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Simple matrix help

  1. Jun 22, 2013 #1
    Question: http://gyazo.com/9c75baf06947bfa9f33a1772e6e6fc03


    I need help from b onwards

    my answers to part a were)

    lambda = 2, eigenvector of [tex]\begin{pmatrix} 1\\-1\\\end{pmatrix}[/tex]

    lambda = 4, eigen vector of [tex]\begin{pmatrix} 1\\1\\\end{pmatrix}[/tex]

    then for part b) I rearranged getting P^(-1)AP = D

    I got orthogonal matrix P to be = [tex]k\begin{pmatrix} 1 & 1\\1 & -1\\\end{pmatrix}[/tex]
    where k = [itex] \frac{1}{\sqrt{2}} [/itex]

    then for D I got [tex]\begin{pmatrix} 4 & 0\\0 & 2\\\end{pmatrix}[/tex]

    so for part c) I done, a rotation of 45 degrees anticlockwise (about (0,0)), followed by a stretch of 4 parallel to x-axis, and x2 parallel to y-axis, followed by another anticlockwise rotation of 45 degrees.

    However, in the answer they put 1. Rotation of pi/4 clockwise, 2. stretch, x4 || to x-axis, x2 || to y-axis, 3. rotation of pi/4 anticlockwise (about(0,0)).

    as you can see this is different from me, and the only that I've done differently is got an eigenvector of [tex]\begin{pmatrix} 1\\-1\\\end{pmatrix}[/tex] instead of [tex]\begin{pmatrix} -1\\1\\\end{pmatrix}[/tex]

    as they did

    any ideas where I went wrong
     
  2. jcsd
  3. Jun 22, 2013 #2

    Ray Vickson

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    Why would you think you are wrong? Your are NOT wrong! Any scalar multiple of an eigenvector is an eigenvector (for the same eigenvalue), so any column vector of the form [c,-c]^T (c ≠ 0) is an eigenvector for λ = 2. You take c = +1, the book takes c = -1. You could take c = -17/sqrt(423) if you wanted to; you would still have an eigenvector.
     
  4. Jun 22, 2013 #3
    but my rotations are different! This is why I assume I'm wrong somewhere, but I can't see where :(. In their rotations, they end back up to the original point (as well as the scale enlargement), I don't end up at the original point. I go 90 degrees away from it.
     
  5. Jun 23, 2013 #4

    vela

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    Your matrix
    $$ P = \begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & -1/\sqrt{2} \end{pmatrix}$$ isn't just a rotation. It's a rotation and a reflection. If it were simply a rotation by 45 degrees, it should send (1,0) to (1/√2, 1/√2) and (0,1) to (-1/√2, 1/√2), but your P actually sends (0,1) to (1/√2,-1/√2).
     
  6. Jun 23, 2013 #5
    oh okay, so would the order be

    rotation pi/4 anticlockwise
    reflect in y=x
    enlargement of scale factor 4 and 2
    rotation pi/4 anticlockwise
    reflect in y=x?
     
  7. Jun 23, 2013 #6

    vela

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    Yup.
     
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