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Linear algebra A^tb=0

  1. Dec 12, 2012 #1
    hi all, i was given a take home exam for my linear algebra course and i cant seem to find the answer to this problem.
    1. The problem statement, all variables and given/known data
    if [itex] A^T\vec{b} = \vec{0}[/itex]
    what can you say about ##\hat{b}## the vector in Col A which is the best approximation of
    ##\vec{b}##.

    2. Relevant equations
    ##A^T A \vec{x} = A^T\vec{b}##



    3. The attempt at a solution
    I don't even know where to start,
    I have a feeling that there is a way to show that ##\hat{b}## is equal to ##\vec{b}##. but I dont know how to go about finding this.
    I've been thinking about it for a day already. any hints or nudges in the right direction would be very helpful please.
     
  2. jcsd
  3. Dec 12, 2012 #2

    lanedance

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    Homework Helper

    start by considering the columns of A. Each component of A^T.b is the same as the dot product of a column of A with b.
     
  4. Dec 12, 2012 #3
    I tried this not sure if its correct. i used this theorem

    ##(col A)^\perp = Nul A^T##
    since i was given ## A^T\vec{b} = \vec{0} ##
    ##Nul A^T = \vec{b}##
    and this means that ##\vec{b}## is already orthogonal to ##col A##
    and because ##\vec{b} - \hat{b} = \perp## to ## Col A##
    ##\hat{b} = \vec{0}##

    right? or am I completely off?
     
  5. Dec 13, 2012 #4

    lanedance

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    Homework Helper

    I think you're heading in the right direction - b is orthogonal to every column vector in A, so they are going to do a pretty poor job when used to approximate b

    this is not quite true

    ## \vec{b} \in Nul(A^T) ##
    is more accurate

    i think you've pretty much got it, but you need to outline how you measure how "good" an approximation is to tie it together
     
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