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Linear Algebra Unique Vectors Help

  1. Dec 5, 2013 #1
    1. (A)The problem statement, all variables and given/known data
    Let the following be A=
    [itex]\left|-1/\sqrt{6} ....... 1/\sqrt{3}\right|[/itex]
    [itex]\left|1/\sqrt{6}....... -1/\sqrt{3}\right|[/itex]
    [itex]\left|2/\sqrt{6}......... 1/\sqrt{3}\right|[/itex]

    ***excuse the "....." on the matrix, I didn't know how to space them out so I used dots instead***

    And the other B=

    Find the unique vector p [itex]\in[/itex] R(A) such that
    [itex]\left\|p-b\right\| < \left\|Ax-b\right\|[/itex]
    for all x [itex]\in[/itex][itex]\Re ^2[/itex]

    (B) Does there exist a vector x[itex]_{0}[/itex][itex]\in[/itex][itex]\Re ^2[/itex] such that Ax[itex]_{0}[/itex]=p? If so, is x[itex]_{0}[/itex] unique? Justify your answer.
    2. Relevant equations

    3. The attempt at a solution
    For part A I basically combined the matrix, and added x1+x2 = B, I then solved for x1 and x2. Needless to say, I got it wrong. Am I going in the right direction for this problem? I'm not very good at abstract algebra and I'm not very sure what it's asking me.

    Could someone please help me solve this problem in order to prepare for my final exam?
    Thank you.
  2. jcsd
  3. Dec 5, 2013 #2


    Staff: Mentor

    Here's your B matrix in a little nicer form. If you Quote this post you can see what I did.

    $$A = \begin{bmatrix} -1/\sqrt{6} & 1/\sqrt{3} \\
    1/\sqrt{6} & -1/\sqrt{3} \\
    2/\sqrt{6} & 1/\sqrt{3}\end{bmatrix} $$

    Your vector b is
    $$\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$$

    I don't have this figured out, but I'll share what I've found. I'm thinking in terms of the geometry here. The columns in B are orthogonal and have magnitude 1, which makes them orthonormal. The expression Ax could also be written as
    $$ x_1\begin{bmatrix}-1/\sqrt{6}\\ 1/\sqrt{6} \\ 2/\sqrt{6}\end{bmatrix} + x_2\begin{bmatrix} 1/\sqrt{3}\\ -1/\sqrt{3} \\ 1/\sqrt{3}\end{bmatrix}$$

    The above is a linear combination of the columns of A, which defines a plane in R3. You could find the normal to this plane by taking the cross product of the two vectors above. Note that you don't have enough information to find the equation of this plane, because you don't have a point in the plane.

    The expression Ax - b, or Ax + -b, can be interpreted as meaning "go out from the origin in the direction of -b, and then go to a point in the plane as specified by the coordinates (x1, x2), where these coordinates are relative to the two column vectors in A.

    If the plane that I have described happens to go through the origin, then the minimum value of ##\left\|Ax-b\right\| ## is zero. If the plane as described doesn't go through the origin, then you are basically finding the point on the plane that is closest to the origin.
  4. Dec 5, 2013 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You are asked to find ##x## which minimizes ##\|Ax - b\|##. This is equivalent to minimizing ##\|Ax - b\|^2##. Assuming the norm is the usual euclidean 2-norm, we have
    $$\|Ax - b\|^2 = (Ax - b)^T (Ax - b) = x^T A^T A x - 2b^T A x + b^T b$$
    I assume that somewhere in your course you must have covered how to minimize such an expression. One way to do this is to differentiate with respect to ##x##.

    Another way is to use the following geometric insight: in order to make ##Ax## as close to ##b## as possible, the error vector ##Ax - b## needs to be orthogonal to ##Ax##. (Draw a triangle to see why.) In other words, you need to find ##x## such that
    $$(Ax - b)^T Ax = 0$$
    But again, I'm assuming this must be covered in your notes somewhere. It would be challenging to solve this sort of problem from scratch without having seen the method (least squares) at some point.
    Last edited: Dec 5, 2013
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