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Thread: finding the distance between a vector and image

  1. #1
    Dec 2010

    finding the distance between a vector and image

    I have a vector b
    and a matrix A
    [1 0 1
    2 -1 1
    -1 2 1
    0 -1 -1]

    it says
    1. the sum of the first two columns is equal to the third and conclude that the rank(A) = 2
    2. replace A by a matrix B with maximal rank and same image
    3. compute the peojection P on Im(A) = Im(B) by u sing the method of least square solution
    4. compute the projection b0 = Pb of b on Im(A)
    5. compute the distance between b and b0

    I get the first two steps but I don't understand how to do the 3rd. I'm pretty sure I can do the last two steps once i get the result from step 3. can anyone clarify on what I'm supposed to do in step 3?
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  2. #2
    MHF Contributor

    Apr 2005
    I presume you have recently been introduced to finding "least squares" approximations by using the "adjoint" matrix.

    If A is a linear operator from vector space U to vector space V, and y is NOT in the image of A, then we cannot solve Ax= y. But we can find x such that Ax is "closest" to y. We argue that Ax will be closest to y when y- Ax is perpendicular to Ax. That means that $\displaystyle <Ax, y- Ax>_V= 0$ where $\displaystyle < , >_V$ is the inner product in V. Since the "adjoint" of A is the operator, $\displaystyle A^T$, from U back to V such that, for all u in U, v, in V, $\displaystyle <Au, v>_V= <u, A^Tv>_U$ where the second $\displaystyle < , >_U$ is the inner product in U.

    Specifically, we have $\displaystyle <x, A^T(y- Ax>_U= 0$. But now, x can be any vector in U so we must have $\displaystyle A^T(y- Ax)= 0$. The "Ax" satisfying $\displaystyle A^Ty= A^TAx$ is the projection of y onto the image of A.
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