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Linear Algebra/Standard matrix of a projection onto a plane

  1. Oct 21, 2011 #1
    1. The problem statement, all variables and given/known data

    Let u=(-1,-2,-2,2) and v=(-1,-2,-2,-1) and let V=span{u,v}. (Just to be clear, u and v are column vectors)

    Find the standard matrix that projects points (orthogonally) onto V.


    2. Relevant equations



    3. The attempt at a solution

    I started by making a matrix A=[u,v], which represents V. From there, however, I don't know where to go.
     
  2. jcsd
  3. Oct 22, 2011 #2

    HallsofIvy

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    I have no idea what you mean by "matrix A= [u,v]" nor in what sense it "represents V". You can't make a 4 by 4 matrix from 2 vectors and a matrix does not "represent" a subspace.

    The projection maps any vector in the plane to itself and any vector perpendicular to the plane to the 0 vector.

    So you want a matrix that maps (-1, -2, 2, 2) to itself and maps (-1, -2, -2, -1) to itself. One vector perpendicular to both of those is (-6, 1, 0. 4) and another, independent, vector perpendicular to both is (-6, 0, 1, 4). You want a matrix that maps that two vectors to 0.

    There are several ways of determining such a matrix. The most basic, and probably hardest, would be to write
    [tex]\begin{bmatrix}a_{11} & a_{21} & a_{31} & a_{41} \\ a_{12} & a_{22} & a_{33} & a_{42} \\ a_{13} & a_{23} & a_{33} & a_{43} \\ a_{14} & a_{24} & a_{34} & a_{44}\end{bmatrix}[/tex]
    Multiply that by each of the four vectors, setting them equal to what I said above. That will give 16 equations for the 16 entries in the matrix.

    A more sophisticated, and hopefully simpler to calculate, method would be this: if we were to use the four vectors as basis vectors, the matrix would be very simple:
    [tex]B= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}[/tex]
    since that maps the first two basis vectors to themselves and the last two to 0.

    To get the form of the matrix in the standard basis, use the fact that the matrix constructed using the four vectors as columns,
    [tex]P= \begin{bmatrix}1 & -1 & -6 & -6 \\ -2 & -2 & 1 & 0\\ 2 & -2 & 0 & 1 \\ 2 & -2 & 4 & 4\end{bmatrix}[/tex]
    maps the standard basis into those four vector and so its inverse, [itex]P^{-1}[/itex] maps the four vectors into the standard basis. That means that [itex]P^{-1}BP[/itex] would map each of the standard basis vectors into the four vectors, apply the transformation there, the map back to the standard basis basis. That is, your matrix is just [tex]A= P^{-1}BP[/tex] where B and P are as above. I'll let you do the tedious part, finding [itex]P^{-1}[/itex] and the multiplications.
     
  4. Oct 22, 2011 #3

    vela

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    Another approach is to use what you know about dot products to construct the linear function f(x) that gives you the projection of x onto V. Then the columns of the matrix will be given by f(ei) where ei are the canonical basis vectors.
     
  5. Oct 22, 2011 #4
    "I have no idea what you mean by "matrix A= [u,v]" nor in what sense it "represents V". You can't make a 4 by 4 matrix from 2 vectors and a matrix does not "represent" a subspace."

    What I meant was that if you took A= [u,v] and then said that Ax=z, where x is any vector in R2, then you have in fact represented V as the multiplication of a vector and a matrix. Sorry, now that I reread it, it does seem foolish, but it was my fault for not explaining myself properly.

    Thank you very much to the both of you, this helped a lot!
     
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