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Math Help - Permutations Matrices Question

  1. #1
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    Permutations Matrices Question

    This is a homework problem i had. Help!

    Explain why the dot product of x and y equals the dot product of Px and Py.
    Then From (Px)^T(Py)=x^Ty deduce that P^TP=I for any permutation. With x=(1,2,3) and y=(1,4,20) choose P to show that Pxdoty is not always xdotPy.
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  2. #2
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    I dare say that you have most of us at a disadvantage.
    I for one have absolutely no idea what that question means.
    You must define the terms and explain the notation.
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  3. #3
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    I think its referring to the Identity Matrix I. and the different permutations which are P. And ^T means to find the dot product of
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  4. #4
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    P is a "permutation" matrix: every row and column contains exactly one "1" and all other entries are 0. For example,
    Px= \begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}y \\ z \\ x\end{bmatrix}
    just permutes the entries x, y, and z.

    If, instead of x= \begin{bmatrix}x \\ y \\ z\end{bmatrix} we use v= \begin{bmatrix}u \\ v \\ w\end{bmatrix} then Pv= \begin{bmatrix}v \\ w \\ u\end{bmatrix}.

    (Px)^TPv= \begin{bmatrix}y & z & x\end{bmatrix}\begin{bmatrix}v \\ w \\ u\end{bmatrix}= vw+ wz+ ux= \begin{bmatrix}x & y & z\end{bmatrix}\begin{bmatrix}u \\ v \\ w\end{bmatrix}= x^Tv
    Essentially since both x and v are permuted in the same way, the "corresponding terms" still match up.

    However,
    (Px)^Tv= \begin{bmatrix}y & z & x\end{bmatrix}\begin{bmatrix}u \\ v \\ w\end{bmatrix}= uy+ vz+ wx
    while
    x^T(Pv)= \begin{bmatrix} x & y & z\end{bmatrix}\begin{bmatrix}v \\ w \\ u\end{bmatrix}= vx+ wy+ uz
    so (Px)^Tv is not, in general, equal to x^T(Pv).
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