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Math Help - find the orthogonal projection

  1. #1
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    Question find the orthogonal projection

    My questions is this:
    How to find the orthogonal projection of vector y= (7,-4,-1,2) on null space
    N(A)

    Where A is a matrix
    A =
    2, 1, 1, 3
    3, 2, 2, 1
    1, 2, 2, -9
    Last edited by yevi; June 5th 2007 at 09:13 AM.
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  2. #2
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    Quote Originally Posted by yevi View Post
    My questions is this:
    How to find the orthogonal projection of vector y= (7,-4,-1,2) on null space
    N(A)

    Where A is a matrix
    A =
    2, 1, 1, 3
    3, 2, 2, 1
    1, 2, 2, -9
    Are you supposed to do this entirely by hand? I used a computer program for half of it.

    First, row reduce A, find general solution to Ax = 0, and thus find basis for N(A). I found parametric solution is [-5t,7t-s,s,t] and thus basis is [-5,7,0,1] and [0,-1,1,0].

    Put basis as columns in a matrix B.

    The orthogonal projection matrix is P = B(B^T B)^-1 B^T.

    The orthogonal projection is v = Py = [5,-5,-2,-1]. Check this is in N(A) since Av = 0 and/or v is linear combination of basis vectors.

    Check y - v = [2,1,1,3] is in A, so is orthogonal to N(A) and thus v is indeed an orthogonal projection of y.
    Last edited by JakeD; June 6th 2007 at 12:37 AM.
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  3. #3
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    Well, I did exactly the same and got the same answer [5,-5,-2,1]
    But the answer should be 3/2(0,-1,1,0)
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  4. #4
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    Quote Originally Posted by yevi View Post
    My questions is this:
    How to find the orthogonal projection of vector y= (7,-4,-1,2) on null space
    N(A)

    Where A is a matrix
    A =
    2, 1, 1, 3
    3, 2, 2, 1
    1, 2, 2, -9
    Quote Originally Posted by JakeD View Post
    Are you supposed to do this entirely by hand? I used a computer program for half of it.

    First, row reduce A, find general solution to Ax = 0, and thus find basis for N(A). I found parametric solution is [-5t,7t-s,s,t] and thus basis is [-5,7,0,1] and [0,-1,1,0].

    Put basis as columns in a matrix B.

    The orthogonal projection matrix is P = B(B^T B)^-1 B^T.

    The orthogonal projection is v = Py = [5,-5,-2,-1]. Check this is in N(A) since Av = 0 and/or v is linear combination of basis vectors.

    Check y - v = [2,1,1,3] is in A, so is orthogonal to N(A) and thus v is indeed an orthogonal projection of y.
    Quote Originally Posted by yevi View Post
    Well, I did exactly the same and got the same answer [5,-5,-2,1]
    But the answer should be 3/2(0,-1,1,0)
    To me that answer is an error because for v = [0,-3/2,3/2,0] to be an orthogonal projection, by definition

    y - v = [7,-4,-1,2] - [0,-3/2,3/2,0] = [7,-5/2,-5/2,2]

    must be orthogonal to N(A). But it is not because the dot product with the basis vector [-5,7,0,1] is not zero.

    v is an orthogonal projection of y on the subspace spanned by [0,-1,1,0], but not on the subspace N(A).
    Last edited by JakeD; June 6th 2007 at 04:30 AM.
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  5. #5
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    Well the following questiob is to find the distance from from vector u to sub space of N(A).

    For this I do: ||y-P(y)||^2

    The answer of this is: sqrt(131/2), and we get this only if we use 3/2(0,-1,1,0) as P(y)
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  6. #6
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    Quote Originally Posted by yevi View Post
    Well the following questiob is to find the distance from from vector u to sub space of N(A).

    For this I do: ||y-P(y)||^2

    The answer of this is: sqrt(131/2), and we get this only if we use 3/2(0,-1,1,0) as P(y)
    Either you have not posted the question correctly or the solution you were given is in error. v = 3/2(0,-1,1,0) is not an orthogonal projection on N(A) because y - v = (7,-5/2,-5/2,2) is obviously not orthogonal to N(A). It also does not have the minimum distance to N(A) as an orthogonal projection must. Using the correct solution v = (5,-5,-2,-1), y - v = (2,1,1,3) and the distance is sqrt(15) < sqrt(131/2).

    The definition of an orthogonal projection of a vector y on a subspace W is a vector v in W such that y - v is orthogonal to W. It has the additional property that the distance ||y - v|| is the minimum distance from y to W. You should check as I did that the correct solution satisfies the definition and that the incorrect solution does not.
    Last edited by JakeD; June 6th 2007 at 04:46 PM.
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  7. #7
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    I gave the question to a friend of mine.
    And he got the same result ([5,-5,-2,1]).
    So I can we can say about 99% that given answer is a mistake.

    Thank you.
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  8. #8
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    Another related question:

    Finding an orthogonal projection of vector y = (4,8,1) on C(A) (columns space)

    on A =
    ( -2 2)
    1/3 ( 1 2)
    ( 2 1)

    For this do I need to find the column space?
    Cause after rref(A) i get "leading" valuse corresponding to 2 columns, so i get C(A) = A!!!
    Why??

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  9. #9
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    Quote Originally Posted by yevi View Post
    Another related question:

    Finding an orthogonal projection of vector y = (4,8,1) on C(A) (columns space)

    on
    Code:
    A = 
        (-2    2)
    1/3 (1     2)
        (2     1)
    For this do I need to find the column space?
    Cause after rref(A) i get "leading" valuse corresponding to 2 columns, so i get C(A) = A!!!
    Why??

    I think you needed to confirm the columns of A are a basis for C(A). And rref is a way to do that.

    Now why are the columns of A a basis for C(A)? Because the columns of A are linearly independent.
    Last edited by JakeD; June 8th 2007 at 02:06 PM.
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  10. #10
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    Quote Originally Posted by JakeD View Post
    I think you needed to confirm C(A) = span(A)
    What do you mean by this? You mean I should get the same projection vector on space of A and space of C(A)? If yes, then something wrong again cause the given answer is: p=1/9(46,52,29)
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  11. #11
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    Quote Originally Posted by JakeD View Post
    I think you needed to confirm the columns of A are a basis for C(A). And rref is a way to do that.

    Now why are the columns of A a basis for C(A)? Because the columns of A are linearly independent.
    I originally said to "confirm C(A) = span(A)." That's what yevi is asking about. But span(A) is the definition of C(A). I edited it to "confirm the columns of A are a basis for C(A)."

    Quote Originally Posted by yevi View Post
    What do you mean by this? You mean I should get the same projection vector on space of A and space of C(A)? If yes, then something wrong again cause the given answer is: p=1/9(46,52,29)
    I meant if the columns of A are a basis for C(A) then you can compute the projection matrix using A. I agree that the projection vector is not the p you give.
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  12. #12
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    Could it be that I made a mistake and A!=C(A)??
    Please check.
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  13. #13
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    Quote Originally Posted by yevi View Post
    Could it be that I made a mistake and A!=C(A)??
    Please check.
    The columns of A are linearly independent, so A is a basis for C(A) and rank(A) = 2. Since rank(A) = rank(A^T A), if rank(A) were less than 2, you would not be able to calculate the (A^T A)^-1 needed for the projection matrix.
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