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Steinitz Replacement Theory

  1. Apr 17, 2007 #1
    Does any one know how to do Steinitz Replacement Theory?

    I have not got a clue how to do it? And can only find theorums on it.

    I need to use Steinitz Replacement Theory to convert B into A step by step.

    A = {(0,2,2)^T, (1,0,1)^T, (1,2,1)^T}

    B= {(1,2,0)^T, (2,0,1)^T, (2,2,0)^T}

    Last edited: Apr 17, 2007
  2. jcsd
  3. Apr 17, 2007 #2


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    the so called steintiz replacement argument is an argument due to riemann 17 years before steintz birth, by whivh ones proves a linearly independent set of vectors in a space cannot have more elements than a spanning set.

    If v1,...,vn is an indept set, and w1,...,wm is a spanning set, then

    the set vnw1,...,wm must be dependent, so some vector depends on the rpevious oens, but it cannot be vn. so it must be some w, say it is w1.

    then we are left with the spanning set vn,w2,w3,...,wm.

    now consider the set vn,vn-1,w2,w3,...,wm. this must be dependent so some vector depends on previous ones, hence some w does. in particular there are some w's, so m is at least 2.

    continuing, until one has the dependent set vn,vn-1,....,v1,...wm, since the v's are independent, there are still some w's left, so m is at least as large as n.
  4. Apr 17, 2007 #3
    Yes I have found that theory in words,

    Can you give me a neumerical example. Such as the ones I have stated, or give clues on the ones I have stated.

  5. Apr 17, 2007 #4


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    i am sure you can apply these words to your vectors, i.e. you can figure out which of your vectors is v, which w, etc... so i presume what you need to know is how to decide which w to eliminate.

    to determine which vector depends on the previous
    ones, put them in a matrix as columns and row reduce the matrix. then the first non pivot column depends on the previous columns.
    Last edited: Apr 17, 2007
  6. Apr 18, 2007 #5
    Yes that is what I need to know, finding out which one to eliminate,

    could you give me an example please with V3(R) if you can not solve the one I have given.

    All we were given in the course notes was the written method, as you stated, and that is also all I can find on the net.

    And this is one of the questions at the end of the tutorial.

  7. Apr 18, 2007 #6
    Right, so I just need someone to show me how to do Steinitz now.

    And example would be great.

  8. Apr 18, 2007 #7
    The replacement argument doesn't convert one set into another set. What it does is takes a linear independent set and a spanning set and creates a basis (a linear independent spanning set). And as a corollary, it shows that every basis has the same size (the dimension of the vector space).

    For instance, if you had
    A = {(0,2,2)^T}
    B= {(1,2,0)^T, (2,0,1)^T, (2,2,0)^T, (1,1,3)^T}

    You would take the set
    {(0,2,2)^T, (1,2,0)^T, (2,0,1)^T, (2,2,0)^T, (1,1,3)^T}
    first remove (1,1,3)^T since it's a linear combination of the others
    (1,1,3)^T = 11/10*(0,2,2)^T - 3/5*(1,2,0)^T + 4/5*(2,0,1)^T

    So you're left with
    {(0,2,2)^T, (1,2,0)^T, (2,0,1)^T, (2,2,0)^T}
    Since {(0,2,2)^T, (1,2,0)^T} are linear independent (since one is not a multiple of the other), we must check them both against (2,0,1)^T

    0 & 1 & 2\\
    2 & 2 & 0\\
    2 & 0 & 1
    \right| = -4
    So these three vectors are linearly independent.

    However, (2,2,0)^T is a linear combination of the others (you can check this)
    so you can remove it, to get the basis

    {(0,2,2)^T, (1,2,0)^T, (2,0,1)^T)}

    I suppose if you want, you can use that technique to replace all of the vectors with the vectors from a basis, but it's really pretty useless since you already know that you have a basis. The theorem is really used (as far as I know, I'll be studying this more next year) to show that any linearly independent set of vectors of a vector space is less than or equal in size to any spanning set. This can also be used to show that all bases have the same size.
  9. Apr 19, 2007 #8


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    it seems to me you should work on understanding explanations given in words a bit. that is our normal medium of communication. i have completely described how to do the procedure you requested. (assuming you know what a pivot column is.)
    Last edited: Apr 19, 2007
  10. Apr 19, 2007 #9
    But the question says to convert B to A.

    Not part of A.


    I do not know what a pivot colum is.

    Steinitz is how I thought you did it though.

    But I do not know how to do reduce mine, I do not know how to find the linear combinations, without it taking me ages.
  11. Apr 19, 2007 #10


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    you should learn to row reduce a matrix. thi is the standard method for reducing a finite set of vectors to a linearly independent set with the same span.

    have you covered gaussian elimination in your course? If not I am puzzled at your exercise assignment.

    what course are you taking and at what level? what is your book? are you up to date in the course or way behind?
    Last edited: Apr 19, 2007
  12. Apr 19, 2007 #11
    I have done it, using the same method as spanning to find the linear combinations.

    I know the methods, I just do not know the names for them
  13. Apr 19, 2007 #12


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    after reducing a matrix, the leftmost non ero term in each row is called a pivot. the column containing a pivot is called a pivot column.

    the original columns located in the pivot positions form an independent subset of the original columns with the same span as the original columns.

    but if you know how to do this, you should be able to carry out steinitz replacement, regardless of what names one calls the columns involved. and yet you could not. so you need to learn something more about what you are studying than you are apparently learning.
  14. Apr 20, 2007 #13
    Like I say I just did it using the same method of spanning and finding linear combinations of the last one to remove it,
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