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What will H(^T)*H be?

  1. Oct 12, 2012 #1
    Problem:

    If P is a n x 1 matrix such that P( ^T)*P = 1 and H = I - 2PP(^T), then H(^T)*H is ?

    Note :

    A(^T) means the transpose of matrix A.

    Confusion:

    What I don't get is are we supposed to do I (identity matrix-I'm assuming of order 1 x 1) - 2 ? If that is the way we're supposed to do the question then are we supposed to treat 2 as a 1 x 1 matrix. If that's the case then is the following correct:

    Since P( ^t)*P = PP(^T) = 1 = [1]

    So,

    H = I - 2PP(^T) = I - 2PP(^T) = I - 2[1] = [1] - [2] =[-1]

    thus,

    H(^T)*H = [-1] * [-1](^T) = [-1] * [-1] = [1]


    Help?
     
  2. jcsd
  3. Oct 12, 2012 #2

    Simon Bridge

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    $$\renewcommand{\vb}[1]{\mathbf{#1}}
    \renewcommand{\hb}[1]{\hat{\mathbf{#1}}}$$

    If P is a n x 1 matrix such that P( ^T)*P = 1 and H = I - 2PP(^T), then H(^T)*H is ?

    If P is nx1, then it is a vector p.
    so, switching to LaTeX, we have ##\vb{p}## so that ##\vb{p}^t\vb{p}=1## so it is a unit vector?

    ##\vb{H}=\vb{I}-2\vb{p}\vb{p}^t##

    Then ##\vb{H}^t\vb{H}=?##

    So does this translate to 1x1 matrices ... then
    ##p_{(1\times 1)}=1##, and ##p^tp=1\times 1=1##, and
    ##H_{(1\times 1)}=I-2p^tp = 1-2=-1## so
    ##H^tH = (-1)(-1)=1=I_{(1\times 1)}##
    ... seems reasonable to me so far.
    The trick is to articulate your confusion.
    The above is consistent with the matrix formulation if ##\vb{H}^t\vb{H}=\vb{I}## for any ##\vb{p}##.

    So can you find a ##\vb{p}## so that:
    ##\vb{H}^t\vb{H}\neq\vb{I}##

    ... but I suspect the central question is whether it is (or "when is it...") sensible to treat a scalar as a 1x1 matrix.

    For instance - you can multiply a nxn matrix by a scalar - but can yout multiply an nxn matrix by a 1x1 matrix?
     
  4. Oct 12, 2012 #3
    ##PP^T## is an n-by-n matrix. (prove: ##P## is n-by-1 and ##P^T## is 1-by-n).

    So ##PP^T\neq P^TP## unless if n = 1.

    ##H## is symmetric.

    ##H^TH=I_{n}## which has dimension n-by-n.
     
  5. Oct 12, 2012 #4
    The linear operator you have here is equivalent to

    [tex]\underline H(a)=a-2(p \cdot a)p[/tex]

    This is a reflection operator, and since it is symmetric, all you have to do is think about what you get when you reflect twice.
     
  6. Oct 12, 2012 #5
    Reflecting twice off the same "surface" gets you the original vector back.
     
  7. Oct 12, 2012 #6

    AlephZero

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    @Simon Bridge, I think you misread the question or made typo part way through your answer. if should be ##H = I - 2pp^t## not ##H = I - 2 p^t p##.

    @the OP, as post #3 said, ##pp^T## is symmetric, so ##H## is symmetric. So ##H^T = H##.

    To answer the question, you just have to multiply out ##H^T H## = ##H.H## and simplify it using ##P^T P = 1##.

    2P means a scalar (2) times a matrix (P). The matrix I has the same dimension as ##P P^T##, which is n by n.
     
  8. Oct 12, 2012 #7
    Thanks.
     
  9. Oct 13, 2012 #8

    Simon Bridge

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    <sigh> typo - thanks Aleph.
     
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