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Orthonormal Sets - Find a projection matrix - Linear Algebra

  1. Jul 21, 2014 #1
    1. The problem statement, all variables and given/known data
    Let A be the 4x2 matrix
    |1/2 -1/2|
    |1/2 -1/2|
    |1/2 1/2|
    |1/2 1/2|

    Find the projection matrix P that projects vectors in R4 onto R(A)

    2. Relevant equations
    projSx = (x * u)u where S is a vector subspace and x is a vector


    3. The attempt at a solution
    v1 = (1/2, 1/2, 1/2, 1/2)T
    v2 = (-1/2, -1/2, 1/2, 1/2)T
    v1v2 = 0, hence the vectors are orthogonal
    ||v1|| = 1
    ||v2|| = 1, hence they form an orthonormal basis for R2

    R(A) = span{(1/2, -1/2)T, (1/2, -1/2)T, (1/2, 1/2)T, (1/2, 1/2)T} = span{(1/2, -1/2)T, (1/2, 1/2)T}

    And from here I am a bit lost. Would I define x as the standard basis for R4 and find projR(A)x ?

    The answer from the book is given as
    [.5 -.5 0 0]
    [-.5 .5 0 0]
    [0 0 .5 -.5]
    [0 0 -.5 .5]

    Thanks

    Note: This is from a section before the Gram-Schmidt orthogonalization process
     
    Last edited: Jul 21, 2014
  2. jcsd
  3. Jul 21, 2014 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The first thing I would do is determine R(A). Given any vector v= [x, y], we have
    [tex]Av= \begin{bmatrix}\frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}\frac{1}{2}x- \frac{1}{2}y \\ \frac{1}{2}x- \frac{1}{2}y \\ \frac{1}{2}x+ \frac{1}{2}y \\ \frac{1}{2}x+ \frac{1}{2}y\end{bmatrix}[/tex].

    We can write that result as
    [tex]\begin{bmatrix}a \\ a \\ b \\ b\end{bmatrix}= \begin{bmatrix}a \\ a \\ 0 \\ 0 \end{bmatrix}+ \begin{bmatrix}0 \\ 0 \\ b \\ b \end{bmatrix}= a\begin{bmatrix}1 \\ 1 \\ 0 \\ 0 \end{bmatrix}+ b\begin{bmatrix}0 \\ 0 \\ 1 \\ 1 \end{bmatrix}[/tex]
    where a= x/2+ y/2 and b= x/2- y/2. That is, a basis for R(A) is [tex]\{\begin{bmatrix}1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1\end{bmatrix}\}[/tex].
     
  4. Jul 22, 2014 #3
    Hmm, I'm still trying to figure out if I'm missing a formula somewhere...
    A =
    [1 0
    1 0
    0 1
    0 1]

    So, projAx = AATx
    =
    [1 1 0 0
    1 1 0 0
    0 0 1 1
    0 0 1 1] x

    I'm now confused as to how to find x? It should be a 4x4, correct?
     
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