, this commutes with D, as you can easily check.) Then Q = PR^{-1} = \begin{bmatrix}0&-\tfrac35&\tfrac45\\ 0&-\tfrac45&-\tfrac35\\ 1&0&0\end{bmatrix}.

    Quote Originally Posted by yakuut View Post
    2)

    Given: matrix(A)=

    -14/25 48/25 0
    48/25 14/25 0
    0 0 0

    Find a diagonal matrix D and an orthogonal matrix Q so that the given matrix A = Q D transpose(Q).
    This is much easier. The diagonal entries of D are the eigenvalues of A, and the columns of Q are the associated normalised eigenvectors.
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  3. #3
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    Thanks Opalg for ur efforts..but the solution for the first question didn't work
    thanks anyways...
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