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Standard Matrix for an orthogonal projection transformation

  1. Nov 20, 2012 #1
    Let T:R^2 -> R^2 be the linear transformation that projects an R^2 vector (x,y) orthogonally onto (-2,4). Find the standard matrix for T.

    I understand how to find a standard transformation matrix, I just don't really know what it's asking for. Is the transformation just (x-2, y+4)? Any clarification would be greatly appreciated, thanks! Also, sorry for the formatting, I'm on my phone.
     
    Last edited: Nov 20, 2012
  2. jcsd
  3. Nov 21, 2012 #2

    haruspex

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    Can you define an orthogonal projection? If (-2, 4) is in the range, what would a vector of its null space look like?
     
  4. Nov 21, 2012 #3

    HallsofIvy

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    You can do this pretty much by "brute strength". Since these vectors are in R2, your matrix is 2 by 2 and looks like
    [tex]\begin{bmatrix}a & b \\ c & d\end{bmatrix}[/tex]
    and, because this is a projection onto [-2, 4], maps [-2, 4] to itself:
    [tex]\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}-2 \\ 4\end{bmatrix}= \begin{bmatrix}-2a+ 4b \\ -2c+ 4d\end{bmatrix}= \begin{bmatrix}-2 \\ 4\end{bmatrix}[/tex]
    so that -2a+ 4b= -2 and -2c+ 4p= 4. We can immediately divide by 2 to get -a+ 2b= -1 and -c+ 2d= 2.

    Also, because this is an orthogonal projection, [4, 2], perpendicular to [-2, 4], must be mapped to 0.
    [tex]\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}4 \\ 2\end{bmatrix}= \begin{bmatrix}4a+ 2b \\ 4a+ 2d\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}[/tex]
    so 4a+ 2b= 0 and 4c+ 2d= 0. We can again divide by 2 to get 2a+ b= 0 and 2c+ d= 0.

    Solve the four equations -a+ 2b= -1, -c+ 2d= 2, 2a+ b= 0, and 2c+ d= 0 for a, b, c, and d.
     
    Last edited: Nov 21, 2012
  5. Nov 21, 2012 #4
    Let's let [itex]v[/itex] be the normalised version of your vector (ie, divide your vector by its norm.) Now, if you have some vector [itex]w[/itex] and you want the projection onto [itex]v[/itex] then (as you probably know) this is just equal to [itex](v^\top w)v[/itex] (since the norm of [itex]v[/itex] is 1.) But this is equal to [itex]v(v^\top w)=(vv^\top)w[/itex]. Now, from this can you figure out how to form the projection?
     
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