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Math Help - Determine ON-base of the column space of matrix

  1. #1
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    Determine ON-base of the column space of matrix

    1 3 1 2 2
    -2 1 3 3 -2
    1 2 1 1 1
    1 2 2 1 0
    3 1 2 -2 -1

    Gauss elimination gives me the pivot positions in colonn number 1,2 and 3.

    I think i have to use a normalised Gram-Smith method but i dont understand how to do it. plz help.
    /Maria
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  2. #2
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    Your (correct) work so far enables you to say that the first three columns span the column space of the matrix. So take the original three columns and apply Gram-Schmidt to them. Do you understand the Gram-Schmidt procedure?
    Last edited by Ackbeet; November 1st 2010 at 09:06 AM. Reason: Better grammar.
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  3. #3
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    I tried looking in my textbook but i didn't really get it, I actually missed the class when we went over the Gram-Schmidt procedure.
    If anyone can explain the steps I would really appreciate it.


    /Maria
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    To produce an ON basis \{e_{1},e_{2},\dots,e_{n}\} by using the Gram-Schmidt procedure on a linearly independent set \{x_{1},x_{2},\dots,x_{n}\}, follow these steps:

    1. Let \displaystyle e_{1}=\frac{x_{n}}{\|x_{n}\|}.

    2. For the remaining vectors in the ON basis, follow a two-step cumulative procedure: first, take the next vector x_{2}, and subtract off the amount of that vector that is in the direction of all previously computed vectors in the ON basis. In this case, set f_{2}=x_{2}-\langle x_{2},e_{1}\rangle e_{1}. Here \langle x,y\rangle is the inner product, or the dot product.

    3. Normalize to get the next ON vector: e_{2}=\dfrac{f_{2}}{\|f_{2}\|}.

    4. Set f_{3}=x_{3}-\langle x_{3},e_{1}\rangle e_{1}-\langle x_{3},e_{2}\rangle e_{2}. (You need to subtract off the amount in the directions of both the previous two ON vectors.)

    5. Then e_{3}=\dfrac{f_{3}}{\|f_{3}\|}.

    Continue. Make sense?
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  5. #5
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    I'm not sure i understand but ill try.
    The first thing I do is divide my first vector [1,2,1,1,3] with the length of the vector (sqrt(16)). This gives me [1/4,2/4,1/4,1/4,3/4], and this is the first vector in my On base right?

    After this step is done (if its right) I am supposed to subtract the inner product from the next vector in my original matrix.
    This is what i don't understand how to do.

    Also I think I am mixing up the terms, in my textbook it says I should use the following formula to calculate a Orthogonal basis. Can i use this and then normalize the answers that i get?

    v1=x1
    v2=x2-((x2v1)/(v1^2))*v1
    v3=x3-((x3v1)/(v1^2))*v1-((x3v2)/(v2^2))*v1

    Thanks in advance

    /Maria
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  6. #6
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    The formulas you wrote down are the orthogonal Gram-Schmidt, not the orthonormal Gram-Schmidt. You can normalize all the vectors afterwards if you like, as you suggested. Either way should work. The amount of work will be about the same in either case. Let's say you continue down the path I laid out. (Incidentally, you have sign error in your first vector. It should be a negative 2/4 in the second position.) Your next calculation would be

    f_{2}=\begin{bmatrix}3\\1\\2\\2\\1\end{bmatrix}-(3/4+1(-1/2)+2/4+2/4+3/4)\begin{bmatrix}1/4\\-1/2\\1/4\\1/4\\3/4\end{bmatrix}<br />
=\begin{bmatrix}3\\1\\2\\2\\1\end{bmatrix}-\begin{bmatrix}1/2\\-1\\1/2\\1/2\\3/2\end{bmatrix}=\begin{bmatrix}5/2\\2\\3/2\\3/2\\-1/2\end{bmatrix}.

    To get the normalized vector, divide by its length, which in this case is \sqrt{15}. Therefore, we have

    e_{2}=\dfrac{1}{\sqrt{15}}\begin{bmatrix}5/2\\2\\3/2\\3/2\\-1/2\end{bmatrix}.

    See how this works?
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  7. #7
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    I think i got it now,

    <br />
e_{3}=\dfrac{1}{\sqrt{5/42}}\begin{bmatrix}-5/4\\19/10\\-9/20\\11/20\\33/20\end{bmatrix}.

    Does this look right to you?

    Im pretty sure I got f_3 right since the inner products of any two bases always amount to 0.
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  8. #8
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    Your vector's direction is correct, but I think you're getting too clever with your multiplication factor. You've obviously got the right idea, but you should multiply by the denominator, or reciprocate that factor one way or another.
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  9. #9
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    Yes I noticed that later. Thank you very much for your help, its much appreciated!
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  10. #10
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    You're welcome! Have a good one.
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